26.11 - Miscellaneous Examples

In the previous section, we completed a discussion on vector algebra. We saw many solved examples also. In this section, we will see some miscellaneous examples.

Solved example 26.64
Write down a unit vector in XY-plane, making an angle of 30^o with the positive direction of x-axis
Solution:
1. Fig.26.36 below shows a unit vector $\mathbf\small{\vec{AB}}$ making an angle of 30^o with the positive direction of x-axis.

Fig.26.36

2. It is clear that, projection of $\mathbf\small{AB}$ on the x-axis is $\mathbf\small{\left|\vec{AB} \right| \cos 30~=~(1)\left(\frac{\sqrt{3}}{2} \right)~=~\frac{\sqrt{3}}{2}}$
• So the x-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\frac{\sqrt{3}}{2} \right)\hat{i}}$   

3. Also, it is clear that, projection of $\mathbf\small{AB}$ on the y-axis is $\mathbf\small{\left|\vec{AB} \right| \cos 60~=~(1)\left(\frac{1}{2} \right)~=~\frac{1}{2}}$
• So the y-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\frac{1}{2} \right)\hat{j}}$

4. So the required vector is:
$\mathbf\small{\left(\frac{\sqrt{3}}{2} \right)\hat{i}~+~\left(\frac{1}{2} \right)\hat{j}}$

Solved example 26.65
A girl walks 4 km towards west, then she walks 3 km in a direction 30^o east of north and stops. Determine the girl's displacement from her initial point of departure.
Solution:
1. Fig.26.37 below shows the rough sketch

Fig.26.37

   ♦ $\mathbf\small{\vec{AB}}$ indicates the initial travel
   ♦ $\mathbf\small{\vec{BC}}$ indicates the final travel
• So $\mathbf\small{\vec{AC}}$ indicates the displacement vector. We are asked to find this vector.

2. Based on the fig., we can write:
• Projection of $\small{\vec{AB}}$ on the x-axis is 4 km.
• This vector does not have any projection on the y-axis.
• So we can write: $\small{\vec{AB} = -4\hat{i}}$
• The −ve sign is required because, this vector is pointing towards the −ve side of the x-axis

3. Also, based on the fig., we can write:
• Projection of $\small{\vec{BC}}$ on the x-axis
= DC = $\small{\left|\vec{BC} \right| \sin 30 = 3 \sin 30 = \frac{3}{2}}$
• Projection of $\small{\vec{BC}}$ on the y-axis
= BD = $\small{\left|\vec{BC} \right| \cos 30 = 3 \cos 30 = \frac{3 \sqrt{3}}{2}}$
So we can write: $\small{\vec{BC} = \frac{3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}}$

4. By triangle law of vector addition, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AC}}    & {~=~}    &{\vec{AB}+\vec{BC}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-4\hat{i} ~+~\frac{3}{2}\hat{i} + \frac{3 \sqrt{3}}{2} \hat{j} }
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-\frac{5}{2}\hat{i} + \frac{3 \sqrt{3}}{2} \hat{j} }
\\ \end{array}}$

Solved example 26.66
Write all the unit vectors in the XY-plane.
Solution:
1. Fig.26.38 below shows a random vector $\mathbf\small{\vec{AB}}$ making an angle of $\small{\theta}$ with the positive direction of x-axis.

Fig.26.38

2. It is clear that, projection of $\mathbf\small{AB}$ on the x-axis is $\mathbf\small{\left|\vec{AB} \right| \cos \theta~=~(1)\left(\cos \theta \right)~=~\cos \theta}$
• So the x-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\cos \theta \right)\hat{i}}$   

3. Also, it is clear that, projection of $\mathbf\small{AB}$ on the y-axis is $\mathbf\small{\left|\vec{AB} \right| \sin \theta~=~(1)\left(\sin \theta \right)~=~\sin \theta}$
• So the y-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\sin \theta \right)\hat{j}}$

4. So we can write:
$\mathbf\small{\vec{AB} = \left(\cos \theta \right)\hat{i} + \left(\sin \theta \right)\hat{j}}$

5. The above result is the general form. For any particular unit vector, all we need to do is to input the value of $\small{\theta}$ of that particular unit vector.
• We know that, $\small{\theta}$ can vary from zero to $\small{2 \pi}$.

6. So we can write:
• All unit vectors in the XY plane, are represented by the vector:
$\mathbf\small{\left(\cos \theta \right)\hat{i} + \left(\sin \theta \right)\hat{j}}$, where $\small{\theta}$ falls in the interval $\small{\left[0, 2 \pi \right]}$

Solved example 26.67
If $\small{\hat{i}+\hat{j}+\hat{k},~2\hat{i}+2\hat{j},~3\hat{i}+5\hat{j}-3\hat{k}}$ and $\small{\hat{i}-6\hat{j}-\hat{k}}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\small{\vec{AB}~\text{and}~\vec{CD}}$. Deduce that $\small{\vec{AB}~\text{and}~\vec{CD}}$ are collinear.
Solution:
1.Based on the given position vectors, we get:
$\small{\vec{AB} = \hat{i}+4\hat{j}-\hat{k}}$
$\small{\vec{CD} = -2\hat{i}-8\hat{j}+2\hat{k}}$
2. Let $\small{\theta}$ be the angle between $\small{\vec{AB}~\text{and}~\vec{CD}}$.
We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{AB}.\vec{CD}}{\left|\vec{AB}\right|\,\left|\vec{CD}\right|} \right)}$
3. So first we need to find $\small{\vec{AB}.\vec{CD}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}.\vec{CD}}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(1)(-2)+(4)(-8)+(-1)(2)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-2-32-2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{-36}
\\ \end{array}}$
4. Next we need to find the magnitudes:
$\small{\left|\vec{AB}\right|=\sqrt{1^2 + 4^2 + (-1)^2}=\sqrt{18}}$
$\small{\left|\vec{CD}\right|=\sqrt{(-2)^2 + (-8)^2 + 2^2}=\sqrt{72}}$
5. Substituting the above values in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{AB}.\vec{CD}}{\left|\vec{AB}\right|\,\left|\vec{CD}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{-36}{(\sqrt{18})(\sqrt{72})} \right) = \cos^{-1}\left(\frac{-36}{(\sqrt{18})(\sqrt{(2)(36)})} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{(-1)\sqrt{(36)(36)}}{(\sqrt{18})(\sqrt{(2)(36)})} \right) = \cos^{-1}\left(\frac{(-1)\sqrt{(36)}}{(\sqrt{18})(\sqrt{(2)})} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(-1 \right)~=~\pi}
\\ \end{array}}$
6. The angle between $\small{\vec{AB}~\text{and}~\vec{CD}}$ is $\small{\pi}$. So the two vectors are collinear.

