26.7 - Scalar Product of Two Vectors

In the previous section, we saw vector joining two points. We also saw section formula. In this section, we will see scalar product of two vectors.

We know that, product of two numbers is a number. Similarly, product of two matrices is a matrix. But in the case of vectors, the product depends on the type of multiplication.
   ♦ Scalar multiplication of two vectors gives a scalar
   ♦ Vector multiplication of two vectors gives a vector
In this section, we will see scalar multiplication. It can be explained in 12 steps:

1. The scalar product of two nonzero vectors $\small{\vec{a}~\text{and}~\vec{b}}$ is defined as the product of three items:
(i) The magnitude of $\small{\vec{a}}$, which is: $\small{\left|\vec{a}\right|}$
(ii) The magnitude of $\small{\vec{b}}$, which is: $\small{\left|\vec{b}\right|}$
(iii) $\small{\cos \theta}$
   ♦ Where $\small{\theta}$ is the angle between the two vectors.
• The angle $\small{\theta}$ must be selected carefully. For any two vectors, there will be two angles between them.
   ♦ One which is less than or equal to $\small{\pi}$ radians.
   ♦ The other,which is greater than or equal to $\small{\pi}$ radians.
• This is shown in fig.26.27 below. We must select only the one which is less than or equal to $\small{\pi}$ radians.

Fig.26.27

2. The scalar product of two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ is denoted as $\small{\vec{a}.\vec{b}}$.
• So we can write: $\small{\vec{a}.\vec{b}~=~\left|\vec{a}\right|\,\left|\vec{b}\right|\,\cos \theta}$
3. If either $\small{\vec{a}=\vec{0}}$ or $\small{\vec{b}=\vec{0}}$, then $\small{\theta}$ is not defined. In such a situation, we define $\small{\vec{a}.\vec{b}~=~\vec{0}}$
4. In (1), we saw the three items which are multiplied to obtain the scalar product. Each of those three items, is a scalar. None of them has direction. So the scalar product is a scalar. In other words, scalar product is a real number. Recall that, cosine of any angle, is a real number. It does not have any unit.
5. Suppose that, $\small{\vec{a}~\text{and}~\vec{b}}$ are perpendicular to each other. Then $\small{\theta = \frac{\pi}{2}}$. Consequently, $\small{\cos \theta = \cos\left(\frac{\pi}{2} \right)=0}$. In such a situation, the scalar product will be zero. In fact, the scalar product is zero if and only if the two vectors are perpendicular to each other. We can write:
$\small{\vec{a}.\vec{b}~=~0\iff\vec{a}\perp\vec{b}}$
6. Suppose that, $\small{\theta = 0}$. Then $\small{\cos \theta = \cos\left(0 \right)=1}$. In such a situation, we get:
$\small{\vec{a}.\vec{b}~=~\left|\vec{a}\right|\,\left|\vec{b}\right|(1)~=~\left|\vec{a}\right|\,\left|\vec{b}\right|}$
• Based on this, we can write: $\small{\vec{a}.\vec{a}~=~\left|\vec{a}\right|^2}$.
This is because, the angle between $\small{\vec{a}}$ and itself is zero.
7. Suppose that, $\small{\theta = \pi}$. Then $\small{\cos \theta = \cos\left(\pi \right)=-1}$. In such a situation, we get:
$\small{\vec{a}.\vec{b}~=~\left|\vec{a}\right|\,\left|\vec{b}\right|(-1)~=~-\left|\vec{a}\right|\,\left|\vec{b}\right|}$
• Based on this, we can write: $\small{\vec{a}.\left(-\vec{a} \right)~=~(-1)\left|\vec{a}\right|^2}$.
This is because, the angle between $\small{\vec{a}~\text{and}~-\vec{a}}$ is $\small{\pi}$.
8. Based on (5), we get three interesting results:
   ♦ $\small{\hat{i}.\hat{j}~=~0}$
   ♦ $\small{\hat{j}.\hat{k}~=~0}$
   ♦ $\small{\hat{k}.\hat{i}~=~0}$
9. Based on (6), we get three interesting results:
   ♦ $\small{\hat{i}.\hat{i}~=~\left|\hat{i}\right|^2~=~1^2~=~1}$
   ♦ $\small{\hat{j}.\hat{j}~=~\left|\hat{j}\right|^2~=~1^2~=~1}$
   ♦ $\small{\hat{k}.\hat{k}~=~\left|\hat{k}\right|^2~=~1^2~=~1}$
10. Scalar product can be used to find the angle between two vectors. The method is shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{\left|\vec{a}\right|\,\left|\vec{b}\right|\,\cos \theta}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\cos \theta}    & {~=~}    &{\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ \end{array}}$
11. Based on (1), we can write:
scalar product involves the multiplication of three real numbers.
• Since they are real numbers, they can be multiplied in any order. So we can write:
$\small{\vec{a}.\vec{b}~=~\vec{b}.\vec{a}}$
• In other words, scalar product is commutative
12. Scalar product of two vectors is also known as dot product of two vectors.

