NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3 - Vector Algebra

Question 2: Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}and3\hat{i}-2\hat{j}+\hat{k}$

Answer:

Given two vectors

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

Now As we know,

The angle between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by

$\theta =co{s}^{-1}\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)$

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

$\theta =co{s}^{-1}\left(\frac{(\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}+\hat{k})}{|\hat{i}-2\hat{j}+3\hat{k}||3\hat{i}-2\hat{j}+\hat{k}|}\right)$

$\theta =co{s}^{-1}\left(\frac{3+4+3}{\sqrt{1^2+(-2{)}^2+3^3}\sqrt{3^2+(-2{)}^2+1^2}}\right)$

$\theta =co{s}^{-1}\frac{10}{14}$

$\theta =co{s}^{-1}\frac{5}{7}$

Question 3: Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$

Answer:

Let

$\overrightarrow{a}=\hat{i}-\hat{j}$

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

Projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$

$\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{(\hat{i}-\hat{j})(\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|}=\frac{1-1}{\sqrt{2}}=0$

Hence, Projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is 0.

Question 4: Find the projection of the vector $\hat{i}+3\hat{j}+7\hat{k}$ on the vector $7\hat{i}-\hat{j}+8\hat{k}$

Answer:

Let

$\overrightarrow{a}=\hat{i}+3\hat{j}+7\hat{k}$

$\overrightarrow{b}=7\hat{i}-\hat{j}+8\hat{k}$

The projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is

$\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{(\hat{i}+3\hat{j}+7\hat{k})(7\hat{i}-\hat{j}+8\hat{k})}{|7\hat{i}-\hat{j}+8\hat{k}|}=\frac{7-3+56}{\sqrt{7^2+(-1{)}^2+8^2}}=\frac{60}{\sqrt{114}}$

Hence, projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is

$\frac{60}{\sqrt{114}}$

Question 5: Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}),\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$ Also, show that they are mutually perpendicular to each other.

Answer:

Given

$$\begin{gathered} \overrightarrow{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}), \\ \overrightarrow{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}), \\ \overrightarrow{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k}) \end{gathered}$$

Now magnitude of $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$

$\left|\overrightarrow{a}\right|=\frac{1}{7}\sqrt{2^2+3^2+6^2}=\frac{\sqrt{49}}{7}=1$

$\left|\overrightarrow{b}\right|=\frac{1}{7}\sqrt{3^2+(-6{)}^2+2^2}=\frac{\sqrt{49}}{7}=1$

$\left|\overrightarrow{c}\right|=\frac{1}{7}\sqrt{6^2+2^2+(-3{)}^2}=\frac{\sqrt{49}}{7}=1$

Hence, they all are unit vectors.

Now,

$\overrightarrow{a}.\overrightarrow{b}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})=\frac{1}{49}(6-18+12)=0$

$\overrightarrow{b}.\overrightarrow{c}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})=\frac{1}{49}(18-12-6)=0$

$\overrightarrow{c}.\overrightarrow{a}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})\frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k})=\frac{1}{49}(12+6-18)=0$

Hence all three are mutually perpendicular to each other.

Question 6: Find $|\overrightarrow{a}|and|\overrightarrow{b}|$ , if $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=8and|\overrightarrow{a}|=8|\overrightarrow{b}|$ .

Answer:

Given in the question

$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=8$

${\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8$

Since $|\overrightarrow{a}|=8|\overrightarrow{b}|$

${\left|\overrightarrow{8b}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8$

${\left|\overrightarrow{63b}\right|}^2=8$

${\left|\overrightarrow{b}\right|}^2=\frac{8}{63}$

$\left|\overrightarrow{b}\right|=\sqrt{\frac{8}{63}}$

So, answer of the question is

$\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|=8\sqrt{\frac{8}{63}}$

Question 7: Evaluate the product $(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})$ .

Answer:

To evaluate the product $(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})$

$(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})=6\overrightarrow{a}.\overrightarrow{a}+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{b}.\overrightarrow{a}-35\overrightarrow{b}.\overrightarrow{b}$

$=6\overrightarrow{a}{.}^2+11\overrightarrow{a}.\overrightarrow{b}-35{\overrightarrow{b}}^2$

$=6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2$

Question 8: Find the magnitude of two vectors $\overrightarrow{a}and\overrightarrow{b}$ , having the same magnitude and such that the angle between them is ${60}^{\circ }$ and their scalar product is 1/2

Answer:

Given two vectors $\overrightarrow{a}and\overrightarrow{b}$

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|$

$\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}$

Now Angle between $\overrightarrow{a}and\overrightarrow{b}$

$\theta ={60}^0$

Now As we know that

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$\frac{1}{2}=\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|cos{60}^0$

