NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2 - Vector Algebra

Question 1: Compute the magnitude of the following vectors:

(2) $\overrightarrow{b}=2\hat{i}-7\hat{j}-3\hat{k}$

Answer:

Here,

$\overrightarrow{b}=2\hat{i}-7\hat{j}-3\hat{k}$

Magnitude of $\overrightarrow{b}$

$\left|\overrightarrow{b}\right|=\sqrt{2^2+(-7{)}^2+(-3{)}^2}=\sqrt{62}$

Question 1: Compute the magnitude of the following vectors:

(3) $\overrightarrow{c}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}-\frac{1}{\sqrt{3}}\hat{k}$

Answer:

Here,

$\overrightarrow{c}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}-\frac{1}{\sqrt{3}}\hat{k}$

Magnitude of $\overrightarrow{c}$

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

Question 2: Write two different vectors having same magnitude

Answer:

Two different Vectors having the same magnitude are

$\overrightarrow{a}=3\hat{i}+6\hat{j}+9\hat{k}$

$\overrightarrow{b}=9\hat{i}+6\hat{j}+3\hat{k}$

The magnitude of both vector

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\sqrt{9^2+6^2+3^2}=\sqrt{126}$

Question 3: Write two different vectors having same direction.

Answer:

Two different vectors having the same direction are:

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

$\overrightarrow{b}=2\hat{i}+4\hat{j}+6\hat{k}$

Question 4: Find the values of x and y so that the vectors $2\hat{i}+3\hat{j}$ and $x\hat{i}+y\hat{j}$ are equal.

Answer:

$2\hat{i}+3\hat{j}$ will be equal to $x\hat{i}+y\hat{j}$ when their corresponding components are equal.

Hence when,

$x=2$ and

$y=3$

Question 5: Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).

Answer:

Let point P = (2, 1) and Q = (– 5, 7).

Now,

$\overrightarrow{PQ}=(-5-2)\hat{i}+(7-1)\hat{j}=-7\hat{i}+6\hat{j}$

Hence scalar components are (-7,6) and the vector is $-7\hat{i}+6\hat{j}$

Question 6: Find the sum of the vectors $\overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k},\overrightarrow{b}=-2\hat{i}+4\hat{j}+5\hat{k}and\overrightarrow{c}=\hat{i}-6\hat{j}-7\hat{k}$

Answer:

Given,

$$\begin{gathered} \overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k}, \\ \overrightarrow{b}=-2\hat{i}+4\hat{j}+5\hat{k}and \\ \overrightarrow{c}=\hat{i}-6\hat{j}-7\hat{k} \end{gathered}$$

Now, The sum of the vectors:

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}+-2\hat{i}+4\hat{j}+5\hat{k}+\hat{i}-6\hat{j}-7\hat{k}$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}$

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=-4\hat{j}-\hat{k}$

Question 7: Find the unit vector in the direction of the vector $\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$

Answer:

Given

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

Magnitude of $\overrightarrow{a}$

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

A unit vector in the direction of $\overrightarrow{a}$

$\overrightarrow{u}=\frac{\hat{i}}{\left|a\right|}+\frac{\hat{j}}{\left|a\right|}+\frac{2\hat{k}}{\left|a\right|}=\frac{\hat{i}}{\sqrt{6}}+\frac{\hat{j}}{\sqrt{6}}+\frac{2\hat{k}}{\sqrt{6}}$

Question 8: Find the unit vector in the direction of vector $\overrightarrow{PQ}$ , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

Answer:

Given P = (1, 2, 3) and Q = (4, 5, 6)

A vector in direction of PQ

$\overrightarrow{PQ}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}$

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

Magnitude of PQ

$\left|\overrightarrow{PQ}\right|=\sqrt{3^2+3^2+3^2}=3\sqrt{3}$

Now, unit vector in direction of PQ

$\hat{u}=\frac{\overrightarrow{PQ}}{\left|\overrightarrow{PQ}\right|}=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}$

$\hat{u}=\frac{\hat{i}}{\sqrt{3}}+\frac{\hat{j}}{\sqrt{3}}+\frac{\hat{k}}{\sqrt{3}}$

Question 9: For given vectors, $\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{b}=-\hat{i}+\hat{j}-\hat{k}$ , find the unit vector in the direction of the vector $\overrightarrow{a}+\overrightarrow{b}$ .