Alternate method to prove collinearity:

1. Take the ratios of the scalar components of $\small{\vec{AB}~\text{and}~\vec{CD}}$:
$\small{\frac{1}{-2},~\frac{4}{-8},~\text{and}~\frac{1}{-2}}$

2. We see that, all ratios are the same, which is $\small{-\frac{1}{2}}$
• So we can write: $\small{\vec{AB}=-\frac{1}{2} \vec{CD}}$
Which implies that, the two vectors are collinear.

Solved example 26.68
Let $\small{\vec{a},~\vec{b}~\text{and}~\vec{c}}$ be three vectors such that $\small{\left|\vec{a} \right|=3,~\left|\vec{b} \right|=4,~\left|\vec{c} \right|=5}$ and each one of them being perpendicular to the sum of the other two, find  $\small{\left|\vec{a}+\vec{b}+\vec{c} \right|}$.
Solution:
1.Given that:
Each vector is perpendicular to the sum of the other two.
• So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\left(\vec{b}+\vec{c} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{b}.\left(\vec{a}+\vec{c} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{c}.\left(\vec{a}+\vec{b} \right)}    & {~=~}    &{0}
\\ \end{array}}$
2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left[\left|\vec{a}+\vec{b}+\vec{c} \right| \right]^2}    & {~=~}    &{\left(\vec{a}+\vec{b}+\vec{c} \right).\left(\vec{a}+\vec{b}+\vec{c} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{~~~~~~~\vec{a}.\vec{a}+\vec{a}.\left(\vec{b}+\vec{c} \right)}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{b}.\vec{b}+\vec{b}.\left(\vec{a}+\vec{c} \right)}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{c}.\vec{c}+\vec{c}.\left(\vec{a}+\vec{b} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{~~~~~~~\vec{a}.\vec{a}+0}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{b}.\vec{b}+0}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{c}.\vec{c}+0}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2+~\left|\vec{b} \right|^2~+~\left|\vec{c} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{9 + 16 + 25 = 50}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\left|\vec{a}+\vec{b}+\vec{c} \right|}    & {~=~}    &{\sqrt{50}~=~5\sqrt{2}}
\\ \end{array}}$

Solved example 26.69
Three vectors $\small{\vec{a},~\vec{b},~\vec{c}}$ satisfy the condition $\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$. Evaluate the quantity $\small{\mu=\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}}$, if $\small{\left|\vec{a} \right|=1,~\left|\vec{b} \right|=4,~\left|\vec{c} \right|=2}$
Solution:
1. Given that:

$\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$
Multiplying both sides by $\small{\vec{a}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{a}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ \end{array}}$

2. Multiplying both sides by $\small{\vec{b}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{b}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{b}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{b}.\vec{a}+\vec{b}.\vec{b}+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{b}.\vec{a}+16+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ \end{array}}$

3. Multiplying both sides by $\small{\vec{c}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{c}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+\vec{c}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+4}    & {~=~}    &{0}
\\ \end{array}}$

4. We have three results:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{}    &{\vec{b}.\vec{a}+16+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+4}    & {~=~}    &{0}
\\ \end{array}}$

5. Adding the L.H.S together, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{21 + 2\vec{a}.\vec{b} + 2\vec{b}.\vec{c} + 2\vec{c}.\vec{a}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{21 + 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{ 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{-21}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{ \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} }    & {~=~}    &{\frac{-21}{2}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\mu}    & {~=~}    &{\frac{-21}{2}}
\\ \end{array}}$

Solved example 26.70
Let $\small{\vec{a}=\hat{i}+4\hat{j}+2\hat{k},~\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}~\text{and}~\vec{c}=2\hat{i}-\hat{j}+4\hat{k}}$. Find a vector $\small{\vec{d}}$, which is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b},~\text{and}~\vec{c}.\vec{d}=15}$
Solution:
1. We want a vector perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$
• We know that $\small{\vec{a}\times\vec{b}}$ is a vector which is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$
• So first, we will find this cross product.

2. The cross product is:
$\small{\vec{a}\times\vec{b}=32\hat{i}-\hat{j}-14\hat{k}}$
• The reader may write all the steps related to this cross product

3. So $\small{32\hat{i}-\hat{j}-14\hat{k}}$ is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$
• Then $\small{\lambda \left(32\hat{i}-\hat{j}-14\hat{k} \right)}$ is also perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$, where $\small{\lambda}$ is a scalar.
• That means, $\small{32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k}}$ is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$

4. We will assume that, the vector written above is the required vector $\small{\vec{d}}$.
• That is., we assume:
$\small{\vec{d} = 32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k}}$

5. But $\small{\vec{d}}$ must satisfy the condition:
$\small{\vec{c}.\vec{d}=15}$
• So we can write:
$\small{\left(2\hat{i}-\hat{j}+4\hat{k} \right).\left(32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k} \right)=15}$
• Therefore we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(2)(32 \lambda)+(-1)(-1)(\lambda)+(4)(-1)(14 \lambda)}    & {~=~}    &{15}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{64\lambda + \lambda - 56\lambda}    & {~=~}    &{15}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{9 \lambda}    & {~=~}    &{15}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{ \lambda}    & {~=~}    &{\frac{15}{9}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{ \lambda}    & {~=~}    &{\frac{5}{3}}
\\ \end{array}}$

6. So based on step (4), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{d}}    & {~=~}    &{32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{32 \left(\frac{5}{3} \right) \hat{i}-\left(\frac{5}{3} \right) \hat{j}-14 \left(\frac{5}{3} \right) \hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{1}{3}\left( 160  \hat{i}- 5\hat{j}-70  \hat{k} \right)}
\\ \end{array}}$

Solved example 26.71
If with reference to the right handed system of mutually perpendicular unit vectors $\small{\hat{i},~\hat{j}~\text{and}~\hat{k},~\vec{\alpha}=3\hat{i}-\hat{j},~\vec{\beta}=2\hat{i}+\hat{j}-3\hat{k}}$, then express $\small{\vec{\beta}}$ in the form $\small{\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}}$, where $\small{\vec{\beta_1}}$ is parallel to $\small{\vec{\alpha}}$ and $\small{\vec{\beta_2}}$ is perpendicular to $\small{\vec{\alpha}}$
Solution:
1. We want $\small{\vec{\beta_1}}$ to be parallel to $\small{\vec{\alpha}}$.
• So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta_1}}    & {~=~}    &{\lambda\,\vec{\alpha}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(3\lambda \right)\hat{i}-\left(\lambda \right)\hat{j}}
\\ \end{array}}$

