Exercise 10.4 (Revised) - Chapter 10 - Vector Algebra - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths: Chapter 10 - Vector Algebra Solutions

Ex 10.4 Question 1.

Find $|\vec{a} \times \vec{b}|$, if $\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$.

Answer.

Given: $\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$
$
\vec{a} \times \vec{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|
$

Expanding along first row,
$
\begin{aligned}
& \quad \vec{a} \times \vec{b}=\hat{i}\left|\begin{array}{cc}
-7 & 7 \\
-2 & 2
\end{array}\right|-\hat{j}\left|\begin{array}{cc}
1 & 7 \\
3 & 2
\end{array}\right|+\hat{k}\left|\begin{array}{cc}
1 & -7 \\
3 & -2
\end{array}\right|=\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21) \\
& =0 \hat{i}+19 \hat{j}+19 \hat{k} \\
& \therefore|\vec{a} \times \vec{b}|=\sqrt{(0)^2+(19)^2+(19)^2}=\sqrt{2(19)^2}=19 \sqrt{2}
\end{aligned}
$

Ex 10.4 Question 2.

Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$

Answer.

Given: $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$
On Adding $\vec{c}=\vec{a}+\vec{b}=3 \hat{i}+2 \hat{j}+2 \hat{k}+\hat{i}+2 \hat{j}-2 \hat{k}=4 \hat{i}+4 \hat{j}+0 \hat{k}$
On Subtracting $\vec{d}=\vec{a}-\vec{b}=3 \hat{i}+2 \hat{j}+2 \hat{k}-\hat{i}-2 \hat{j}+2 \hat{k}=2 \hat{i}+0 \hat{j}+4 \hat{k}$

Therefore, $\vec{n}=\vec{c} \times \vec{d}=\left|\begin{array}{lll}i & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4\end{array}\right|$
Expanding along first row $=\hat{i}(16-0)-\hat{i}(16-0)+\hat{k}(0-8)$
$
\begin{aligned}
& \Rightarrow \vec{n}=16 \hat{i}-16 \hat{j}-8 \hat{k} \\
& \therefore|\vec{n}|=\sqrt{(16)^2+(-16)^2+(-8)^2}=\sqrt{256+256+64}=\sqrt{576}=24
\end{aligned}
$

Therefore, a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is
$
\begin{aligned}
\hat{n} & = \pm \frac{\vec{n}}{|\vec{n}|}= \pm \frac{(16 \hat{i}-16 \hat{j}-8 \hat{k})}{24}= \pm\left(\frac{16}{24} \hat{i}-\frac{16}{24} \hat{j}-\frac{8}{24} \hat{k}\right)= \pm\left(\frac{2}{3} \hat{i}-\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right) \\
& = \pm \frac{2}{3} \hat{i} \mp \frac{2}{3} \hat{j} \mp \frac{2}{3} \hat{k}
\end{aligned}
$

Ex 10.4 Question 3.

If a unit vector $\hat{a}$ makes an angle $\frac{\pi}{3}$ with $\hat{i}, \frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence, the components of $\hat{a}$.

Answer.

Let $\hat{a}=x \hat{i}+y \hat{j}+z \hat{k}$ be a unit vector. (i)
$
\Rightarrow|\hat{a}|=1 \Rightarrow \sqrt{x^2+y^2+z^2}=1
$

Squaring both sides, $x^2+y^2+z^2=1$
Given: Angle between vectors $\hat{a}$ and $\hat{i}$ is $\frac{\pi}{3}$.
$
\therefore \cos \frac{\pi}{3}=\frac{\hat{a} \cdot \hat{i}}{|\hat{a}| \cdot|\hat{i}|} \Rightarrow \frac{1}{2}=\frac{x(1)+y(0)+z(0)}{(1)(1)}
$