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Two important properties of scalar product

Property I: Distributivity of scalar 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:
$\small{\vec{a}\left(\vec{b}+\vec{c} \right)~=~\vec{a}.\vec{b}+\vec{a}.\vec{c}}$

Property II: Distributivity of scalar 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:
$\small{\lambda \left(\vec{a}.\vec{b} \right) ~=~\left(\lambda \vec{a}\right).\vec{b} ~=~\vec{a}.\left(\lambda\vec{b} \right)}$

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Scalar 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:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{\left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \right).\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{a_1\hat{i}\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2\hat{j}\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3\hat{k}\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\hat{i}.\hat{i}\right)+a_1 b_2\left(\hat{i}.\hat{j}\right)+a_1 b_3\left(\hat{i}.\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(\hat{j}.\hat{i}\right)+a_2 b_2\left(\hat{j}.\hat{j}\right)+a_2 b_3\left(\hat{j}.\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{k}.\hat{i}\right)+a_3 b_2\left(\hat{k}.\hat{j}\right)+a_3 b_3\left(\hat{k}.\hat{k}\right)}
\\ {~\color{magenta}    4   }    &{}    &{}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ \end{array}}$

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Now we will see some solved examples.

Solved example 26.35
Find the angle between two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ with magnitudes 1 and 2 respectively and when $\small{\vec{a}.\vec{b}}$ = 1.
Solution:
• We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{1}{(1)(2)} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{1}{2} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{\pi}{3}}
\\ \end{array}}$

• Note:
$\small{\theta=\cos^{-1}\left(\frac{1}{2} \right)}$ will give infinite number of solutions. But we want that angle which lies in the interval $\small{\left(0,\pi \right)}$ 

Solved example 26.36
Find the angle between two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ with magnitudes $\small{\sqrt{3}}$ and 2, respectively having $\small{\vec{a}.\vec{b}=\sqrt{6}}$.
Solution:
• We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{\sqrt{6}}{(\sqrt{3})(2)} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{(\sqrt{3})(\sqrt{2})}{(\sqrt{3})(2)} \right)~=~\cos^{-1}\left(\frac{1}{\sqrt{2}} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\frac{\pi}{4}}
\\ \end{array}}$
• Note:
$\small{\theta=\cos^{-1}\left(\frac{1}{\sqrt{2}} \right)}$ will give infinite number of solutions. But we want that angle which lies in the interval $\small{\left(0,\pi \right)}$ 

Solved example 26.37
Find the angle $\small{\theta}$ between the vectors $\small{\vec{a}=\hat{i}+\hat{j}-\hat{k}}$ and $\small{\vec{b}=\hat{i}-\hat{j}+\hat{k}}$.
Solution:
1. We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}$
2. So first we need to find $\small{\vec{a}.\vec{b}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(1)(1)+(1)(-1)+(-1)(1)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{1-1-1}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{-1}
\\ \end{array}}$
3. Next we need to find the magnitudes:
$\small{\left|\vec{a}\right|=\sqrt{1^2 + 1^2 + (-1)^2}=\sqrt{3}}$
$\small{\left|\vec{b}\right|=\sqrt{1^2 + (-1)^2 + 1^2}=\sqrt{3}}$
4. Substituting the above values in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{-1}{(\sqrt{3})(\sqrt{3})} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{-1}{3} \right)}
\\ \end{array}}$

Solved example 26.38
Find the angle between the vectors $\small{\vec{a}=\hat{i}-2\hat{j}+3\hat{k}}$ and $\small{\vec{b}=3\hat{i}-2\hat{j}+\hat{k}}$.
Solution:
1. We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}$

2. So first we need to find $\small{\vec{a}.\vec{b}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\vec{b}}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(1)(3)+(-2)(-2)+(3)(1)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{3+4+3}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{10}
\\ \end{array}}$

3. Next we need to find the magnitudes:
$\small{\left|\vec{a}\right|=\sqrt{1^2 + (-2)^2 + 3^2}=\sqrt{14}}$

$\small{\left|\vec{b}\right|=\sqrt{3^2 + (-2)^2 + 1^2}=\sqrt{14}}$

4. Substituting the above values in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{a}.\vec{b}}{\left|\vec{a}\right|\,\left|\vec{b}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{10}{(\sqrt{14})(\sqrt{14})} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{10}{14} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{5}{7} \right)}
\\ \end{array}}$

Solved example 26.39
If $\small{\vec{a}=5\hat{i}-\hat{j}-3\hat{k}}$ and $\small{\vec{b}=\hat{i}+3\hat{j}-5\hat{k}}$, then show that the vectors $\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$ are perpendicular.
Solution:
1. First we write the sum and difference:
$\small{\vec{c}~=~\vec{a}+\vec{b}~=~6\hat{i}+2\hat{j}-8\hat{k}}$
$\small{\vec{d}~=~\vec{a}-\vec{b}~=~4\hat{i}-4\hat{j}+2\hat{k}}$

