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

NCERT solutions for exercise 10.4 Class 12 Maths chapter 10 discuss the questions related to cross product of vectors and their applications. Exercise 10.4 Class 12 Maths have 12 questions. All these 12 questions of NCERT solutions for Class 12 Maths chapter 10 exercise 10.4 are explained with all necessary steps. The Class 12 Maths chapter 10 exercise 10.4 solutions are designed by a mathematics expert and are in accordance with Class 12 CBSE patterns. Concepts of cross products are also explained in Class 11 NCERT Physics TextBooks in the chapter system of particles and rotational motion and are used in other chapters of Class 11 and 12 too. One question may be expected from the Class 12 Maths chapter 10 exercise 10.4 for Class 12 CBSE Class 12 Board Exam. Along with Class 12th Maths chapter 10 exercise 10.4 students can go through the following exercise for practice.

• Vector Algebra Exercise 10.1

• Vector Algebra Exercise 10.2

• Vector Algebra Exercise 10.3

• Vector Algebra Miscellaneous Exercise

## Vector Algebra Class 12 Chapter 10 Exercise 10.4

Question:1 Find $|\vec a \times \vec b |, if \vec a = \hat i - 7 \hat j + 7 \hat k \: \: and \: \: \vec b = 3 \hat i - 2 \hat j + 2 \hat k$

Given in the question,

$\\ \vec a = \hat i - 7 \hat j + 7 \hat k \: \: and \: \: \\\vec b = 3 \hat i - 2 \hat j + 2 \hat k$

and we need to find $\dpi{100} |\vec a \times \vec b |$

Now,

$|\vec a \times \vec b | =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &-7 &7 \\ 3& -2 &2 \end{vmatrix}$

$|\vec a \times \vec b | =\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)$

$|\vec a \times \vec b | =19\hat j+19\hat k$

So the value of $\dpi{100} |\vec a \times \vec b |$ is $19\hat j+19\hat k$

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

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 $\vec a$ .

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

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

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 $\vec a \: \:and \: \: \vec b$ ?

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

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.

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).

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 $\dpi{100} \frac{\sqrt{61}}{2}$ 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 $\vec b = 2\hat i -7 \hat j + \hat k$ .

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 $\vec a \: \:and \: \: \vec b$

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

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

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.

## More About NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4

Exercise 10.4 Class 12 Maths throughout explains about the cross products. The applications of cross products include finding the area of a triangle, parallelogram etc. Question number 9 of the Class 12th Maths chapter 10 exercise 10.4 is to find the area of a triangle. And questions 10 and 12 of NCERT Solutions for Class 12 Maths chapter 10 exercise 10.4 is to find the area of parallelogram and rectangle respectively.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4

• Going through the NCERT syllabus exercise 10.4 Class 12 Maths give more conceptual clarifications about the topics covered just before the exercise.

• Class 12th Maths chapter 10 exercise 10.4 is designed for a better understanding of the concepts of vector products and their applications.

The topics covered under NCERT Solutions for class 12 maths chapter 10 exercise 10.4 will also be useful for the exams like JEE Main also.

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