$
\Rightarrow \frac{1}{2}=x
$

Again, given Angel between vectors $\hat{a}$ and $\hat{j}$ is $\frac{\pi}{4}$.
$
\begin{aligned}
& \therefore \cos \frac{\pi}{4}=\frac{\hat{a} \cdot \hat{j}}{|\hat{a}| \cdot|\hat{j}|} \Rightarrow \frac{1}{\sqrt{2}}=\frac{x(0)+y(1)+z(0)}{(1)(1)} \\
& \Rightarrow \frac{1}{\sqrt{2}}=y \ldots \ldots \ldots . \text { (iv) }
\end{aligned}
$

Again, given Angel between vectors $\hat{a}$ and $\hat{k}$ is $\theta$, where $\theta$ is acute angle.
$
\begin{aligned}
& \therefore \cos \theta=\frac{\hat{a} \hat{k}}{|\hat{a}| \cdot|\hat{k}|} \Rightarrow \cos \theta=\frac{x(0)+y(0)+z(1)}{(1)(1)} \\
& \Rightarrow \cos \theta=z \ldots \ldots \ldots . .(\mathrm{v})
\end{aligned}
$

Putting the values of $x, y$ and $z$ in eq. (ii),
$
\begin{aligned}
& \frac{1}{4}+\frac{1}{2}+\cos ^2 \theta=1 \Rightarrow \cos ^2 \theta=1-\frac{1}{4}-\frac{1}{2} \\
& \Rightarrow \cos ^2 \theta=\frac{4-1-2}{4}=\frac{1}{4} \Rightarrow \cos \theta= \pm \frac{1}{2}
\end{aligned}
$

Since $\theta$ is acute angle, therefore $\cos \theta$ is positive and hence $\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}$
From eq. (v), $z=\cos \theta=\frac{1}{2}$

$\Rightarrow \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \text { and angle } \theta=\frac{\pi}{3}$

Ex 10.4 Question 4.

Show that $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$
$
\begin{aligned}
& \text { Ans. L.H.S. }=(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b} \\
& {[\because \vec{a} \times \vec{a}=\overrightarrow{0}, \vec{b} \times \vec{b}=\overrightarrow{0}, \vec{b} \times \vec{a}=-\vec{a} \times \vec{b}]} \\
& =\overrightarrow{0}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-\overrightarrow{0}=2(\vec{a} \times \vec{b})=\text { R.H.S. }
\end{aligned}
$

Ex 10.4 Question 5.

Find $\lambda$ and $\mu$ if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$.

Answer.

Given: $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=0$
$
\Rightarrow\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|=\overrightarrow{0}
$

Expanding along first row,
$
\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}
$

Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides, we have
$
\begin{aligned}
& 6 \mu-27 \lambda=0 \\
& 2 \mu-27=0
\end{aligned}
$
$\qquad$
$\qquad$
And $2 \lambda-6=0$ $\qquad$
From eq. (ii), $2 \mu-27=0 \Rightarrow \mu=\frac{27}{2}$

From eq. (iii), $2 \lambda-6=0 \Rightarrow \lambda=\frac{6}{2}=3$
Putting the values of $\mu$ and $\lambda$ in eq. (i),
$
6\left(\frac{27}{2}\right)-27(3)=0 \Rightarrow 3(27)-27(3)=0 \Rightarrow 0=0
$

Therefore, $\mu=\frac{27}{2}$ and $\lambda=3$.

Ex 10.4 Question 6.

Given that $\vec{a} \cdot \vec{b}=0$ and $\vec{a} \times \vec{b}=\overrightarrow{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b} ?$

Answer.

Given: $\vec{a} \cdot \vec{b}=0 \Rightarrow|\vec{a}| \cdot|\vec{b}| \cos \theta=0$
$\therefore|\vec{a}|=0$ or $|\vec{b}|=0$ or $\cos \theta=0 \Rightarrow \theta=90^{\circ}$
$\Rightarrow \vec{a}=0$ or $\vec{b}=0$ or vector $\vec{a}$ is perpendicular to $\vec{b}$.
Again, given $\vec{a} \times \vec{b}=0 \Rightarrow|\vec{a} \times \vec{b}|=0 \Rightarrow|\vec{a}| \cdot|\vec{b}| \sin \theta=0$
$\therefore|\vec{a}|=0$ or $|\vec{b}|=0$ or $\sin \theta=0 \Rightarrow \theta=0^{\circ}$
$\Rightarrow \vec{a}=0 \quad$ or $\vec{b}=0$ or vector $\vec{a}$ and $\vec{b}$ are collinear or parallel.
Since, vectors $\vec{a}$ \& $\vec{b}$ are perpendicular to each other as well as parallel are not possible... (iii)

Therefore, from eq. (i), (ii) and (iii), either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$
$\therefore \vec{a} \cdot \vec{b}=0$ and $\vec{a} \times \vec{b}=0$

Ex 10.4 Question 7. 

$\text {Let the vectors } \vec{a}, \vec{b}, \vec{c} \text { be given as } a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}, c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k} \text {, }$ then show that $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$.
Answer.

Given: Vector $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ and $\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$
$
\begin{aligned}
& \therefore \vec{b}+\vec{c}=\left(b_1+c_1\right) \hat{i}+\left(b_2+c_2\right) \hat{j}+\left(b_3+c_3\right) \hat{k} \\
& \text { Now L.H.S. }=\vec{a} \times(\vec{b}+\vec{c})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1+c_1 & b_2+c_2 & b_3+c_3
\end{array}\right| \\
& =\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|+\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
c_1 & c_2 & c_3
\end{array}\right|
\end{aligned}
$
[By Property of Determinants]
$
=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}=\text { R.H.S. }
$

Ex 10.4 Question 8.

If either $\vec{a}=\overrightarrow{0}$ and $\vec{b}=\overrightarrow{0}$, then $\vec{a} \times \vec{b}=0$. Is the converse true? Justify your answer with an example.

Answer.

Given: Either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$
$
\begin{aligned}
& \therefore|\vec{a}|=|\overrightarrow{0}|=0 \text { or }|\vec{b}|=|\overrightarrow{0}|=0 \ldots \ldots \ldots . . \text { (i) } \\
& \therefore|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=0 \\
& \Rightarrow 0 . \sin \theta=0 \text { [Using eq. (i)] } \\
& \therefore \vec{a} \times \vec{b}=\overrightarrow{0} \text { [By definition of zero vector] }
\end{aligned}
$

But the converse is not true.

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k} \Rightarrow|\vec{a}|=\sqrt{1+1+1}=\sqrt{3} \neq 0$
$\therefore \vec{a}$ is a non-zero vector.
Let $\vec{b}=2(\hat{i}+\hat{j}+\hat{k})=2 \hat{i}+2 \hat{j}+2 \hat{k}$
$\Rightarrow|\vec{b}|=\sqrt{4+4+4}=\sqrt{12}=2 \sqrt{3} \neq 0$
$\therefore \vec{b}$ is a non-zero vector.
But $\vec{a} \times \vec{b}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2\end{array}\right|$
Taking 2 common from $R_3=\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|=\overrightarrow{0}\left[\because \mathrm{R}_2\right.$ and $\mathrm{R}_3$ are identical $]$

Ex 10.4 Question 9.

Find the area of the triangle with vertices $A(1,1,2), B(2,3,5)$ and $C(1,5,5)$.

Answer.

Vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(1,1,2), \mathrm{B}(2,3,5)$ and $\mathrm{C}(1,5,5)$.
$\therefore$ Position vector of point $\mathrm{A}=(1,1,2)=\hat{i}+\hat{j}+2 \hat{k}$
Position vector of point $\mathrm{B}=(2,3,5)=2 \hat{i}+3 \hat{j}+5 \hat{k}$
Position vector of point $\mathrm{C}=(1,5,5)=\hat{i}+5 \hat{j}+5 \hat{k}$
Now $\overrightarrow{\mathrm{AB}}=$ Position vector of point $B-$ Position vector of point $A$
$
\begin{aligned}
& =2 \hat{i}+3 \hat{j}+5 \hat{k}-(\hat{i}+\hat{j}+2 \hat{k}) \\
& =2 \hat{i}+3 \hat{j}+5 \hat{k}-\hat{i}-\hat{j}-2 \hat{k}
\end{aligned}
$

