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

Question 1: Compute the magnitude of the following vectors:

(2) $\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Answer:

Here,

$\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Magnitude of $\vec b$

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

Question 1: Compute the magnitude of the following vectors:

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

Answer:

Here,

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

Magnitude of $\vec c$

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

Question 2: Write two different vectors having same magnitude

Answer:

Two different Vectors having the same magnitude are

$\vec a= 3\hat i+6\hat j+9\hat k$

$\vec b= 9\hat i+6\hat j+3\hat k$

The magnitude of both vector

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

Question 3: Write two different vectors having same direction.

Answer:

Two different vectors having the same direction are:

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

$\vec b=2\hat i+4\hat j+6\hat k$

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,

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

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

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

Answer:

Given,

$\\ \vec a = \hat i - 2 \hat j + \hat k ,\\ \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \\\vec c = \hat i - 6 \hat j - 7 \hat k$

Now, The sum of the vectors:

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

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

$\vec a +\vec b+\vec c =-4\hat j-\hat k$

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

Answer:

Given

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

Magnitude of $\vec a$

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

A unit vector in the direction of $\vec a$

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

Question 8: Find the unit vector in the direction of vector $\vec { PQ}$ , 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

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

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

Magnitude of PQ

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

Now, unit vector in direction of PQ

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

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

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

Answer:

Given

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

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

Now,

$\vec a + \vec b=(2-1)\hat i+(-1+1)\hat j+ (2-1)\hat k$

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

Now a unit vector in the direction of $\vec a + \vec b$

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

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

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

$\vec a=5 \hat i - \hat j + 2 \hat k$

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

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

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

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

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

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

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k$

It can be seen that

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

Hence here $\vec b=-2\vec a$

As we know

Whenever we have $\vec b=\lambda \vec a$ , the vector $\vec a$ and $\vec b$ 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

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

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

Hence direction cosine of $\vec a$ are

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

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

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

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

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

Hence Direction cosines of vector AB are

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

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

Answer:

Let

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

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

Hence direction cosines of this vectors is

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

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

Now as we know,

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

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 $i + 2 j - k$ 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:

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

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

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

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

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

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:

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

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

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

$\vec r = -3\hat i +3\hat k$

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:

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

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

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

Answer:

Given

the position vector of A, B, and C are

$\\\vec a = 3 \hat i - 4 \hat j - 4 \hat k ,\\\vec b = 2 \hat i - \hat j + \hat k \: \: and \\\vec c = \hat i - 3 \hat j - 5 \hat k$

Now,

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

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

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

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

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

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

As we can see

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

Hence, ABC is a right angle triangle.

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

$A ) \overline{AB}+ \overline{BC}+ \overline{CA} = \vec 0 \\\\ B ) \overline{AB}+ \overline{BC}- \overline{AC} = \vec 0 \\\\ C ) \overline{AB}+ \overline{BC}- \overline{CA} = \vec 0 \\\\ D ) \overline{AB}- \overline{CB}+ \overline{CA} = \vec 0$

Answer:

From triangles law of addition we have,

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

From here

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

also

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

Also

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

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) $\vec b = \lambda \vec a$ for some saclar $\lambda$
(B) $\vec a = \pm \vec b$
(C) the respective components of $\vec a \: \:and \: \: \vec b$ are not proportional
(D) both the vectors $\vec a \: \:and \: \: \vec b$ 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 $\vec{b}=\lambda \vec{a}$ 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).