2. Next we find the angle $\small{\theta}$ between $\small{\vec{c}~\text{and}~\vec{d}}$

We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{c}.\vec{d}}{\left|\vec{c}\right|\,\left|\vec{d}\right|} \right)}$

3. So we need to find $\small{\vec{c}.\vec{d}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}.\vec{d}}    & {~=~}    &{c_1 d_1 + c_2 d_2 + c_3 d_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(6)(4)+(2)(-4)+(-8)(2)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{24-8-16}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$

4. There is no need to find the denominator.
Since $\small{\vec{c}.\vec{d}=0}$, we can directly write:
$\small{\theta=\cos^{-1}(0)}$

5. Solving this equation, we get: $\small{\theta~=~\frac{\pi}{2}}$
• Therefore, $\small{\vec{c}~\text{and}~\vec{d}}$ are perpendicular.
• That is., $\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$ are perpendicular.

Solved example 26.40
If $\small{\vec{a}}$ is a unit vector and $\small{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right) = 8}$, then find $\small{\left|\vec{x}\right|}$.
Solution:
1. First we write the scalar product:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right)}    & {~=~}    &{\left(\vec{x}-\vec{a} \right).\vec{x}~+~\left(\vec{x}-\vec{a} \right).\vec{a}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\vec{x}.\vec{x}-\vec{a}.\vec{x}+\vec{x}.\vec{a}-\vec{a}.\vec{a}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2-\left|\vec{a} \right|\left|\vec{x} \right|\cos \theta + \left|\vec{x} \right|\left|\vec{a} \right|\cos \theta - \left|\vec{a} \right|^2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2 - \left|\vec{a} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\left|\vec{x} \right|^2 - 1}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we assume that, the angle between $\small{\vec{x}~\text{and}~\vec{a}}$ is $\small{\theta}$
• 5 (magenta color): Here we use the given information that, $\small{\vec{a}}$ is a unit vector

2. Given that, this product is equal to 8. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{x}-\vec{a} \right).\left(\vec{x}+\vec{a} \right)}    & {~=~}    &{8}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\vec{x} \right|^2 - 1}    & {~=~}    &{8}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\vec{x} \right|^2}    & {~=~}    &{9}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\left|\vec{x} \right|}    & {~=~}    &{3}
\\ \end{array}}$

◼ Remarks:
• 4 (magenta color): Here we discard the −ve root because, length cannot be −ve.

Solved example 26.41
Find $\small{\left|\left(\vec{a}-\vec{b} \right) \right|}$ if two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ are such that $\small{\left|\vec{a}\right|}$ = 2, $\small{\left|\vec{b}\right|}$ = 3 and $\small{\vec{a}.\vec{b}=4}$.
Solution:
1. First let us write $\small{\left(\vec{a}-\vec{b} \right).\left(\vec{a}-\vec{b} \right)}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}-\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{\vec{a}.\left(\vec{a}-\vec{b} \right)~-~\vec{b}.\left(\vec{a}-\vec{b} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\vec{a}.\vec{a} - \vec{a}.\vec{b}-\vec{b}.\vec{a}+\vec{b}.\vec{b}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{a}.\vec{a} - 4-4+\vec{b}.\vec{b}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2 - 8+\left|\vec{b} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{2^2 - 8+3^2}
\\ {~\color{magenta}    6    }    &{}    &{}    & {~=~}    &{5}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we use the given information that,
$\small{\vec{a}.\vec{b}}$ = 4
• 5 (magenta color): Here we use the given information that,
$\small{\left|\vec{a}\right|}$ = 2, $\small{\left|\vec{b}\right|}$ = 3

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}-\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{5}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left|\left(\vec{a}-\vec{b} \right) \right|^2}    & {~=~}    &{5}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\left(\vec{a}-\vec{b} \right) \right|}    & {~=~}    &{\sqrt{5}}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we discard the −ve root because, length cannot be −ve.

Solved example 26.42
Find $\small{\left|\vec{a}\right|}$ and $\small{\left|\vec{b}\right|}$, if $\small{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)=8}$ and $\small{\left|\vec{a}\right|=8\left|\vec{b} \right|}$.
Solution:
1. First we write the dot product:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{\vec{a}\left(\vec{a}-\vec{b} \right)+\vec{b}\left(\vec{a}-\vec{b} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2-\vec{a}.\vec{b}+\vec{b}.\vec{a}-\left|\vec{b} \right|^2}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2-\left|\vec{b} \right|^2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{64\left|\vec{b} \right|^2-\left|\vec{b} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{63\left|\vec{b} \right|^2}
\\ \end{array}}$

◼ Remarks:
• 4 (magenta color): Here we use the given information that, $\small{\left|\vec{a}\right|=8\left|\vec{b} \right|}$.

2. We are also given that: $\small{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)=8}$
• Substituting this in the above result, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}-\vec{b} \right)}    & {~=~}    &{63\left|\vec{b} \right|^2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{63\left|\vec{b} \right|^2}    & {~=~}    &{8}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left|\vec{b} \right|^2}    & {~=~}    &{\frac{8}{63}~=~\frac{(2)2^2}{(7)3^2}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\left|\vec{b} \right|}    & {~=~}    &{\frac{2 \sqrt{2}}{3 \sqrt{7}}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\left|\vec{a} \right|}    & {~=~}    &{8\,\left|\vec{b} \right|~=~\frac{16 \sqrt{2}}{3 \sqrt{7}}}
\\ \end{array}}$

◼ Remarks:
• 4 (magenta color): Here we discard the −ve root because, length cannot be −ve.

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In the next section, we will see a few more solved examples. We will also see the Cauchy-Schwartz inequality.

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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.

---

The link below gives a few more solved examples:

Exercise 26.2

---

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

---

In the next section, we will see scalar product.

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26.5 - Solved Examples

In the previous section, we saw components of a vector. We saw a solved example also. In this  section, we will see a few more solved examples.

Solved example 26.7
Find the values of x and y so that the vectors $\small{2\hat{i} + 3\hat{j}}$ and $\small{x\hat{i} + y\hat{j}}$ are equal
Solution:
1. If two vectors are equal, their corresponding components will be equal.
2. Let us equate the corresponding components:
• Equating the vector components along the x-axis, we get: $\small{2\hat{i} = x\hat{i}}$. Therefore, x = 2
• Equating the vector components along the y-axis, we get: $\small{3\hat{j} = y\hat{i}}$. Therefore, y = 3

Solved example 26.8
Let $\small{\vec{a} = \hat{i} + 2\hat{j}}$ and $\small{\vec{b} = 2\hat{i} + \hat{j}}$.
Is $\small{\left|\vec{a} \right| = \left|\vec{b} \right|}$?
Are the vectors $\small{\vec{a}~\text{and}~\vec{b}}$ equal?
Solution:
Part (a): Comparing the magnitudes
1. $\small{\left|\vec{a} \right| = \sqrt{1^2 + 2^2} = \sqrt{5}}$
2. $\small{\left|\vec{b} \right| = \sqrt{2^2 + 1^2} = \sqrt{5}}$
3. The magnitudes are equal. So $\small{\left|\vec{a} \right| = \left|\vec{b} \right|}$

Part (b): Checking equality of vectors
1. If two vectors are equal, their corresponding components will be equal.
2. Let us compare the corresponding components:
• Comparing the vector components along the x-axis, we see that: $\small{\hat{i} ~\ne~ 2\hat{i}}$.
• Comparing the vector components along the y-axis, we see that: $\small{2\hat{j} ~\ne~ \hat{j}}$
3. Two vectors cannot be equal if even one of the corresponding components are not equal. So the given two vectors are not equal.

Solved example 26.9
Compute the magnitude of the following vectors:
$\small{\vec{a} = \hat{i} + \hat{j} + \hat{k}}$
$\small{\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}}$
$\small{\vec{c} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} - \frac{1}{\sqrt{3}} \hat{k}}$
Solution:
Part (a):
$\small{\left|\vec{a} \right| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}}$
Part (b):
$\small{\left|\vec{b} \right| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}}$
Part (c):
$\small{\left|\vec{c} \right| = \sqrt{\left(\frac{1}{\sqrt{3}} \right)^2 + \left(\frac{1}{\sqrt{3}} \right)^2 + \left(-\frac{1}{\sqrt{3}} \right)^2} =\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1}$

Solved example 26.10
Write two different vectors having same magnitude
Solution:
We want two different vectors having the same magnitude. It can be done in 2 steps:
1. First write any convenient vector, say: $\small{\vec{a} = 3\hat{i}+4\hat{j}}$
2. Now we write the second vector by changing the order/sign of the scalar components. So the two vectors are:
$\small{\vec{a} = 3\hat{i}+4\hat{j}}$
$\small{\vec{b} = 4\hat{i}+3\hat{j}}$
• We have the magnitudes:
$\small{\left|\vec{a} \right| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5}$
$\small{\left|\vec{b} \right| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5}$
• Fig.26.20 below shows the difference in directions:

Fig.26.20

◼ Note:
There are other possibilities also:
$\small{\vec{a} = 3\hat{i}+4\hat{j}}$
$\small{\vec{c} = 3\hat{i}-4\hat{j}}$
$\small{\vec{d} = -4\hat{i}+3\hat{j}}$
$\small{\vec{e} = -3\hat{i}-4\hat{j}}$
• All these vectors have different directions. But they have the same magnitude. We can pick any two from them.