Answer:

Given

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

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

Now,

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

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

Now a unit vector in the direction of $\overrightarrow{a}+\overrightarrow{b}$

$\overrightarrow{u}=\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}=\frac{\hat{i}+\hat{j}}{\sqrt{1^2+1^2}}$

$\overrightarrow{u}=\frac{\hat{i}}{\sqrt{2}}+\frac{\hat{j}}{\sqrt{2}}$

Question 10: Find a vector in the direction of vector $5\hat{i}-\hat{j}+2\hat{k}$ which has magnitude 8 units.

Answer:

Given a vector

$\overrightarrow{a}=5\hat{i}-\hat{j}+2\hat{k}$

the unit vector in the direction of $5\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{u}=\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{5^2+(-1{)}^2+2^2}}=\frac{5\hat{i}}{\sqrt{30}}-\frac{\hat{j}}{\sqrt{30}}+\frac{2\hat{k}}{\sqrt{30}}$

A vector in direction of $5\hat{i}-\hat{j}+2\hat{k}$ and whose magnitude is 8 =

$8\overrightarrow{u}=\frac{40\hat{i}}{\sqrt{30}}-\frac{8\hat{j}}{\sqrt{30}}+\frac{16\hat{k}}{\sqrt{30}}$

Question 11: Show that the vectors $2\hat{i}-3\hat{j}+4\hat{k}$ and $-4\hat{i}+6\hat{j}-8\hat{k}$ are collinear.

Answer:

Let

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

$\overrightarrow{b}=-4\hat{i}+6\hat{j}-8\hat{k}$

It can be seen that

$\overrightarrow{b}=-4\hat{i}+6\hat{j}-8\hat{k}=-2(2\hat{i}-3\hat{j}+4\hat{k})=-2\overrightarrow{a}$

Hence here $\overrightarrow{b}=-2\overrightarrow{a}$

As we know

Whenever we have $\overrightarrow{b}=\lambda \overrightarrow{a}$ , the vector $\overrightarrow{a}$ and $\overrightarrow{b}$ will be colinear.

Here $\lambda =-2$

Hence vectors $2\hat{i}-3\hat{j}+4\hat{k}$ and $-4\hat{i}+6\hat{j}-8\hat{k}$ are collinear.

Question 12: Find the direction cosines of the vector $\hat{i}+2\hat{j}+3\hat{k}$

Answer:

Let

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

$\left|\overrightarrow{a}\right|=\sqrt{1^2+2^2+3^2}=\sqrt{14}$

Hence direction cosine of $\overrightarrow{a}$ are

$(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}})$

Question 13: Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed from A to B.

Answer:

Given

point A=(1, 2, –3)

point B=(–1, –2, 1)

Vector joining A and B Directed from A to B

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

$\overrightarrow{AB}=-2\hat{i}+-4\hat{j}+4\hat{k}$

$\left|\overrightarrow{AB}\right|=\sqrt{(-2{)}^2+(-4{)}^2+4^2}=\sqrt{36}=6$

Hence Direction cosines of vector AB are

$(\frac{-2}{6},\frac{-4}{6},\frac{4}{6})=(\frac{-1}{3},\frac{-2}{3},\frac{2}{3})$

Question 14: Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes OX, OY and OZ.

Answer:

Let

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

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

Hence direction cosines of this vectors is

$(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

Let $\alpha$ , $\beta$ and $\gamma$ be the angle made by x-axis, y-axis and z- axis respectively

Now as we know,

$cos\alpha =\frac{1}{\sqrt{3}}$ , $cos\beta =\frac{1}{\sqrt{3}}$ $andcos\gamma =\frac{1}{\sqrt{3}}$

Hence Given vector is equally inclined to axis OX,OY and OZ.