${\left|a\right|}^2=1$

Hence, the magnitude of two vectors $\overrightarrow{a}and\overrightarrow{b}$

$\left|a\right|=\left|b\right|=1$

Question 9: Find $|\overrightarrow{x}|$ , if for a unit vector $\overrightarrow{a},(\overrightarrow{x}-\overrightarrow{a}).(\overrightarrow{x}+\overrightarrow{a})=12$

Answer:

Given in the question that

$(\overrightarrow{x}-\overrightarrow{a}).(\overrightarrow{x}+\overrightarrow{a})=12$

And we need to find $\left|\overrightarrow{x}\right|$

${\left|\overrightarrow{x}\right|}^2-{\left|\overrightarrow{a}\right|}^2=12$

${\left|\overrightarrow{x}\right|}^2-1=12$

${\left|\overrightarrow{x}\right|}^2=13$

$\left|\overrightarrow{x}\right|=\sqrt{13}$

So the value of $\left|\overrightarrow{x}\right|$ is $\sqrt{13}$

Question 10: If $\overrightarrow{a}=2\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b}=-\hat{i}+2\hat{j}+\hat{k}and\overrightarrow{c}=3\hat{i}+\hat{j}$ are such that $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$ , then find the value of $\lambda$

Answer:

Given in the question is

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

and $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$

and we need to find the value of $\lambda$ ,

so the value of $\overrightarrow{a}+\lambda \overrightarrow{b}$ -

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

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

As $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$

$(\overrightarrow{a}+\lambda \overrightarrow{b}).\overrightarrow{c}=0$

$((2-\lambda )\hat{i}+(2+2\lambda )\hat{j}+(3+\lambda )\hat{k})(3\hat{i}+\hat{j})=0$

$3(2-\lambda )+2+2\lambda =0$

$6-3\lambda +2+2\lambda =0$

$\lambda =8$

the value of $\lambda =8$ ,

Question 11: Show that $|\overrightarrow{a}|\overrightarrow{b}+|\overrightarrow{b}|\overrightarrow{a}$ is perpendicular to $|\overrightarrow{a}|\overrightarrow{b}-|\overrightarrow{b}|\overrightarrow{a}$ , for any two nonzero vectors $\overrightarrow{a}and\overrightarrow{b}$ .

Answer:

Given in the question that -

$\overrightarrow{a}and\overrightarrow{b}$ are two non-zero vectors

According to the question

$(|\overrightarrow{a}|\overrightarrow{b}+|\overrightarrow{b}|\overrightarrow{a})(|\overrightarrow{a}|\overrightarrow{b}-|\overrightarrow{b}|\overrightarrow{a})$

$=|\overrightarrow{a}{|}^2|\overrightarrow{b}{|}^2-|\overrightarrow{b}{|}^2|\overrightarrow{a}{|}^2+|\overrightarrow{b}||\overrightarrow{a}|\overrightarrow{a}.\overrightarrow{b}-|\overrightarrow{a}||\overrightarrow{b}|\overrightarrow{b}.\overrightarrow{a}=0$

Hence $|\overrightarrow{a}|\overrightarrow{b}+|\overrightarrow{b}|\overrightarrow{a}$ is perpendicular to $|\overrightarrow{a}|\overrightarrow{b}-|\overrightarrow{b}|\overrightarrow{a}$ .

Question 12: If $\overrightarrow{a}.\overrightarrow{a}=0and\overrightarrow{a}.\overrightarrow{b}=0$ , then what can be concluded about the vector $\overrightarrow{b}$ ?

Answer:

Given in the question

$$\begin{gathered} \overrightarrow{a}.\overrightarrow{a}=0 \\ |\overrightarrow{a}{|}^2=0 \end{gathered}$$

$|\overrightarrow{a}|=0$

Therefore $\overrightarrow{a}$ is a zero vector. Hence any vector $\overrightarrow{b}$ will satisfy $\overrightarrow{a}.\overrightarrow{b}=0$

Question 13: If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ , find the value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}$

Answer:

Given in the question

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors $\Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$

and $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$

and we need to find the value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}$

$(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{)}^2=\overrightarrow{0}$

${\overrightarrow{a}}^2+{\overrightarrow{b}}^2+{\overrightarrow{c}}^2+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

$|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

$1+1+1+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

$\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=\frac{-3}{2}$

Answer- the value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}$ is $\frac{-3}{2}$

Question 14: If either vector $\overrightarrow{a}=0or\overrightarrow{b}=0then\overrightarrow{a}.\overrightarrow{b}=0$ . But the converse need not be true. Justify your answer with an example

Answer:

Let

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

$\overrightarrow{b}=5\hat{i}+4\hat{j}+1\hat{k}$

we see that

$\overrightarrow{a}.\overrightarrow{b}=(\hat{i}-2\hat{j}+3\hat{k})(5\hat{i}+4\hat{j}+1\hat{k})=5-8+3=0$

we now observe that

$|\overrightarrow{a}|=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{14}$

$|\overrightarrow{b}|=\sqrt{5^2+4^2+1^2}=\sqrt{42}$

Hence here converse of the given statement is not true.