Solved example 26.11
Find the unit vector in the direction of the vector $\small{\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}}$
Solution:
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{a}}    & {~=~}    &{\frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{2^2 + 3^2 + 1^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{2}{\sqrt{14}} \right) \hat{i}+\left(\frac{3}{\sqrt{14}} \right) \hat{j}+\left(\frac{1}{\sqrt{14}} \right) \hat{k}}
\\ \end{array}}$

Solved example 26.12
Find the unit vector in the direction of the vector $\small{\vec{a} = \hat{i}+\hat{j}+2\hat{k}}$
Solution:
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{a}}    & {~=~}    &{\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{1^2 + 1^2 + 2^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{1}{\sqrt{6}} \right) \hat{i}+\left(\frac{1}{\sqrt{6}} \right) \hat{j}+\left(\frac{2}{\sqrt{6}} \right) \hat{k}}
\\ \end{array}}$

Solved example 26.13
Find a vector in the direction of the vector $\small{\vec{a} = \hat{i}-2\hat{j}}$ that has magnitude 7 units.
Solution:
1. First we will write the unit vector $\small{\hat{a}}$
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{a}}    & {~=~}    &{\frac{\hat{i}-2\hat{j}}{\sqrt{1^2 + (-2)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{\hat{i}-2\hat{j}}{\sqrt{5}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{1}{\sqrt{5}} \right) \hat{i}-\left(\frac{2}{\sqrt{5}} \right) \hat{j}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{a}}$

2. So the required vector is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{7\hat{a}}    & {~=~}    &{7\left[\left(\frac{1}{\sqrt{5}} \right) \hat{i}-\left(\frac{2}{\sqrt{5}} \right) \hat{j} \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(\frac{7}{\sqrt{5}} \right) \hat{i}-\left(\frac{14}{\sqrt{5}} \right) \hat{j}}
\\ \end{array}}$

Solved example 26.14
Find a vector in the direction of the vector $\small{\vec{a} = 5\hat{i}-\hat{j}+2\hat{k}}$ that has magnitude 8 units.
Solution:
1. First we will write the unit vector $\small{\hat{a}}$
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{a}}    & {~=~}    &{\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{5^2 + (-1)^2 + 2^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{5}{\sqrt{30}} \right) \hat{i}-\left(\frac{1}{\sqrt{30}} \right) \hat{j}+\left(\frac{2}{\sqrt{30}} \right) \hat{j}}
\\ \end{array}}$

• This unit vector has the same direction as $\small{\vec{a}}$

2. So the required vector is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{8\hat{a}}    & {~=~}    &{8\left[\left(\frac{5}{\sqrt{30}} \right) \hat{i}-\left(\frac{1}{\sqrt{30}} \right) \hat{j}+\left(\frac{2}{\sqrt{30}} \right) \hat{j} \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(\frac{40}{\sqrt{30}} \right) \hat{i}-\left(\frac{8}{\sqrt{30}} \right) \hat{j}+\left(\frac{16}{\sqrt{30}} \right) \hat{j}}
\\ \end{array}}$

Solved example 26.15
Write two different vectors having same direction
Solution:
We want two different vectors having the same direction. It can be done in 6 steps:
1. First write any convenient vector, say: $\small{\vec{a} = 2\hat{i}+3\hat{j}}$
2. Now write $\small{\hat{a}}$. We get:
$\small{\hat{a}=\left(\frac{2}{\sqrt{13}} \right) \hat{i}+\left(\frac{3}{\sqrt{13}} \right) \hat{j}}$
• The reader may write all steps involved in finding the unit vector
3. Multiply the unit vector in (2), by any convenient scalar, to get a new vector $\small{\vec{b}}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{b}}    & {~=~}    &{\sqrt{13}\,\hat{a}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{13}\left[\left(\frac{2}{\sqrt{13}} \right) \hat{i}+\left(\frac{3}{\sqrt{13}} \right) \hat{j} \right]}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{b}}    & {~=~}    &{2 \hat{i}+3 \hat{j}}
\\ \end{array}}$
• Note that: $\small{\vec{b}=\vec{a}}$
4. Multiply the unit vector in (2), by any other convenient scalar, to get a third vector $\small{\vec{c}}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}}    & {~=~}    &{2\sqrt{13}\,\hat{a}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2\sqrt{13}\left[\left(\frac{2}{\sqrt{13}} \right) \hat{i}+\left(\frac{3}{\sqrt{13}} \right) \hat{j} \right]}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{c}}    & {~=~}    &{4 \hat{i}+6 \hat{j}}
\\ \end{array}}$
5. $\small{\vec{b}~\text{and}~\vec{c}}$ are the required vectors
6. Fig.26.21 shows the two vectors:

Fig.26.21

Solved example 26.16
Find the sum of the vectors
$\small{\vec{a} = \hat{i}-2 \hat{j} + \hat{k}}$
$\small{\vec{b} = -2\hat{i}+4 \hat{j} + 5\hat{k}}$
$\small{\vec{c} = \hat{i}-6 \hat{j} -7 \hat{k}}$
Solution:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}+\vec{b}+\vec{c}}    & {~=~}    &{(1-2+1)\hat{i}+(-2+4-6) \hat{j} + (1+5-7)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(0)\hat{i}+(-4) \hat{j} + (-1)\hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-4 \hat{j}-\hat{k}}
\\ \end{array}}$

Solved example 26.17
Find unit vector in the direction of the sum of the vectors
$\small{\vec{a} = 2\hat{i}+2 \hat{j} -5 \hat{k}}$
$\small{\vec{b} = 2\hat{i}+ \hat{j} + 3\hat{k}}$
Solution:
1. First we will find the sum:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c} = \vec{a}+\vec{b}}    & {~=~}    &{(2+2)\hat{i}+(2+1) \hat{j} + (-5+3)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(4)\hat{i}+(3) \hat{j} + (-2)\hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{4\hat{i}+3 \hat{j} -2\hat{k}}
\\ \end{array}}$