2. Given that: $\small{\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}}$
• So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta}}    & {~=~}    &{\vec{\beta_1}+\vec{\beta_2}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{\beta_2}}    & {~=~}    &{\vec{\beta}~-~\vec{\beta_1}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{\beta_2}}    & {~=~}    &{2\hat{i}+\hat{j}-3\hat{k}~-~\left[\left(3\lambda \right)\hat{i}-\left(\lambda \right)\hat{j} \right]}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\vec{\beta_2}}    & {~=~}    &{\left(2 - 3\lambda \right)\hat{i}+\left(1+\lambda \right)\hat{j}-3\hat{k}}
\\ \end{array}}$

3. $\small{\vec{\beta_2}}$ should be perpendicular to $\small{\vec{\alpha}}$. So their dot product will be zero.
• We can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\alpha}.\vec{\beta_2}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left[\left(2 - 3\lambda \right)\hat{i}+\left(1+\lambda \right)\hat{j}-3\hat{k} \right].\left[3\hat{i}-\hat{j} \right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left(2 - 3\lambda \right)(3)~+~\left(1+\lambda \right)(-1)}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{6 - 9\lambda - 1 - \lambda}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{5 - 10\lambda}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\lambda}    & {~=~}    &{\frac{1}{2}}
\\ \end{array}}$

4. So from step (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta_1}}    & {~=~}    &{\left(3\lambda \right)\hat{i}-\left(\lambda \right)\hat{j}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(\frac{3}{2} \right)\hat{i}-\left(\frac{1}{2} \right)\hat{j}}
\\ \end{array}}$

5. Also, from step (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta_2}}    & {~=~}    &{\left(2 - 3\lambda \right)\hat{i}+\left(1+\lambda \right)\hat{j}-3\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(2 - 3\left(\frac{1}{2} \right) \right)\hat{i}+\left(1+\left(\frac{1}{2} \right) \right)\hat{j}-3\hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{1}{2}  \right)\hat{i}+\left(\frac{3}{2}  \right)\hat{j}-3\hat{k}}
\\ \end{array}}$

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The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

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In the next chapter, we will see Three dimensional geometry.

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26.10 - Cross Product of Two Vectors

In the previous section, we completed a discussion on dot product of two vectors. We saw projection of a vector also. In this section, we will see cross product of two vectors.

Vector (or cross) product of two vectors

We have already seen the basic details about vector product in chapter 7 of our physics classes. Let us recall:
1. We have seen right handed screws and left handed screws in section 7.10.
2. We have seen the direction of the cross product in section 7.11.
3. We have seen the magnitude of the cross product in section 7.12.
4. So we can write:
$\mathbf\small{\vec{a}\times \vec{b}~=~\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta \,\hat{n}}$

---

Let us write the important properties of cross product:
1. $\mathbf\small{\vec{a}\times\vec{b}}$ is a vector

2. If $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are two nonzero vectors, then $\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}}$ if and only if $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are parallel (or collinear) to each other. We can write:
$\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}~\Leftrightarrow~\vec{a}{\parallel} \vec{b}}$
Let us see two specific cases:
Case 1:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \vec{a}}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{a} \right|\sin(0) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{0}}
\\ \end{array}}$

Case 2:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \left(-\vec{a} \right)}    & {~=~}    &{\left|\vec{a} \right|\left|-\vec{a} \right|\sin(\pi) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|-\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{0}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\frac{\pi}{2}}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \vec{b}}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|\sin(\frac{\pi}{2}) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|(1) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right| \,\hat{n}}
\\ \end{array}}$

4. Based on (2) and (3), we get some interesting results:
• Based on (2), we get:
$\mathbf\small{\hat{i}\times\hat{i}~=~\hat{j}\times\hat{j}~=~\hat{k}\times\hat{k}~=~\vec{0}}$
• Based on (3), we get:
    ♦ $\mathbf\small{\hat{i}\times\hat{j}~=~\hat{k}}$
    ♦ $\mathbf\small{\hat{j}\times\hat{k}~=~\hat{i}}$
    ♦ $\mathbf\small{\hat{k}\times\hat{i}~=~\hat{j}}$
Also see fig.7.63 and fig.7.64(a) of section 7.12 of physics notes

5. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\theta}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times \vec{b} \right|}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\sin \theta}    & {~=~}    &{\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{ \theta}    & {~=~}    &{\sin^{-1}\left(\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|} \right)}
\\ \end{array}}$

6. It is always true that, vector product is not commutative. This is because, $\mathbf\small{\vec{a}\times \vec{b}~=~-\left(\vec{b}\times \vec{a} \right)}$
• Assume that, both $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ lie on the plane of the paper.
• Also assume that, $\mathbf\small{\vec{a}\times\vec{b}}$ is directed towards the upper side of the plane of the paper.
• Then $\mathbf\small{\vec{b}\times\vec{a}}$ will be directed towards the bottom side of the plane of the paper.
• This is because:
    ♦ For $\mathbf\small{\vec{a}\times\vec{b}}$, we rotate the right handed screw from $\mathbf\small{\vec{a}~\text{to}~\vec{b}}$
    ♦ For $\mathbf\small{\vec{b}\times\vec{a}}$, we rotate the right handed screw from $\mathbf\small{\vec{b}~\text{to}~\vec{a}}$

7. Based on (4) and (6), we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{i}\times \hat{j} }    & {~=~}    &{\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{j}\times \hat{i}}    & {~=~}    &{-\left(\hat{i}\times \hat{j} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{j}\times \hat{i}}    & {~=~}    &{-\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{j}\times \hat{k} }    & {~=~}    &{\hat{i}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{k}\times \hat{j}}    & {~=~}    &{-\left(\hat{j}\times \hat{k} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{k}\times \hat{j}}    & {~=~}    &{-\hat{i}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{k}\times \hat{i} }    & {~=~}    &{\hat{j}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{i}\times \hat{k}}    & {~=~}    &{-\left(\hat{k}\times \hat{i} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{i}\times \hat{k}}    & {~=~}    &{-\hat{j}}
\\ \end{array}}$

8. The fig.26.30 below, shows triangle ABC.

Fig.26.30

• $\mathbf\small{\vec{AC}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the triangle ABC
• We know that, area of the triangle ABC = $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(CD \right)}$
• But $\mathbf\small{CD = \left|\vec{AC} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the triangle ABC
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the triangle is equal to half of the magnitude of the cross product

