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 30o 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 30o 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 30o 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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{\beta_2}} & {~=~} &{\left(2 - 3\lambda \right)\hat{i}+\left(1+\lambda \right)\hat{j}-3\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\left(2 - 3\left(\frac{1}{2} \right) \right)\hat{i}+\left(1+\left(\frac{1}{2} \right) \right)\hat{j}-3\hat{k}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{3}{2} \right)\hat{j}-3\hat{k}}
\\ \end{array}}$
The link below gives a few more miscellaneous examples:
Miscellaneous Exercise
In the next chapter, we will see Three dimensional geometry.
Previous
Contents
Next
Copyright©2026 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment