NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 - Vector Algebra

Question 2: Find a unit vector perpendicular to each of the vector $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ , 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 in the question

$\vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k$

$\vec a + \vec b =3\hat i +2\hat j+2\hat k+\hat i +2\hat j-2\hat k=4\hat i +4\hat j$

$\vec a - \vec b =3\hat i +2\hat j+2\hat k-\hat i -2\hat j+2\hat k=2\hat i +4\hat k$

Now , A vector which perpendicular to both $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is $(\vec a + \vec b) \times (\vec a - \vec b)$

$(\vec a + \vec b) \times (\vec a - \vec b)=\begin{vmatrix} \hat i &\hat j &\hat k \\ 4&4 &0 \\ 2& 0& 4 \end{vmatrix}$

$(\vec a + \vec b) \times (\vec a - \vec b)= \hat i (16-0)-\hat j(16-0)+\hat k(0-8)$

$(\vec a + \vec b) \times (\vec a - \vec b)= 16\hat i -16\hat j-8\hat k$

And a unit vector in this direction :

$\vec u =\frac{16\hat i-16\hat j-8\hat k}{|16\hat i-16\hat j-8\hat k|}=\frac{16\hat i-16\hat j-8\hat k}{\sqrt{16^2+(-16)^2+(-8)^2}}$

$\vec u =\frac{16\hat i-16\hat j-8\hat k}{24}=\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$

Hence Unit vector perpendicular to each of the vector $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is $\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$ .

Question 3: If a unit vector $\vec a$ makes angles $\frac{\pi }{3}$ 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 $\vec a$ .

Answer:

Given in the question,

angle between $\vec a$ and $\hat i$ :

$\alpha =\frac{\pi}{3}$

angle between $\vec a$ and $\hat j$

$\beta =\frac{\pi}{4}$

angle with $\vec a$ and $\hat k$ :

$\gamma =\theta$

Now, As we know,

$cos^2\alpha+cos^2\beta+cos^2\gamma=1$

$cos^2\frac{\pi}{3}+cos^2\frac{\pi }{4}+cos^2\theta=1$

$\left ( \frac{1}{2} \right )^2+\left ( \frac{1}{\sqrt{2}} \right )^2+cos^2\theta=1$

$cos^2\theta=\frac{1}{4}$

$cos\theta=\frac{1}{2}$

$\theta=\frac{\pi}{3}$

Now components of $\vec a$ are:

$\left ( cos\frac{\pi}{3},cos\frac{\pi}{2},cos\frac{\pi}{3} \right )=\left ( \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2} \right )$

Question 4: Show that $( \vec a - \vec b ) \times (\vec a + \vec b ) = 2 ( \vec a \times \vec b )$

Answer:

To show that $( \vec a - \vec b ) \times (\vec a + \vec b ) = 2 ( \vec a \times \vec b )$

LHS=

$\\( \vec a - \vec b ) \times (\vec a + \vec b )=( \vec a - \vec b ) \times (\vec a)+( \vec a - \vec b ) \times (\vec b)$

$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times \vec a-\vec b \times\vec a+\vec a \times \vec b-\vec b \times \vec b$

As product of a vector with itself is always Zero,

$( \vec a - \vec b ) \times (\vec a + \vec b )= 0-\vec b \times\vec a+\vec a \times \vec b-0$

As cross product of a and b is equal to negative of cross product of b and a.

$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times\vec b+\vec a \times \vec b$

$( \vec a - \vec b ) \times (\vec a + \vec b )= 2(\vec a \times\vec b)$ = RHS

LHS is equal to RHS, Hence Proved.

Question 5: Find $\lambda$ and $\mu$ if $( 2 \hat i + 6 \hat j + 27 \hat k ) \times ( \hat i + \lambda j + \mu \hat k ) = \vec 0$

Answer:

Given in the question

$( 2 \hat i + 6 \hat j + 27 \hat k ) \times ( \hat i + \lambda j + \mu \hat k ) = \vec 0$

and we need to find values of $\lambda$ and $\mu$

$\begin{vmatrix} \hat i &\hat j & \hat k\\ 2& 6&27 \\ 1& \lambda &\mu \end{vmatrix}=0$

$\hat i (6\mu-27\lambda)-\hat j(2\mu-27)+\hat k(2\lambda-6)=0$

From Here we get,

$6\mu-27\lambda=0$

$2\mu-27=0$

$2\lambda -6=0$

From here, the value of $\lambda$ and $\mu$ is

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

Question 6: Given that $\vec a . \vec b = 0 \: \:and \: \: \vec a \times \vec b = 0$ and . What can you conclude about the vectors $\vec a \: \:and \: \: \vec b$ ?

Answer:

Given in the question

$\vec a . \vec b = 0$ and $\vec a \times \vec b = 0$

When $\vec a . \vec b = 0$ , either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$ are perpendicular to each other

When $\vec a \times \vec b = 0$ either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$ are parallel to each other

Since two vectors can never be both parallel and perpendicular at same time,we conclude that

$|\vec a| =0\:or\: |\vec b|=0$

Question 7: Let the vectors $\vec a , \vec b , \vec c$ be given as $\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k , \vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k , \vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k$ Then show that $\vec a \times ( \vec b + \vec c ) = \vec a \times \vec b + \vec a \times \vec c$

Answer:

Given in the question

$\\\vec a=\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k , \\\vec b=\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k , \\\vec c=\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k$

We need to show that $\vec a \times ( \vec b + \vec c ) = \vec a \times \vec b + \vec a \times \vec c$

Now,

$\vec a \times ( \vec b + \vec c ) =(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times(\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k +\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k)$

$=(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times((\vec b_ 1+\vec c_1) \hat i + (\vec b_ 2+\vec c_2) \hat j +( \vec b_3 +\vec c_3)\hat k)$

$=\begin{vmatrix} \hat i &\hat j &\hat k \\ a_1&a_2 &a_3 \\ (b_1+c_1)&(b_2+c_2) &(b_3+c_3) \end{vmatrix}$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)- a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

Now

$\vec a \times \vec b + \vec a \times \vec c=\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{vmatrix}+\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ c_1&c_2 &c_3 \end{vmatrix}$

$\vec a \times \vec b + \vec a \times \vec c=\hat i(a_2b_3-a_3b_2)-\hat j (a_1b_3-a_3b_1)+\hat k(a_1b_2-b_1a_2)+\hat i(a_2c_3-a_3c_2)-\hat j (a_1c_3-a_3c_1)+\hat k(a_1c_2-c_1a_2)$

$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$

Hence they are equal.

Question 8: If either $\vec a = \vec 0 \: \: or \: \: \vec b = \vec 0$ then $\vec a \times \vec b = \vec 0$ . Is the converse true? Justify your answer with an example.

Answer:

No, the converse of the statement is not true, as there can be two non zero vectors, the cross product of whose are zero. they are colinear vectors.

Consider an example

$\vec a=\hat i +\hat j + \hat k$

$\vec b =2\hat i +2\hat j + 2\hat k$

Here $|\vec a| =\sqrt{1^2+1^2+1^2}=\sqrt{3}$

$|\vec b| =\sqrt{2^2+2^2+2^2}=2\sqrt{3}$

$\vec a \times \vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 1&1 &1 \\ 2&2 &2 \end{vmatrix}=\hat i(2-2)-\hat j(2-2)+\hat k(2-2)=0$

Hence converse of the given statement is not true.