9. The fig.26.31 below, shows parallelogram ABCD.

Fig.26.31

• $\mathbf\small{\vec{AD}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the parallelogram ABCD.
• We know that, area of the parallelogram ABCD = $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left(DE \right)}$
• But $\mathbf\small{DE = \left|\vec{AD} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the parallelogram ABCD
=  $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left( \left|\vec{a} \right| \sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left( \left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the parallelogram is equal to the magnitude of the cross product

10. Vector product of two vectors is also known as cross product of two vectors.

---

Two important properties of cross product

Property I: Distributivity of cross product over addition
This can be explained as below:
Let $\small{\vec{a},~\vec{b},~\vec{c}}$ be any three vectors. Then we can write:
$\small{\vec{a}\times\left(\vec{b}+\vec{c} \right)~=~\vec{a}\times\vec{b}+\vec{a}\times\vec{c}}$

Property II: Distributivity of cross product over multiplication
This can be explained as below:
Let $\small{\vec{a}~\text{and}~\vec{b}}$ be any two vectors and $\small{\lambda}$ be any scalar. Then we can write:
$\small{\lambda \left(\vec{a}\times\vec{b} \right) ~=~\left(\lambda \vec{a}\right)\times\vec{b} ~=~\vec{a}\times\left(\lambda\vec{b} \right)}$

---

Cross product when vectors are given in component form

• Let the two vectors be:
$\small{\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$
$\small{\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}}$
• Then we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \right)\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{a_1\hat{i}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2\hat{j}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3\hat{k}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\hat{i}\times\hat{i}\right)+a_1 b_2\left(\hat{i}\times\hat{j}\right)+a_1 b_3\left(\hat{i}\times\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(\hat{j}\times\hat{i}\right)+a_2 b_2\left(\hat{j}\times\hat{j}\right)+a_2 b_3\left(\hat{j}\times\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{k}\times\hat{i}\right)+a_3 b_2\left(\hat{k}\times\hat{j}\right)+a_3 b_3\left(\hat{k}\times\hat{k}\right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\vec{0}\right)+a_1 b_2\left(\hat{k}\right)+a_1 b_3\left(-\hat{j}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(-\hat{k}\right)+a_2 b_2\left(\vec{0}\right)+a_2 b_3\left(\hat{i}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{j}\right)+a_3 b_2\left(-\hat{i}\right)+a_3 b_3\left(\vec{0}\right)}
\\ {~\color{magenta}    5   }    &{}    &{}    & {~=~}    &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} + \left(a_3 b_1 - a_1 b_3 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ {~\color{magenta}    6   }    &{}    &{}    & {~=~}    &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} - \left(a_1 b_3 - a_3 b_1 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ \end{array}}$

• The above result is equivalent to the determinant as shown below:

$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ a_{1}      &{    a_{2}     }    &{   a_{3}      }    
\\ b_{1}      &{    b_{2}     }    &{ b_{3}        }    
\\ \end{array}\right|$

---

Now we will see some solved examples.

Solved example 26.55
Find $\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if $\mathbf\small{\vec{a}=2\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 2      &{    1     }    &{   3      }    
\\ 3      &{    5     }    &{ -2        }    
\\ \end{array}\right|$

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-2 - 15 \right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$

3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-17)^2 + (13)^2 + (7)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{507}}
\\ \end{array}}$

Solved example 26.56
Find $\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if $\mathbf\small{\vec{a}=\hat{i}-7\hat{j}+7\hat{k}~\text{and}~\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    -7     }    &{   7      }    
\\ 3      &{    -2     }    &{ 2        }    
\\ \end{array}\right|$

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-2 - 15 \right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$

3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(19)^2 + (19)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{19\sqrt{2}}
\\ \end{array}}$

Solved example 26.57
Find a unit vector perpendicular to each of the vectors $\mathbf\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where $\mathbf\small{\vec{a}=\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}+3\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=2\hat{i}+3\hat{j}+4\hat{k}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-\hat{j}-2\hat{k}}$

2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 2      &{    3     }    &{   4      }    
\\ 0      &{    -1     }    &{ -2        }    
\\ \end{array}\right|$

• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}\times\vec{d}}    & {~=~}    &{\left(-6 + 4 \right)\hat{i} - \left(-4 - 0 \right)\hat{j} + \left(-2 - 0 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-2\hat{i}+4\hat{j}-2\hat{k}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{(-2)^2 + (4)^2+(-2)^2}}~=~\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{24}}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{2\sqrt{6}}~=~\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{6}}}$

4. Note:
We have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie. The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane. Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of $\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)

Solved example 26.58
Find a unit vector perpendicular to each of the vectors $\mathbf\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where $\mathbf\small{\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}-2\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=4\hat{i}+4\hat{j}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-2\hat{i}+4\hat{k}}$

2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 4      &{    4     }    &{   0      }    
\\ -2      &{    0     }    &{ 4        }    
\\ \end{array}\right|$
• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}\times\vec{d}}    & {~=~}    &{\left(16 - 0 \right)\hat{i} - \left(16 - 0 \right)\hat{j} + \left(0+8 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{16\hat{i}-16\hat{j}+8\hat{k}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required unit vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{16\hat{i}-16\hat{j}+8\hat{k}}{\sqrt{(16)^2 + (-16)^2+(8)^2}}~=~\frac{16\hat{i}-16\hat{j}+8\hat{k}}{24}}$
= $\mathbf\small{\frac{2\hat{i}-2\hat{j}+\hat{k}}{3}}$

4. Note:
We have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie. The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane. Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of $\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)

Solved example 26.59
Find the area of a triangle having the points A(1,1,1), B(1,2,3) and C(2,3,1) as it's vertices
Solution:
1. Fig.26.32 below shows the rough sketch:

Fig.26.32

2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{j}+2\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=\hat{i}+2\hat{j}}$

3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 0      &{    1     }    &{   2      }    
\\ 1      &{    2     }    &{ 0        }    
\\ \end{array}\right|$

4. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(0 - 4 \right)\hat{i} - \left(0 - 2 \right)\hat{j} + \left(0 - 1 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-4\hat{i}+2\hat{j}-\hat{k}}
\\ \end{array}}$

5. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-4)^2 + (2)^2 + (-1)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{21}}
\\ \end{array}}$

6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{21}}{2}}$

Solved example 26.60
Find the area of a triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5).
Solution:
1. Fig.26.33 below shows the rough sketch:

Fig.26.33

2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{i}+2\hat{j}+3\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=4\hat{j}+3\hat{k}}$

3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    2     }    &{   3      }    
\\ 0      &{    4     }    &{ 3        }    
\\ \end{array}\right|$

4. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(6 - 12 \right)\hat{i} - \left(3 - 0 \right)\hat{j} + \left(4 - 0 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-6\hat{i}-3\hat{j}+4\hat{k}}
\\ \end{array}}$

5. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-6)^2 + (-3)^2 + (4)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{61}}
\\ \end{array}}$

6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{61}}{2}}$

Solved example 26.61
Find the area of the parallelogram whose adjacent sides are given by the vectors $\small{\vec{a}=3\hat{i}+\hat{j}+4\hat{k}~\text{and}~\vec{b}=\hat{i}-\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.34 below shows the rough sketch:

Fig.26.34

2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 3      &{    1     }    &{   4      }    
\\ 1      &{    -1     }    &{ 1        }    
\\ \end{array}\right|$

3. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(1 + 4 \right)\hat{i} - \left(3 - 4 \right)\hat{j} + \left(-3 - 1 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{5\hat{i}+\hat{j}-4\hat{k}}
\\ \end{array}}$

4. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(5)^2 + (1)^2 + (-4)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{42}}
\\ \end{array}}$

5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{42}}$

Solved example 26.62
Find the area of the parallelogram whose adjacent sides are determined by the vectors $\small{\vec{a}=\hat{i}-\hat{j}+3\hat{k}~\text{and}~\vec{b}=2\hat{i}-7\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.35 below shows the rough sketch:

Fig.26.35

2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    -1     }    &{   3      }    
\\ 2      &{    -7     }    &{ 1        }    
\\ \end{array}\right|$

3. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-1 + 21 \right)\hat{i} - \left(1 - 6 \right)\hat{j} + \left(-7 + 2 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{20\hat{i}+5\hat{j}-5\hat{k}}
\\ \end{array}}$

4. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(20)^2 + (5)^2 + (-5)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{450}}
\\ \end{array}}$

5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{450}~=~15\sqrt{2}}$

Solved example 26.63
If a unit vector $\small{\vec{a}}$ makes angles $\small{\frac{\pi}{3}~\text{with}~\hat{i},~~\frac{\pi}{4}~\text{with}~\hat{j}}$ and an acute angle $\small{\theta}$ with $\small{\hat{k}}$, then find $\small{\theta}$ and hence, the components of $\small{\vec{a}}$.
Solution:
1. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}~\text{with}~\hat{i}}$.
• $\small{\hat{i}}$ lies along the x-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}}$ with the x-axis.
• So the x-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{3} \right)\hat{i}}$ 
= $\small{(1)\left(\frac{1}{2} \right)\hat{i}}$ 
= $\small{\left(\frac{1}{2} \right)\hat{i}}$  

2. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}~\text{with}~\hat{j}}$.
• $\small{\hat{j}}$ lies along the y-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}}$ with the y-axis.
• So the y-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{4} \right)\hat{j}}$ 
= $\small{(1)\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$ 
= $\small{\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$  

3. $\small{\vec{a}}$ makes an acute angle $\small{\theta~\text{with}~\hat{k}}$.
• $\small{\hat{k}}$ lies along the z-axis. That means, $\small{\vec{a}}$ makes an acute angle $\small{\theta}$ with the z-axis.
• So the z-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\theta \right)\hat{k}}$ 
= $\small{(1)\cos\left(\theta \right)\hat{k}}$ 
= $\small{\cos\left(\theta \right)\hat{k}}$

4. We wrote the three components. So we can write:
$\small{\vec{a}=\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\theta \right)\hat{k}}$

5. Given that, $\small{\vec{a}}$ is a unit vector. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a} \right|}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\sqrt{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2}}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2}    & {~=~}    &{1}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\cos^2\left(\theta \right)}    & {~=~}    &{1-\frac{1}{4}-\frac{1}{2}~=~\frac{1}{4}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\cos\left(\theta \right)}    & {~=~}    &{\pm\frac{1}{2}}
\\ \end{array}}$

6. So we have two equations:
(i) $\small{\cos \theta~=~\frac{1}{2}}$ 
(ii) $\small{\cos \theta~=~-\frac{1}{2}}$

• Solving the first equation, we get: $\small{\theta~=~\frac{\pi}{3}}$ 
• Solving the second equation, we get: $\small{\theta~=~\frac{2\pi}{3}}$  

• Given that, $\small{\theta}$ is an acute angle. So we can write:
$\small{\theta~=~\frac{\pi}{3}}$

7. So based on step (4), we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}}    & {~=~}    &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\frac{\pi}{3} \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{a}}    & {~=~}    &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\left(\frac{1}{2} \right)\hat{k}}
\\ \end{array}}$

8. Therefore, the components of $\small{\vec{a}}$ are:
$\small{\frac{1}{2},~\frac{1}{\sqrt{2}}~\text{and}~\frac{1}{2}}$

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The link below gives a few more solved examples:

Exercise 10.4

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In the next section, we will see some miscellaneous examples.

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26.6 - Section Formula

Exercise 26.2In the previous section, we completed a discussion on components of a vector. In this  section, we will see vector joining two points. Later in this section, we will see section formula also.

Vector joining two points

This can be explained in 6 steps:
1. In fig.26.22 below, P₁ and P₂ are any two points in space. We want $\small{\vec{P_1 P_2}}$ in component form.

Fig.26.22

2. Consider the three vectors $\small{\vec{OP_1},~\vec{OP_2}~\vec{P_1 P_2}}$.
They form the sides of a triangle.
3. Applying triangle law, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OP_1}~+~\vec{P_1 P_2}}    & {~=~}    &{\vec{OP_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{OP_1}~+~\vec{P_1 P_2}~+~\left(-\vec{OP_1} \right)}    & {~=~}    &{\vec{OP_2}~+~\left(-\vec{OP_1} \right)}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\vec{P_1 P_2}}    & {~=~}    &{\vec{OP_2}~-~\vec{OP_1}}
\\ \end{array}}$
◼ Remarks:
2 (magenta color): Here we add  the −ve of $\small{\vec{OP_1}}$ on both sides
4. We have the component form of $\small{\vec{OP_1}~\text{and}~\vec{OP_2}}$:
$\small{\vec{OP_1}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}}$
$\small{\vec{OP_2}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}}$
5. Therefore:
$\small{\vec{P_1 P_2}=\left(x_2 - x_1 \right)\hat{i}~+~\left(y_2 - y_1 \right)\hat{j}~+~\left(z_2 - z_1 \right)\hat{k}}$
6. We can write the magnitude also:
$\small{\left| \vec{P_1 P_2}\right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left(z_2 - z_1 \right)^2}}$

---

Now we will see a solved example.

Solved example 26.28
Find the vector joining the points P(2,3,0) and Q(−1,−2,−4) directed from P to Q
Solution:
• We want the vector directed from P to Q. So P is the initial point and Q is the terminal point.
• Then we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{PQ}}    & {~=~}    &{(-1-2)\hat{i}+(-2-3)\hat{j}+(-4-0)\hat{k}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{PQ}}    & {~=~}    &{-3\hat{i}-5\hat{j}-4\hat{k}}
\\ \end{array}}$

---

Section formula

• P and Q are two points in space.
    ♦ $\small{\vec{OP}}$ is the position vector of P.
    ♦ $\small{\vec{OQ}}$ is the position vector of Q
• We know that, a line of infinite length can be drawn connecting P and Q. Consider a point R on this line. We want the position vector of R
• Two cases can arise in this situation.
Case I: R is within the line segment PQ
This can be analyzed in 6 steps:
1. In fig.26.23 below, point R is within PQ such that:
    ♦ Length PR = $\small{m\left|\vec{PQ} \right|}$
    ♦ Length QR = $\small{n\left|\vec{PQ} \right|}$
• $\small{m~\text{and}~n}$ are +ve scalars

Fig.26.23

2. We can write:
$\small{\frac{\left|\vec{PR} \right|}{\left|\vec{RQ} \right|} = \frac{m\left|\vec{PQ} \right|}{n\left|\vec{PQ} \right|} = \frac{m}{n}}$
• That means, R divides PQ internally in the ratio m:n
3. In the above step, all quantities are scalars because, we took the ratio of magnitudes. Let us try to bring vectors also into the equation.
• In the fig.26.23 above, $\small{\vec{PR}~\text{and}~\vec{RQ}}$ have the same direction. So their corresponding unit vectors will be equal.
We get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{PR}}    & {~=~}    &{\hat{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{\left|\vec{PR} \right|}}    & {~=~}    &{\frac{\vec{RQ}}{\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{mn\left|\vec{PR} \right|}}    & {~=~}    &{\frac{\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(n\left|\vec{PR} \right| \right)}}    & {~=~}    &{\frac{\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(m\left|\vec{RQ} \right| \right)}}    & {~=~}    &{\frac{\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m}}    & {~=~}    &{\frac{\vec{RQ}}{n}}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{n\,\vec{PR}}    & {~=~}    &{m\,\vec{RQ}}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we divide both sides by mn
• 5 (magenta color): Here we use the result
$\small{n\left|\vec{PR} \right| = m\left|\vec{RQ} \right|}$, which can be obtained from (2)

4. From triangle ORP, we get:
$\small{\vec{PR} = \vec{OR} - \vec{OP}}$
5. From triangle ORQ, we get:
$\small{\vec{RQ} = \vec{OQ} - \vec{OR}}$
6. From (3), we have: $\small{n\,\vec{PR} = m\,\vec{RQ}}$
• Substituting from (4) and (5), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{n\,\vec{PR}}    & {~=~}    &{m\,\vec{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{n\left(\vec{OR} - \vec{OP} \right)}    & {~=~}    &{m\left(\vec{OQ} - \vec{OR} \right)}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{n\,\vec{OR} - n\,\vec{OP}}    & {~=~}    &{m\,\vec{OQ} - m\,\vec{OR}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(m+n)\vec{OR}}    & {~=~}    &{m\,\vec{OQ}+n\,\vec{OP}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\vec{OR}}    & {~=~}    &{\frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}
\\ \end{array}}$

Case II: R is outside the line segment PQ, on the extension of PQ
This can be analyzed in 6 steps:
1. In fig.26.24 below, point R is outside PQ such that:
    ♦ Length PR = $\small{m\left|\vec{PQ} \right|}$
    ♦ Length QR = $\small{n\left|\vec{PQ} \right|}$
• $\small{m~\text{and}~n}$ are +ve scalars

Fig.26.24

2. We can write:
$\small{\frac{\left|\vec{PR} \right|}{\left|\vec{RQ} \right|} = \frac{m\left|\vec{PQ} \right|}{n\left|\vec{PQ} \right|} = \frac{m}{n}}$
• That means, R divides PQ externally in the ratio m:n
3. In the above step, all quantities are scalars because, we took the ratio of magnitudes. Let us try to bring vectors also into the equation.
• In the fig.26.24 above, $\small{\vec{PR}~\text{and}~\vec{RQ}}$ have opposite directions. So their corresponding unit vectors will differ by sign only.
We get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{PR}}    & {~=~}    &{-\hat{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{\left|\vec{PR} \right|}}    & {~=~}    &{\frac{-\vec{RQ}}{\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{mn\left|\vec{PR} \right|}}    & {~=~}    &{\frac{-\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(n\left|\vec{PR} \right| \right)}}    & {~=~}    &{\frac{-\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m\left(m\left|\vec{RQ} \right| \right)}}    & {~=~}    &{\frac{-\vec{RQ}}{mn\left|\vec{RQ} \right|}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{\vec{PR}}{m}}    & {~=~}    &{\frac{\vec{-RQ}}{n}}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{n\,\vec{PR}}    & {~=~}    &{-m\,\vec{RQ}}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we divide both sides by mn
• 5 (magenta color): Here we use the result
$\small{n\left|\vec{PR} \right| = m\left|\vec{RQ} \right|}$, which can be obtained from (2)

4. From triangle ORP, we get:
$\small{\vec{PR} = \vec{OR} - \vec{OP}}$
5. From triangle ORQ, we get:
$\small{\vec{RQ} = \vec{OQ} - \vec{OR}}$
6. From (3), we have: $\small{n\,\vec{PR} = -m\,\vec{RQ}}$
• Substituting from (4) and (5), we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{n\,\vec{PR}}    & {~=~}    &{-m\,\vec{RQ}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{n\left(\vec{OR} - \vec{OP} \right)}    & {~=~}    &{-m\left(\vec{OQ} - \vec{OR} \right)}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{n\,\vec{OR} - n\,\vec{OP}}    & {~=~}    &{-m\,\vec{OQ} + m\,\vec{OR}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(m-n)\vec{OR}}    & {~=~}    &{m\,\vec{OQ}-n\,\vec{OP}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\vec{OR}}    & {~=~}    &{\frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}
\\ \end{array}}$

---

Now we will see a special case. It can be written in 2 steps:
1. Let R be the midpoint of PQ. Then we can apply case I because, R will be between P and Q
• So we have: $\small{\vec{OR} = \frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}$
2. Since R is the midpoint, we can write: m = n = 1
• Substituting these values of m and n in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(1)\,\vec{OQ}~+~(1)\,\vec{OP}}{1+1}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{OR}}    & {~=~}    &{\frac{\vec{OQ}~+~\vec{OP}}{2}}
\\ \end{array}}$

---

Now we will see some solved examples.