2. Now we will write the unit vector $\small{\hat{c}}$
• We have: $\small{\hat{c}=\frac{\vec{c}}{\left|\vec{c} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{c}}    & {~=~}    &{\frac{4\hat{i}+3 \hat{j} -2\hat{k}}{\sqrt{4^2 + 3^2 + (-2)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{4\hat{i}+3 \hat{j} -2\hat{k}}{\sqrt{29}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{4}{\sqrt{29}} \right) \hat{i}+\left(\frac{3}{\sqrt{29}} \right) \hat{j}-\left(\frac{2}{\sqrt{29}} \right) \hat{k}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{c}}$ 

Solved example 26.18
Find unit vector in the direction of the sum of the vectors
$\small{\vec{a} = 2\hat{i}- \hat{j} +2 \hat{k}}$
$\small{\vec{b} = -\hat{i}+ \hat{j} - \hat{k}}$
Solution:
1. First we will find the sum:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c} = \vec{a}+\vec{b}}    & {~=~}    &{(2-1)\hat{i}+(-1+1) \hat{j} + (2-1)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(1)\hat{i}+(0) \hat{j} + (1)\hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\hat{i}+\hat{k}}
\\ \end{array}}$

2. Now we will write the unit vector $\small{\hat{c}}$
• We have: $\small{\hat{c}=\frac{\vec{c}}{\left|\vec{c} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{c}}    & {~=~}    &{\frac{\hat{i}+\hat{k}}{\sqrt{1^2 + 1^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{\hat{i}+\hat{k}}{\sqrt{2}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{1}{\sqrt{2}} \right) \hat{i}+\left(\frac{1}{\sqrt{2}} \right) \hat{k}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{c}}$

Solved example 26.19
Write the direction ratios of the vector $\small{\vec{a} = \hat{i} + \hat{j} -2 \hat{k}}$ and hence calculate its direction cosines.
Solution:
1. Any vector is the resultant of three component vectors:
    ♦ Component along the OX axis, which is $\small{x \hat{i}}$
    ♦ Component along the OY axis, which is $\small{y \hat{j}}$
    ♦ Component along the OZ axis, which is $\small{z \hat{k}}$
• So any given vector can be written as: $\small{x\hat{i}+y\hat{j}+z\hat{k}}$
2. The vector given to us is: $\small{\vec{a}=\hat{i}+\hat{j}-2\hat{k}}$
• Comparing the corresponding components, we get: x = 1, y = 1 and z = −2
3. $\small{r = \left|\vec{a} \right|=\sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}}$
4. The three direction ratios are:
    ♦ $\small{lr = x = 1}$
    ♦ $\small{mr = y = 1}$
    ♦ $\small{nr = z = -2}$
Where $\small{l,~m,~n}$ are the direction cosines
5. So the three direction cosines are:
    ♦ $\small{l = \frac{x}{r} = \frac{1}{\sqrt{6}}}$
    ♦ $\small{m = \frac{y}{r} = \frac{1}{\sqrt{6}}}$
    ♦ $\small{n = \frac{z}{r} = \frac{-2}{\sqrt{6}}}$

Solved example 26.20
Find the direction cosines of the vector $\small{\vec{a} = \hat{i} + 2\hat{j} +3 \hat{k}}$.
Solution:
1. Any vector is the resultant of three component vectors:
    ♦ Component along the OX axis, which is $\small{x \hat{i}}$
    ♦ Component along the OY axis, which is $\small{y \hat{j}}$
    ♦ Component along the OZ axis, which is $\small{z \hat{k}}$
• So any given vector can be written as: $\small{x\hat{i}+y\hat{j}+z\hat{k}}$
2. The vector given to us is: $\small{\vec{a}=\hat{i}+2\hat{j}+3\hat{k}}$
• Comparing the corresponding components, we get: x = 1, y = 2 and z = 3
3. $\small{r = \left|\vec{a} \right|=\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}}$
4. The three direction ratios are:
    ♦ $\small{lr = x = 1}$
    ♦ $\small{mr = y = 2}$
    ♦ $\small{nr = z = 3}$
Where $\small{l,~m,~n}$ are the direction cosines
5. So the three direction cosines are:
    ♦ $\small{l = \frac{x}{r} = \frac{1}{\sqrt{14}}}$
    ♦ $\small{m = \frac{y}{r} = \frac{2}{\sqrt{14}}}$
    ♦ $\small{n = \frac{z}{r} = \frac{3}{\sqrt{14}}}$