Question 9: Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Answer:

Given in the question

vertices A=(1, 1, 2), B=(2, 3, 5) and C=(1, 5, 5). and we need to find the area of the triangle

$AB=(2-1)\hat i+(3-1)\hat j+(5-2)\hat k=\hat i+2\hat j+3\hat k$

$BC=(1-2)\hat i+(5-3)\hat j+(5-5)\hat k=-\hat i+2\hat j$

Now as we know

Area of triangle

$A=\frac{1}{2}|\vec {AB}\times\vec {BC}|=\frac{1}{2}|(\hat i+2\hat j +3\hat k)\times(-\hat i+2\hat j)|$

$\\A=\frac{1}{2}\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ -1 &2 &0 \end{vmatrix}=\frac{1}{2}|\hat i(0-6)-\hat j(0-(-3))+\hat k(2-(-2))| \\A=\frac{1}{2}|-6\hat i-3\hat j+4\hat k|$

$A=\frac{1}{2}*\sqrt{(-6)^2+(-3)^2+(4)^2}=\frac{\sqrt{61}}{2}$

The area of the triangle is $\frac{\sqrt{61}}{2}$ square units

Question 10: Find 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 in the question

$\vec a = \hat i - \hat j + 3 \hat k$

$\vec b = 2\hat i -7 \hat j + \hat k$

Area of parallelogram with adjescent side $\vec a$ and $\vec b$ ,

$A=|\vec a\times\vec b|=|(\vec i-\vec j+3\vec k)\times (2\hat i-7\hat j+\hat k)|$

$A=\begin{vmatrix} \hat i& \hat j & \hat k\\ 1&-1 &3 \\ 2&-7 &1 \end{vmatrix}=|\hat i(-1+21)-\hat j (1-6)+\hat k (-7+2)|$

$A=|\hat i(20)-\hat j (-5)+\hat k (-5)|=\sqrt{20^2+5^2+(-5)^2}$

$A=\sqrt{450}=15\sqrt{2}$

The area of the parallelogram whose adjacent sides are determined by the vectors $\vec a = \hat i - \hat j + 3 \hat k$ and $\vec b = 2\hat i -7 \hat j + \hat k$ is $A=\sqrt{450}=15\sqrt{2}$

Question 11: Let the vectors $\vec a \: \: and\: \: \vec b$ be such that $|\vec a| = 3 \: \: and\: \: |\vec b | = \frac{\sqrt 2 }{3}$ , then $\vec a \times \vec b$ is a unit vector, if the angle between is $\vec a \: \:and \: \: \vec b$

$\\A ) \pi /6 \\\\ B ) \pi / 4 \\\\ C ) \pi / 3 \\\\ D ) \pi /2$

Answer:

Given in the question,

$|\vec a| = 3 \: \: and\: \: |\vec b | = \frac{\sqrt 2 }{3}$

As given $\vec a \times \vec b$ is a unit vector, which means,

$|\vec a \times \vec b|=1$

$|\vec a| | \vec b|sin\theta=1$

$3*\frac{\sqrt{2}}{3}sin\theta=1$

$sin\theta=\frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{4}$

Hence the angle between two vectors is $\frac{\pi}{4}$ . Correct option is B.

Question 12: Area of a rectangle having vertices A, B, C and D with position vectors

$- \hat i + \frac{1}{2} \hat j + 4 \hat k , \hat i + \frac{1}{2} \hat j + 4 \hat k , \hat i - \frac{1}{2}\hat j + 4 \hat k \: \: and \: \: - \hat i - \frac{1}{2} \hat j + 4 \hat k$

(A)1/2

(B) 1

(C) 2

(D) 4

Answer:

Given 4 vertices of rectangle are

$\\\vec a=- \hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec b=\hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec c= \hat i - \frac{1}{2}\hat j + 4 \hat k \: \: and \: \: \\\vec d= - \hat i - \frac{1}{2} \hat j + 4 \hat k$

$\vec {AB}=\vec b-\vec a=(1+1)\hat i+(\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=2\hat i$

$\vec {BC}=\vec c-\vec b=(1-1)\hat i+(-\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=-\hat j$

Now,

Area of the Rectangle

$A=|\vec {AB}\times\vec {BC}|=|2\hat i \times (-\hat j)|=2$

Hence option C is correct.