Question 15 (1): Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $i+2j-k$ and $-i+j+k$ respectively, in the ratio 2 : 1 internally

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n internally:

$\overrightarrow{r}=\frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}$

Here

position vector os P = $\overrightarrow{a}$ = $i+2j-k$

the position vector of Q = $\overrightarrow{b}=-i+j+k$

m:n = 2:1

And Hence

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

$\overrightarrow{r}=\frac{-2\hat{i}+2\hat{j}+2\hat{k}+\hat{i}+2\hat{j}-\hat{k}}{3}=\frac{-\hat{i}+4\hat{j}+\hat{k}}{3}$

$\overrightarrow{r}=\frac{-\hat{i}}{3}+\frac{4\hat{j}}{3}+\frac{\hat{k}}{3}$

Question 15 (2): Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i}+2\hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio 2 : 1 externally

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n externally:

$\overrightarrow{r}=\frac{m\overrightarrow{b}-n\overrightarrow{a}}{m-n}$

Here

position vector os P = $\overrightarrow{a}$ = $i+2j-k$

the position vector of Q = $\overrightarrow{b}=-i+j+k$

m:n = 2:1

And Hence

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

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

Question 16: Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

Answer:

Given

The position vector of point P = $2\hat{i}+3\hat{j}+4\hat{k}$

Position Vector of point Q = $4\hat{i}+\hat{j}-2\hat{k}$

The position vector of R which divides PQ in half is given by:

$\overrightarrow{r}=\frac{2\hat{i}+3\hat{j}+4\hat{k}+4\hat{i}+\hat{j}-2\hat{k}}{2}$

$\overrightarrow{r}=\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}=3\hat{i}+2\hat{j}+\hat{k}$

Question 17: Show that the points A, B and C with position vectors, $\overrightarrow{a}=3\hat{i}-4\hat{j}-4\hat{k},\overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}and\overrightarrow{c}=\hat{i}-3\hat{j}-5\hat{k}$ , respectively form the vertices of a right angled triangle.

Answer:

Given

the position vector of A, B, and C are

$$\begin{gathered} \overrightarrow{a}=3\hat{i}-4\hat{j}-4\hat{k}, \\ \overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}and \\ \overrightarrow{c}=\hat{i}-3\hat{j}-5\hat{k} \end{gathered}$$

Now,

$\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=-\hat{i}+3\hat{j}+5\hat{k}$

$\overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}=-\hat{i}-2\hat{j}-6\hat{k}$

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

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

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

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

As we can see

${\left|\overrightarrow{BC}\right|}^2={\left|\overrightarrow{CA}\right|}^2+{\left|\overrightarrow{AB}\right|}^2$

Hence, ABC is a right angle triangle.

Question 18: In triangle ABC (Fig 10.18), which of the following is not true:

$$\begin{gathered} A)\overline{AB}+\overline{BC}+\overline{CA}=\overrightarrow{0} \\ B)\overline{AB}+\overline{BC}-\overline{AC}=\overrightarrow{0} \\ C)\overline{AB}+\overline{BC}-\overline{CA}=\overrightarrow{0} \\ D)\overline{AB}-\overline{CB}+\overline{CA}=\overrightarrow{0} \end{gathered}$$

Answer:

From triangles law of addition we have,

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

From here

$\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=0$

also

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0$

Also

$\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}=0$

Hence options A,B and D are true SO,

Option C is False.

Question 19: If are two collinear vectors, then which of the following are incorrect:
(A) $\overrightarrow{b}=\lambda \overrightarrow{a}$ for some saclar $\lambda$
(B) $\overrightarrow{a}=\pm \overrightarrow{b}$
(C) the respective components of $\overrightarrow{a}and\overrightarrow{b}$ are not proportional
(D) both the vectors $\overrightarrow{a}and\overrightarrow{b}$ have same direction, but different magnitudes.

Answer:

(A) If two vectors are collinear then, they have the same direction or are parallel or antiparallel.Therefore, They can be expressed in the form $\overrightarrow{b}=\lambda \overrightarrow{a}$ where a and b are vectors and $\lambda$ is some scalar quantity. Therefore, (A) is true.
(B) $\lambda$ is a scalar quantity so its value may be equal to $\pm 1$ is the obvious result. Therefore, ( B ) is also true.
(C) The vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are proportional, Therefore, (C) is not true.
(D) The vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ can have different magnitudes as well as different directions. Therefore, (D) is not true. Therefore, the incorrect options are (C) and (D).