Solved example 26.29
Consider two points P and Q with position vectors $\small{\vec{OP} = 3\vec{a}-2\vec{b}}$ and $\small{\vec{OQ} = \vec{a}+\vec{b}}$. Find the position vector of a point R which divides the line joining P and Q in the ratio 2:1, (I) internally and (ii) externally.
Solution:
Part (i):
• For internal division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(\vec{a}+\vec{b} \right)~+~(1)\left(3\vec{a}-2\vec{b} \right)}{2+1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{2\vec{a} + 2\vec{b}+3\vec{a}-2\vec{b}}{3}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{5\vec{a}}{3}}
\\ \end{array}}$

Part (ii):
• For external division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(\vec{a}+\vec{b} \right)~-~(1)\left(3\vec{a}-2\vec{b} \right)}{2-1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{2\vec{a} + 2\vec{b}-3\vec{a}+2\vec{b}}{1}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{4\vec{b}-\vec{a}}
\\ \end{array}}$

Solved example 26.30
Show that the points
$\small{A\left(2\hat{i}-\hat{j}+\hat{k} \right)}$
$\small{B\left(\hat{i}-3\hat{j}-5\hat{k} \right)}$
$\small{C\left(3\hat{i}-4\hat{j}-4\hat{k} \right)}$
are the vertices of a right angled triangle
Solution:
1. Let us write the vectors connecting the points
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{\vec{OB} - \vec{OA}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\hat{i}-3\hat{j}-5\hat{k}~-~\left[2\hat{i}-\hat{j}+\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{\vec{OC} - \vec{OB}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{3\hat{i}-4\hat{j}-4\hat{k}~-~\left[\hat{i}-3\hat{j}-5\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{\vec{OA} - \vec{OC}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}~-~\left[3\hat{i}-4\hat{j}-4\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ \end{array}}$

2. Now we can write the squares of the magnitudes of the above vectors:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{AB} \right|^2}    & {~=~}    &{(-1)^2 + (-2)^2 + (-6)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{41}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{BC} \right|^2}    & {~=~}    &{(2)^2 + (-1)^2 + (1)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{6}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{CA} \right|^2}    & {~=~}    &{(-1)^2 + (3)^2 + (5)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{35}
\\ \end{array}}$

3. We see that:
$\small{\left|\vec{AB} \right|^2 = \left|\vec{BC} \right|^2 + \left|\vec{CA} \right|^2}$

4. Applying Pythagoras theorem, we can say that:
    ♦ AB is the hypotenuse
    ♦ BC and CA form base and altitude
• So the three points are the vertices of a right angled triangle.

Solved example 26.31
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are:
$\small{\vec{OP}=\hat{i}+2\hat{j}-\hat{k}}$ and $\small{\vec{OQ} = -\hat{i}+\hat{j}+\hat{k}}$, respectively in the ratio 2:1
(i) internally (ii) externally.
Solution:
Part (i):
• For internal division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~+~n\,\vec{OP}}{m+n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(-\hat{i}+\hat{j}+\hat{k} \right)~+~(1)\left(\hat{i}+2\hat{j}-\hat{k} \right)}{2+1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{-2\hat{i}+2\hat{j}+2\hat{k}~+~\left(\hat{i}+2\hat{j}-\hat{k} \right)}{3}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{-\hat{i}+4\hat{j}+\hat{k}}{3}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left(\frac{-1}{3} \right)\hat{i}+\left(\frac{4}{3} \right)\hat{j}+\left(\frac{1}{3} \right)\hat{k}}
\\ \end{array}}$

Part (ii):
• For external division, we have the formula:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}$
• Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{(2)\left(-\hat{i}+\hat{j}+\hat{k} \right)~-~(1)\left(\hat{i}+2\hat{j}-\hat{k} \right)}{2-1}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{-2\hat{i}+2\hat{j}+2\hat{k} ~-~\left(\hat{i}+2\hat{j}-\hat{k} \right)}{1}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-3\hat{i}+3\hat{k}}
\\ \end{array}}$

Solved example 26.32
Show that the points A, B and C with position vectors
$\small{3\hat{i}-4\hat{j}-4\hat{k}}$
$\small{2\hat{i}-\hat{j}+\hat{k}}$
$\small{\hat{i}-3\hat{j}-5\hat{k}}$,
respectively form the vertices of a right angled triangle
Solution:
1. Let us write the vectors connecting the points
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{\vec{OB} - \vec{OA}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}~-~\left[3\hat{i}-4\hat{j}-4\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{\vec{OC} - \vec{OB}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\hat{i}-3\hat{j}-5\hat{k}~-~\left[2\hat{i}-\hat{j}+\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{\vec{OA} - \vec{OC}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{3\hat{i}-4\hat{j}-4\hat{k}~-~\left[\hat{i}-3\hat{j}-5\hat{k} \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ \end{array}}$

2. Now we can write the squares of the magnitudes of the above vectors:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{-\hat{i}+3\hat{j}+5\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{AB} \right|^2}    & {~=~}    &{(-1)^2 + 3^2 + 5^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{35}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{-\hat{i}-2\hat{j}-6\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{BC} \right|^2}    & {~=~}    &{(-1)^2 + (-2)^2 + (-6)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{41}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{CA}}    & {~=~}    &{2\hat{i}-\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{CA} \right|^2}    & {~=~}    &{(2)^2 + (-1)^2 + (1)^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{6}
\\ \end{array}}$

3. We see that:
$\small{\left|\vec{BC} \right|^2 = \left|\vec{AB} \right|^2 + \left|\vec{CA} \right|^2}$

4. Applying Pythagoras theorem, we can say that:
    ♦ BC is the hypotenuse
    ♦ AB and CA form base and altitude
• So the three points are the vertices of a right angled triangle.

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The link below gives a few more solved examples:

Exercise 26.2

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After completing the above exercise, the reader may attempt the two problems given below:

Solved example 26.33
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are:
$\small{\left(2\vec{a}+\vec{b} \right)}$ and $\small{\left(\vec{a}-3\vec{b} \right)}$, externally in the ratio 1:2. Also show that P is the midpoint of the line segment RQ
Solution:
Part (i):
1. Given that, P is to be the midpoint of RQ. So we draw the rough sketch in such a way that, P is some where between R and Q. It is shown in fig.26.25 below:

Fig.26.25

2. OP and OQ are the original vectors. They are shown in magenta color. Point R divides QP externally into two parts: QR and PR

3. For this problem, we make the following changes:
(i) We consider $\small{\vec{QP}}$ instead of the usual $\small{\vec{PQ}}$
(ii) In the usual case, 'm' is related to the end P of $\small{\vec{PQ}}$. In the present case, 'm' is related to end Q of $\small{\vec{QP}}$
• The segment related to end Q is QR. In this problem, for external division, we consider QR:PR. Segment QR is larger than segment PR. So we can write:
QR:PR = m:n = 2:1
(iii) In the usual case, the formula that we use is:
$\small{\vec{OR} = \frac{m\,\vec{OQ}~-~n\,\vec{OP}}{m-n}}$
• So for the present case, we must change the formula to:
$\small{\vec{OR} = \frac{m\,\vec{OP}~-~n\,\vec{OQ}}{m-n}}$

4. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OR}}    & {~=~}    &{\frac{m\,\vec{OP}~-~n\,\vec{OQ}}{m-n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{(2)\left(2\vec{a}+\vec{b} \right)~-~(1)\left(\vec{a}-3\vec{b} \right)}{2-1}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{4\vec{a}+2\vec{b} ~-~\left(\vec{a}-3\vec{b} \right)}{2-1}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{3\vec{a}+5\vec{b}}
\\ \end{array}}$

Part (ii):
If P is the midpoint of RQ, we can say that, P divides RQ in the ratio 1:1.
1. We can treat $\small{\vec{OQ}~\text{and}~\vec{OR}}$ as the original vectors. Then we can find the position vector $\small{\vec{OS}}$ of the "assumed midpoint" S of QR
2. The original formula is:
$\small{\vec{OR} = \frac{\vec{OQ}~+~\vec{OP}}{2}}$
• In this original case, R is the midpoint of PQ
3. For the present case, we must change the formula to:
$\small{\vec{OS} = \frac{\vec{OR}~+~\vec{OQ}}{2}}$
4. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{OS}}    & {~=~}    &{\frac{\vec{OR}~+~\vec{OQ}}{2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{\left(3\vec{a}+5\vec{b} \right)~+~\left(\vec{a}-3\vec{b} \right)}{2}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{4\vec{a}+2\vec{b}}{2}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{2\vec{a}+\vec{b}}
\\ \end{array}}$
5. We see that: $\small{\vec{OP}=\vec{OS}}$
• That means, points P and S are the same.
• That means, P is the midpoint of QR

Solved example 26.34
Show that the points A(1,−2,−8), B(5,0,−2) and C(11,3,7) are collinear, and find the ratio in which B divides AC
Solution:
Part (i):
1. Fig.26.26(i) below shows the rough sketch

Fig.26.26

2. First we write $\small{\vec{AB}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}}    & {~=~}    &{(5-1)\hat{i}+(0-(-2))\hat{j}+(-2-(-8))\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{4\hat{i}+2\hat{j}+6\hat{k}}
\\ \end{array}}$

• Now we write the unit vector:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{AB}}    & {~=~}    &{\frac{\vec{AB}}{\left|\vec{AB} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{4\hat{i}+2\hat{j}+6\hat{k}}{\sqrt{16+4+36}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{4\hat{i}+2\hat{j}+6\hat{k}}{\sqrt{56}}~=~\frac{4\hat{i}+2\hat{j}+6\hat{k}}{\sqrt{4(14)}}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{4\hat{i}+2\hat{j}+6\hat{k}}{2\sqrt{14}}}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}}
\\ \end{array}}$

3. Next we write $\small{\vec{BC}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{BC}}    & {~=~}    &{(11-5)\hat{i}+(3-0)\hat{j}+(7-(-2))\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{6\hat{i}+3\hat{j}+9\hat{k}}
\\ \end{array}}$

• Now we write the unit vector:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{BC}}    & {~=~}    &{\frac{\vec{BC}}{\left|\vec{BC} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{6\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{36+9+81}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{6\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{126}}~=~\frac{6\hat{i}+3\hat{j}+9\hat{k}}{\sqrt{9(14)}}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{6\hat{i}+3\hat{j}+9\hat{k}}{3\sqrt{14}}}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}}
\\ \end{array}}$

4. We see that: $\small{\hat{AB}=\hat{BC}}$
• So $\small{\vec{AB}~\text{and}~\vec{BC}}$ are parallel.
• But B is a common point. So the three points are collinear.

Part (ii):
1. Fig.26.26(ii) above shows the rough sketch. From part (i), we know that A, B and C are collinear. We are asked to find the ratio $\small{\left|\vec{CB} \right|:\left|\vec{BA} \right|}$. We will denote this ratio as m:n

2. When CA is divided in this way, we get:
    ♦ $\small{\left|\vec{CB} \right|=m\left|\vec{CA} \right|}$
    ♦ $\small{\left|\vec{BA} \right|=n\left|\vec{CA} \right|}$
• So we want: $\small{\left|\vec{CB} \right|,~\left|\vec{BA} \right|~\text{and}~\left|\vec{CA} \right|}$

3. From the coordinates of A and C, we get:
$\small{\vec{CA} = -10\hat{i}-5\hat{j}-15\hat{k}}$
• Therefore, $\small{\left|\vec{CA} \right|=\sqrt{350}=5\sqrt{14}}$

4. From Part (i), we have: $\small{\vec{CB} = -6\hat{i}-3\hat{j}-9\hat{k}}$
• Therefore, $\small{\left|\vec{CB} \right|=\sqrt{126}=3\sqrt{14}}$ 

5. From Part (i), we have: $\small{\vec{BA} = -4\hat{i}-2\hat{j}-6\hat{k}}$
• Therefore, $\small{\left|\vec{BA} \right|=\sqrt{56}=2\sqrt{14}}$

6. So from (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{CB} \right|}    & {~=~}    &{m\left|\vec{CA} \right|}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{3\sqrt{14}}    & {~=~}    &{m\left(5\sqrt{14} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{m}    & {~=~}    &{\frac{3}{5}}
\\ \end{array}}$

7. Also from (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{BA} \right|}    & {~=~}    &{n\left|\vec{CA} \right|}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{2\sqrt{14}}    & {~=~}    &{n\left(5\sqrt{14} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{n}    & {~=~}    &{\frac{2}{5}}
\\ \end{array}}$

8. Thus the ratio m:n is $\small{\frac{3}{5}:\frac{2}{5}}$
• This is same as 3:2

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In the next section, we will see scalar product.

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