$
=\hat{i}+2 \hat{j}+3 \hat{k}
$

And $\overrightarrow{\mathrm{AC}}=$ Position vector of point $\mathrm{C}-$ Position vector of point $\mathrm{A}$
$
\begin{aligned}
& =\hat{i}+5 \hat{j}+5 \hat{k}-(\hat{i}+\hat{j}+2 \hat{k}) \\
& =\hat{i}+5 \hat{j}+5 \hat{k}-\hat{i}-\hat{j}-2 \hat{k} \\
& =0 \hat{i}+4 \hat{j}+3 \hat{k} \\
& \therefore \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
0 & 4 & 3
\end{array}\right| \\
& =\hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0)=-6 \vec{i}-3 \vec{j}+4 \vec{k}
\end{aligned}
$

Now Area of triangle $\mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\frac{1}{2} \sqrt{36+9+16}$
$
=\frac{1}{2} \sqrt{61} \text { sq. units }
$

Ex 10.4 Question Ex 10.4 Question F

ind the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$

Answer.

Given: Vectors representing two adjacent sides of a parallelogram are
$
\begin{aligned}
& \vec{a}=\hat{i}-\hat{j}+3 \hat{k} \text { and } \vec{b}=2 \hat{i}-7 \hat{j}+\hat{k} \\
& \therefore \bar{a} \times \bar{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 3 \\
2 & -7 & 1
\end{array}\right|=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)=20 \hat{i}+5 \hat{j}-5 \hat{k}
\end{aligned}
$

Now Area of parallelogram $=|\vec{a} \times \vec{b}|$

$=\sqrt{400+25+25}=\sqrt{450}=15 \sqrt{2} \text { sq. units }$

Ex 10.4 Question 11.

Let the vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}|=3,|\vec{b}|=\frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is:
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{2}$

Answer.

Given: $|\vec{a}|=3,|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector.
$\Rightarrow|\vec{a} \times \vec{b}|=1 \Rightarrow|\vec{a}| \cdot|\vec{b}| \sin \theta=1$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$\Rightarrow 3\left(\frac{\sqrt{2}}{3}\right) \sin \theta=1$
$\Rightarrow \sqrt{2} \sin \theta=1$
$\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}$
$\Rightarrow \sin \theta=\sin \frac{\pi}{4}$
$\Rightarrow \theta=\frac{\pi}{4}$
Therefore, option (B) is correct.

Ex 10.4 Question 12.

$\text {Area of a rectangle having vertices } A, B, C \text { and } D \text { with position vectors }$

$-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is:
(A) $\frac{1}{2}$
(B) 1
(C) 2
(D) 4

Answer.

Given: $\mathrm{ABCD}$ is a rectangle.
Now $\overrightarrow{\mathrm{AB}}=$ Position vector of point $\mathrm{B}-$ Position vector of point $\mathrm{A}$
$
\begin{aligned}
& =\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}-\left(-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right) \\
& =\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}+\hat{i}-\frac{1}{2} \hat{j}-4 \hat{k} \\
& =2 \hat{i}+0 \hat{j}+0 \hat{k} \\
& \therefore \mathrm{AB}=|\overrightarrow{\mathrm{AB}}|=\sqrt{4+0+0}=\sqrt{4}=2
\end{aligned}
$

And $\overrightarrow{\mathrm{AD}}=$ Position vector of point $\mathrm{D}-$ Position vector of point $\mathrm{A}$
$
\begin{aligned}
& =-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}-\left(-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right) \\
& =-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}+\hat{i}-\frac{1}{2} \hat{j}-4 \hat{k} \\
& =0 \hat{i}-\hat{j}+0 \hat{k} \\
& \therefore \mathrm{AD}=|\mathrm{AD}|=\sqrt{0+1+0}=\sqrt{1}=1
\end{aligned}
$
$\therefore$ Area of rectangle $\mathrm{ABCD}=$ Length $\mathrm{x}$ Breadth $=\mathrm{AB} \times \mathrm{AD}=2 \times 1=2$ sq. units Therefore, option (C) is correct.