Solved example 26.21
Show that the vector $\small{\vec{a} = \hat{i} + \hat{j} + \hat{k}}$ is equally inclined to the axes OX, OY and OZ.
Solution:
1. Any vector is the resultant of three component vectors:
    ♦ Component along the OX axis, which is $\small{x \hat{i}}$
    ♦ Component along the OY axis, which is $\small{y \hat{j}}$
    ♦ Component along the OZ axis, which is $\small{z \hat{k}}$
• So any given vector can be written as: $\small{x\hat{i}+y\hat{j}+z\hat{k}}$
2. The vector given to us is: $\small{\vec{a}=\hat{i}+\hat{j}+\hat{k}}$
• Comparing the corresponding components, we get: x = 1, y = 1 and z = 1
3. $\small{r = \left|\vec{a} \right|=\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}}$
4. The three direction ratios are:
    ♦ $\small{lr = x = 1}$
    ♦ $\small{mr = y = 1}$
    ♦ $\small{nr = z = 1}$
Where $\small{l,~m,~n}$ are the direction cosines
5. So the three direction cosines are:
    ♦ $\small{l = \frac{x}{r} = \frac{1}{\sqrt{3}}}$
    ♦ $\small{m = \frac{y}{r} = \frac{1}{\sqrt{3}}}$
    ♦ $\small{n = \frac{z}{r} = \frac{1}{\sqrt{3}}}$
6. The three direction cosines are equal. That means, the three angles are equal.

Solved example 26.22
Show that the vectors $\small{\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}}$ and $\small{\vec{b}=-4\hat{i}+6\hat{j}-8\hat{k}}$ are collinear.
Solution:
1. Unit vector in the direction of $\small{\vec{a}}$
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{a}}    & {~=~}    &{\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{2^2 + (-3)^2 + (4)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{2}{\sqrt{29}} \right) \hat{i}-\left(\frac{3}{\sqrt{29}} \right) \hat{j}+\left(\frac{4}{\sqrt{29}} \right) \hat{k}}
\\ \end{array}}$

2. Unit vector in the direction of $\small{\vec{b}}$
• We have: $\small{\hat{b}=\frac{\vec{b}}{\left|\vec{b} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{b}}    & {~=~}    &{\frac{-4\hat{i}+6\hat{j}-8\hat{k}}{\sqrt{(-4)^2 + 6^2 + (-8)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{-4\hat{i}+6\hat{j}-8\hat{k}}{\sqrt{116}}~=~\frac{-4\hat{i}+6\hat{j}-8\hat{k}}{2\sqrt{29}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{-2}{\sqrt{29}} \right) \hat{i}+\left(\frac{3}{\sqrt{29}} \right) \hat{j}-\left(\frac{4}{\sqrt{29}} \right) \hat{k}}
\\ \end{array}}$

3. We see that, $\small{\hat{a}=-\hat{b}}$
• That means: $\small{\hat{a}}$ has the exact opposite direction of $\small{\hat{b}}$
• That means: $\small{\vec{a}}$ has the exact opposite direction of $\small{\vec{b}}$
• That means: $\small{\vec{a}}$ and $\small{\vec{b}}$ are parallel.
• Therefore  $\small{\vec{a}}$ and $\small{\vec{b}}$ are collinear.

Solved example 26.23
If $\small{\vec{a}~\text{and}~\vec{b}}$ are two collinear vectors, then which of the following are incorrect:
(a) $\small{\vec{b} = \lambda\vec{a}}$ for some scalar $\small{\lambda}$
(b) $\small{\vec{a} = \pm \vec{b}}$
(c) the respective components of $\small{\vec{a}~\text{and}~\vec{b}}$ are proportional
(d) both the vectors $\small{\vec{a}~\text{and}~\vec{b}}$ have same direction, but different magnitudes
Solution:
Part (a):
Given that $\small{\vec{a}~\text{and}~\vec{b}}$ are collinear. That means, they are parallel. So (a) is true.
Part (b):
$\small{\vec{a}~\text{and}~\vec{b}}$ are parallel. But they need not have the same direction. The directions may be opposite to each other. So (b) is true.
Part (c):
Since (a) is true, we can multiply each component of $\small{\vec{a}}$ by $\small{\lambda}$. That means, corresponding components are proportional. So (c) is true.
Part (d):
$\small{\vec{a}~\text{and}~\vec{b}}$ are parallel. But they need not have the same direction. The directions may be opposite to each other. So (d) is false.

Therefore, the correct option is (d)

Solved example 26.24
Find the value of x for which $\small{x\left(\hat{i}+\hat{j}+\hat{k} \right)}$ is a unit vector
Solution:
1. Let $\small{\vec{a}=x\left(\hat{i}+\hat{j}+\hat{k} \right)}$.
Then we can write: $\small{\vec{a}=x\hat{i}+x\hat{j}+x\hat{k}}$
2. Now we can calculate the magnitude:
$\small{\left|\vec{a} \right| = \sqrt{x^2 + x^2 + x^2} = \sqrt{3x^2} = \pm\sqrt{3}\,x}$
3. $\small{\vec{a}}$ is to be a unit vector. So it's magnitude must be 1. We can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a} \right|}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\pm\sqrt{3}\,x}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{x}    & {~=~}    &{\pm\frac{1}{\sqrt{3}}}
\\ \end{array}}$

