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 $\overrightarrow{a}+\overrightarrow{b}and\overrightarrow{a}-\overrightarrow{b}$ , where $\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}and\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$

Answer:

Given in the question

$\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}and\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$

$\overrightarrow{a}+\overrightarrow{b}=3\hat{i}+2\hat{j}+2\hat{k}+\hat{i}+2\hat{j}-2\hat{k}=4\hat{i}+4\hat{j}$

$\overrightarrow{a}-\overrightarrow{b}=3\hat{i}+2\hat{j}+2\hat{k}-\hat{i}-2\hat{j}+2\hat{k}=2\hat{i}+4\hat{k}$

Now , A vector which perpendicular to both $\overrightarrow{a}+\overrightarrow{b}and\overrightarrow{a}-\overrightarrow{b}$ is $(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})$

$(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 4& 0\\ 2& 0& 4\end{array}\right|$

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

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

And a unit vector in this direction :

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

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

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

Question 3: If a unit vector $\overrightarrow{a}$ 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 $\overrightarrow{a}$ .

Answer:

Given in the question,

angle between $\overrightarrow{a}$ and $\hat{i}$ :

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

angle between $\overrightarrow{a}$ and $\hat{j}$

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

angle with $\overrightarrow{a}$ and $\hat{k}$ :

$\gamma =\theta$

Now, As we know,

$co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =1$

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

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

$co{s}^2\theta =\frac{1}{4}$

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

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

Now components of $\overrightarrow{a}$ are:

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

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

Answer:

To show that $(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})=2(\overrightarrow{a}\times \overrightarrow{b})$

LHS=

$(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})=(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a})+(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{b})$

$(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{a}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{b}\times \overrightarrow{b}$

As product of a vector with itself is always Zero,

$(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})=0-\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}-0$

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

$(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}$

$(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})=2(\overrightarrow{a}\times \overrightarrow{b})$ = RHS

LHS is equal to RHS, Hence Proved.

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

Answer:

Given in the question

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

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

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 6& 27\\ 1& \lambda & \mu \end{array}\right|=0$

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

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

Answer:

Given in the question

$\overrightarrow{a}.\overrightarrow{b}=0$ and $\overrightarrow{a}\times \overrightarrow{b}=0$

When $\overrightarrow{a}.\overrightarrow{b}=0$ , either $|\overrightarrow{a}|=0,or|\overrightarrow{b}|=0,\overrightarrow{a}and\overrightarrow{b}$ are perpendicular to each other

When $\overrightarrow{a}\times \overrightarrow{b}=0$ either $|\overrightarrow{a}|=0,or|\overrightarrow{b}|=0,\overrightarrow{a}and\overrightarrow{b}$ are parallel to each other

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

$|\overrightarrow{a}|=0or|\overrightarrow{b}|=0$

Question 7: Let the vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be given as ${\overrightarrow{a}}_1\hat{i}+{\overrightarrow{a}}_2\hat{j}+{\overrightarrow{a}}_3\hat{k},{\overrightarrow{b}}_1\hat{i}+{\overrightarrow{b}}_2\hat{j}+{\overrightarrow{b}}_3\hat{k},{\overrightarrow{c}}_1\hat{i}+{\overrightarrow{c}}_2\hat{j}+{\overrightarrow{c}}_3\hat{k}$ Then show that $\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}$

Answer:

Given in the question

$$\begin{gathered} \overrightarrow{a}={\overrightarrow{a}}_1\hat{i}+{\overrightarrow{a}}_2\hat{j}+{\overrightarrow{a}}_3\hat{k}, \\ \overrightarrow{b}={\overrightarrow{b}}_1\hat{i}+{\overrightarrow{b}}_2\hat{j}+{\overrightarrow{b}}_3\hat{k}, \\ \overrightarrow{c}={\overrightarrow{c}}_1\hat{i}+{\overrightarrow{c}}_2\hat{j}+{\overrightarrow{c}}_3\hat{k} \end{gathered}$$

We need to show that $\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}$

Now,

$\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})=({\overrightarrow{a}}_1\hat{i}+{\overrightarrow{a}}_2\hat{j}+{\overrightarrow{a}}_3\hat{k})\times ({\overrightarrow{b}}_1\hat{i}+{\overrightarrow{b}}_2\hat{j}+{\overrightarrow{b}}_3\hat{k}+{\overrightarrow{c}}_1\hat{i}+{\overrightarrow{c}}_2\hat{j}+{\overrightarrow{c}}_3\hat{k})$

$=({\overrightarrow{a}}_1\hat{i}+{\overrightarrow{a}}_2\hat{j}+{\overrightarrow{a}}_3\hat{k})\times (({\overrightarrow{b}}_1+{\overrightarrow{c}}_1)\hat{i}+({\overrightarrow{b}}_2+{\overrightarrow{c}}_2)\hat{j}+({\overrightarrow{b}}_3+{\overrightarrow{c}}_3)\hat{k})$

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

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

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

Now

$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|+\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ c_1& c_2& c_3\end{array}\right|$

$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=\hat{i}(a_2{b}_3-a_3{b}_2)-\hat{j}(a_1{b}_3-a_3{b}_1)+\hat{k}(a_1{b}_2-b_1{a}_2)+\hat{i}(a_2{c}_3-a_3{c}_2)-\hat{j}(a_1{c}_3-a_3{c}_1)+\hat{k}(a_1{c}_2-c_1{a}_2)$

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

Hence they are equal.

Question 8: If either $\overrightarrow{a}=\overrightarrow{0}or\overrightarrow{b}=\overrightarrow{0}$ then $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0}$ . 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

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b}=2\hat{i}+2\hat{j}+2\hat{k}$

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

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

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& 2& 2\end{array}\right|=\hat{i}(2-2)-\hat{j}(2-2)+\hat{k}(2-2)=0$

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

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

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

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

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 $\overrightarrow{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\overrightarrow{b}=2\hat{i}-7\hat{j}+\hat{k}$ .

Answer:

Given in the question

$\overrightarrow{a}=\hat{i}-\hat{j}+3\hat{k}$

$\overrightarrow{b}=2\hat{i}-7\hat{j}+\hat{k}$

Area of parallelogram with adjescent side $\overrightarrow{a}$ and $\overrightarrow{b}$ ,

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

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

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

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

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

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

$$\begin{gathered} A)\pi /6 \\ B)\pi /4 \\ C)\pi /3 \\ D)\pi /2 \end{gathered}$$

Answer:

Given in the question,

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

As given $\overrightarrow{a}\times \overrightarrow{b}$ is a unit vector, which means,

$|\overrightarrow{a}\times \overrightarrow{b}|=1$

$|\overrightarrow{a}||\overrightarrow{b}|sin\theta =1$

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

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

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

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

$$\begin{gathered} \overrightarrow{a}=-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}, \\ \overrightarrow{b}=\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}, \\ \overrightarrow{c}=\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}and \\ \overrightarrow{d}=-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k} \end{gathered}$$

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

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

Now,

Area of the Rectangle

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

Hence option C is correct.