Question 15: If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find $\mathrm{\angle }ABC,[\mathrm{\angle }ABC$ is the angle between the vectors $\overline{BA}and\overline{BC}]$ .

Answer:

Given points,

A=(1, 2, 3),

B=(–1, 0, 0),

C=(0, 1, 2),

As need to find Angle between $\overline{BA}and\overline{BC}]$

$\overrightarrow{BA}=(1-(-1))\hat{i}+(2-0)\hat{j}+(3-0)\hat{k}=2\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{BC}=(0-(-1))\hat{i}+(1-0)\hat{j}+(2-0)\hat{k}=\hat{i}+\hat{j}+2\hat{k}$

Hence angle between them ;

$\theta =co{s}^{-1}(\frac{\overrightarrow{BA}.\overrightarrow{BC}}{\left|\overrightarrow{BA}\right|\left|\overrightarrow{BC}\right|})$

$\theta =co{s}^{-1}\frac{2+2+6}{\sqrt{17}\sqrt{6}}$

$\theta =co{s}^{-1}\frac{10}{\sqrt{102}}$

Answer - Angle between the vectors $\overline{BA}and\overline{BC}$ is $\theta =co{s}^{-1}\frac{10}{\sqrt{102}}$

Question 16: Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

Answer:

Given in the question

A=(1, 2, 7), B=(2, 6, 3) and C(3, 10, –1)

To show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear

$\overrightarrow{AB}=(2-1)\hat{i}+(6-2)\hat{j}+(3-7)\hat{k}$

$\overrightarrow{AB}=\hat{i}+4\hat{j}-4\hat{k}$

$\overrightarrow{BC}=(3-2)\hat{i}+(10-6)\hat{j}+(-1-3)\hat{k}$

$\overrightarrow{BC}=\hat{i}+4\hat{j}-4\hat{k}$

$\overrightarrow{AC}=(3-1)\hat{i}+(10-2)\hat{j}+(-1-7)\hat{k}$

$\overrightarrow{AC}=2\hat{i}+8\hat{j}-8\hat{k}$

$|\overrightarrow{AB}|=\sqrt{1^2+4^2+(-4{)}^2}=\sqrt{33}$

$|\overrightarrow{BC}|=\sqrt{1^2+4^2+(-4{)}^2}=\sqrt{33}$

$|\overrightarrow{AC}|=\sqrt{2^2+8^2+(-8{)}^2}=2\sqrt{33}$

As we see that

$|\overrightarrow{AC}|=|\overrightarrow{AB}|+|\overrightarrow{BC}|$

Hence point A, B , and C are colinear.

Question 17: Show that the vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}and3\hat{i}-4\hat{j}-4\hat{k}$ form the vertices of a right angled triangle.

Answer:

Given the position vector of A, B , and C are

$2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}and3\hat{i}-4\hat{j}-4\hat{k}$

To show that the vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k}and3\hat{i}-4\hat{j}-4\hat{k}$ form the vertices of a right angled triangle

$\overrightarrow{AB}=(1-2)\hat{i}+(-3-(-1))\hat{j}+(-5-1)\hat{k}=-1\hat{i}-2\hat{j}-6\hat{k}$

$\overrightarrow{BC}=(3-1)\hat{i}+(-4-(-3))\hat{j}+(-4-(-5))\hat{k}=-2\hat{i}-\hat{j}+\hat{k}$

$\overrightarrow{AC}=(3-2)\hat{i}+(-4-(-1))\hat{j}+(-4-(1))\hat{k}=\hat{i}-3\hat{j}-5\hat{k}$

$|\overrightarrow{AB}|=\sqrt{(-1{)}^2+(-2{)}^2+(-6{)}^2}=\sqrt{41}$

$|\overrightarrow{BC}|=\sqrt{(-2{)}^2+(-1{)}^2+(1{)}^2}=\sqrt{6}$

$|\overrightarrow{AC}|=\sqrt{(1{)}^2+(-3{)}^2+(-5{)}^2}=\sqrt{35}$

Here we see that

$|\overrightarrow{AC}{|}^2+|\overrightarrow{BC}{|}^2=|\overrightarrow{AB}{|}^2$

Hence A,B, and C are the vertices of a right angle triangle.

Question 18: If $\overrightarrow{a}$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar, then $\lambda \overrightarrow{a}$ is unit vector if

$$\begin{gathered} A)\lambda =1 \\ B)\lambda =-1 \\ C)a=|\lambda | \\ D)a=1/|\lambda | \end{gathered}$$

Answer:

Given $\overrightarrow{a}$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar

$\lambda \overrightarrow{a}$ is a unit vector when

$|\lambda \overrightarrow{a}|=1$

$|\lambda ||\overrightarrow{a}|=1$

$|\overrightarrow{a}|=\frac{1}{|\lambda |}$

Hence the correct option is D.