Solved example 26.25
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
$\small{\vec{a}=2\hat{i}+3\hat{j}-\hat{k}~\text{and}~\vec{b}=\hat{i}-2\hat{j}+\hat{k}}$
Solution:
1. First we find the resultant $\small{\vec{a}+\vec{b}}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}}    & {~=~}    &{\vec{a}+\vec{b}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{2\hat{i}+3\hat{j}-\hat{k}~+~\hat{i}-2\hat{j}+\hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{3\hat{i}+\hat{j}}
\\ \end{array}}$

2. Now we will write the unit vector $\small{\hat{c}}$
• We have: $\small{\hat{c}=\frac{\vec{c}}{\left|\vec{c} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{c}}    & {~=~}    &{\frac{3\hat{i}+\hat{j}}{\sqrt{3^2 + 1^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{3\hat{i}+\hat{j}}{\sqrt{10}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{3}{\sqrt{10}} \right) \hat{i}+\left(\frac{1}{\sqrt{10}} \right) \hat{j}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{c}}$

3. So the required vector is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{5\hat{c}}    & {~=~}    &{5\left[\left(\frac{3}{\sqrt{10}} \right) \hat{i}+\left(\frac{1}{\sqrt{10}} \right) \hat{j} \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(\frac{15}{\sqrt{10}} \right) \hat{i}+\left(\frac{5}{\sqrt{10}} \right) \hat{j}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{15 \sqrt{10}}{10} \right) \hat{i}+\left(\frac{5 \sqrt{10}}{10} \right) \hat{j}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left(\frac{3 \sqrt{10}}{2} \right) \hat{i}+\left(\frac{\sqrt{10}}{2} \right) \hat{j}}
\\ \end{array}}$

Solved example 26.26
If
$\small{\vec{a}=\hat{i}+\hat{j}+\hat{k}}$
$\small{\vec{b}=2\hat{i}-\hat{j}+3\hat{k}}$
$\small{\vec{c}=\hat{i}-2\hat{j}+\hat{k}}$
find a unit vector parallel to the vector $\small{2\vec{a}-\vec{b}+3\vec{c}}$
Solution:
1. First we find the vector $\small{\vec{d}=2\vec{a}-\vec{b}+3\vec{c}}$.
We get: $\small{\vec{d}=3\hat{i}-3\hat{j}+2\hat{k}}$
2. Now we will write the unit vector $\small{\hat{d}}$
• We have: $\small{\hat{d}=\frac{\vec{d}}{\left|\vec{d} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{d}}    & {~=~}    &{\frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{3^2 +(-3)^2 + 2^2}}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{22}}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{3}{\sqrt{22}} \right) \hat{i}-\left(\frac{3}{\sqrt{22}} \right) \hat{j}+\left(\frac{2}{\sqrt{22}} \right) \hat{k}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{d}}$.

Solved example 26.27
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\small{\frac{1}{\sqrt{3}},~\frac{1}{\sqrt{3}},~\frac{1}{\sqrt{3}}}$
Solution:
1. Any vector is the resultant of three component vectors:
    ♦ Component along the OX axis, which is $\small{x \hat{i}}$
    ♦ Component along the OY axis, which is $\small{y \hat{j}}$
    ♦ Component along the OZ axis, which is $\small{z \hat{k}}$
• So any given vector can be written as: $\small{\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}}$
2. Let $\small{r = \left|\vec{a} \right|=\sqrt{x^2 + y^2 + z^2}}$
3. The three direction ratios are:
    ♦ $\small{lr = x}$
    ♦ $\small{mr = y}$
    ♦ $\small{nr = z}$
Where $\small{l,~m,~n}$ are the direction cosines
4. So the three direction cosines are:
    ♦ $\small{l = \frac{x}{r}}$
    ♦ $\small{m = \frac{y}{r}}$
    ♦ $\small{n = \frac{z}{r}}$
5. The vector is  equally inclined to the axes OX, OY and OZ. So we can write:
$\small{l=m=n}$
• Then from (4), we get:
$\small{\Rightarrow x=y=z}$
6. So from (2), we get:
• $\small{r=\sqrt{x^2 + x^2 + x^2}=\sqrt{3x^2}=\sqrt{3}\,x }$
• $\small{r=\sqrt{y^2 + y^2 + y^2}=\sqrt{3y^2}=\sqrt{3}\,y }$
• $\small{r=\sqrt{z^2 + z^2 + z^2}=\sqrt{3z^2}=\sqrt{3}\,z }$
• Note: Here we take only the +ve roots because, r is a distance.
7. Based on (4) and (6), we get:
• $\small{l = \frac{x}{r}=\frac{x}{\sqrt{3}\,x}=\frac{1}{\sqrt{3}}}$
• $\small{m = \frac{y}{r}=\frac{y}{\sqrt{3}\,y}=\frac{1}{\sqrt{3}}}$
• $\small{n = \frac{z}{r}=\frac{z}{\sqrt{3}\,z}=\frac{1}{\sqrt{3}}}$

 

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In the next section, we will see vector joining two points.

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