Class 12 NCERT Solutions – Mathematics Part ii – Chapter 10 – Vector Algebra Exercise 10.2

In the article, we will solve Exercise 10.2 from Chapter 10, “Vector Algebra” in the NCERT. Exercise 10.2 covers the basics of vectors like scalar and vector components of vectors, section formulas, Multiplication of a Vector by a scalar, etc.

Vector Algebra Formula to Solve Exercise 10.2

The basic vector algebra formulas to solve this exercise are mentioned below:

1. Magnitude of vector

let a vector [Tex]\overrightarrow{a}[/Tex],

if [Tex]\overrightarrow{a} = x\hat{i}+y\hat{j}+z\hat{k} [/Tex]

then [Tex]|\overrightarrow{a}| = \sqrt{x^2+y^2+z^2}[/Tex]

2. Section formula

(a) internally: let the position vector of a point R which divides the line joining P and Q in the ratio m:n internally.

[Tex]\overrightarrow{r} \space = \frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}[/Tex]

where [Tex]\overrightarrow{OP} = \overrightarrow{a}[/Tex] and [Tex]\overrightarrow{OQ} = \overrightarrow{b}[/Tex]

(b) Externally: let the position vector of a point R which divides the line joining P and Q in the ratio m:n externally.

[Tex]\overrightarrow{r} \space = \frac{m\overrightarrow{b}-n\overrightarrow{a}}{m-n}[/Tex]

3. Direction Cosine: let a vector [Tex]\overrightarrow{a} = x\hat{i}+y\hat{j}+z\hat{k}[/Tex]

then Direction Cosines [Tex]\ell , m , n [/Tex] will be:

[Tex]\ell = \frac{x}{|\overrightarrow{a}|}[/Tex] ; [Tex]m = \frac{y}{|\overrightarrow{a}|}[/Tex] ; [Tex]n = \frac{z}{|\overrightarrow{a}|}[/Tex]

4. Unit Vector

let a vector [Tex]\overrightarrow{a}[/Tex]

Unit vector [Tex]\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|}[/Tex]

Exercise 10.2 Solutions

Q1. Compute the magnitude of the following vectors:

[Tex]\overrightarrow{a} = \hat{i}+\hat{j}+\hat{k}; [/Tex] [Tex]\overrightarrow{b} = 2\hat{i}-7\hat{j}-3\hat{k}; [/Tex] [Tex]\overrightarrow{c} = \frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}-\frac{1}{\sqrt{3}}\hat{k} [/Tex]

Solution:

using formula of magnitude of vector

[Tex]|\overrightarrow{a}| = \sqrt{1^2+1^2+1^2} \space = \sqrt{3}[/Tex]

[Tex]|\overrightarrow{b}| = \sqrt{(2)^2+(-7)^2+(-3)^2} =\space \sqrt{4+49+9}\space = \sqrt{62}[/Tex]

[Tex]|\overrightarrow{c}| = \sqrt{(\frac{1}{\sqrt{3}})^2+(\frac{1}{\sqrt{3}})^2+(-\frac{1}{\sqrt{3}})^2} \space= \sqrt{\frac{1}{{3}}+\frac{1}{{3}}+\frac{1}{3}}\space = 1[/Tex]

Q2. Write two different vectors having same magnitude.

Solution:

An infinite number of possible answers. An example is given below:

let vectors as [Tex]\overrightarrow{a} = \hat{i}+2\hat{j}-3\hat{k};[/Tex] and [Tex]\overrightarrow{a} = 2\hat{i}-\hat{j}-3\hat{k};[/Tex]

their magnitudes

[Tex]|\overrightarrow{a}| = \sqrt{1^2+(-2)^2+3^2} \space = \sqrt{1+4+9} \space = \sqrt{14}[/Tex]

[Tex]|\overrightarrow{b}| = \sqrt{(2)^2+(-1)^2+(-3)^2} \space = \sqrt{4+1+9} \space = \sqrt{14}[/Tex]

we can see that [Tex]|\overrightarrow{a}| = |\overrightarrow{b}|[/Tex] but [Tex]\overrightarrow{a} \neq \overrightarrow{b}[/Tex]

Q3. Write two different vectors having same direction.

Solution:

An infinite number of possible answers. An example is given below:

let [Tex]\overrightarrow{a} = \hat{i}+\hat{j}+\hat{k};[/Tex] and [Tex]\overrightarrow{b} = 2\hat{i}+2\hat{j}+2\hat{k};[/Tex]

we are using Direction Cosines (D.C) to find direction of a vector

Direction Cosine of [Tex]\overrightarrow{a}[/Tex]

[Tex]\ell = \frac{1}{\sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}} ; [/Tex]

[Tex]m = \frac{1}{\sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}} ; [/Tex]

[Tex]n = \frac{1}{\sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}} ; [/Tex]

and Direction Cosine of [Tex]\overrightarrow{b}[/Tex]

[Tex]\ell = \frac{2}{\sqrt{2^2+2^2+2^2}} = \frac{2}{2\sqrt{3}} =\frac{1}{\sqrt{3}} ; [/Tex]

[Tex]m = \frac{2}{\sqrt{2^2+2^2+2^2}} = \frac{2}{2\sqrt{3}} =\frac{1}{\sqrt{3}} ; [/Tex]

[Tex]n = \frac{2}{\sqrt{2^2+2^2+2^2}} = \frac{2}{2\sqrt{3}} =\frac{1}{\sqrt{3}} ; [/Tex]

This proved that direction of [Tex]\overrightarrow{a}[/Tex] and [Tex]\overrightarrow{b}[/Tex] is same whereas [Tex]\overrightarrow{a}\neq\overrightarrow{b}[/Tex]

Q4. Find the values of x and y so that the vectors [Tex]2\hat{i}+3\hat{j} [/Tex]and [Tex]x\hat{i}+y\hat{j} [/Tex]are equal

Solution:

As it is given that [Tex]2\hat{i}+3\hat{j}[/Tex] and [Tex]x\hat{i}+y\hat{j}[/Tex] are equal

i.e. [Tex] x\hat{i}+y\hat{j} = 2\hat{i}+3\hat{j}[/Tex]

The corresponding components will also be equal. By Comparing corresponding components

we will get x = 2 and y = 3

Q5. Find the scalar and vector components of the vector with initial point (2 , 1) and terminal point (-5 , 7).

Solution:

Let p(2 , 1) and q(-5 , 7) and position vector [Tex]\overrightarrow{OP}[/Tex] and [Tex]\overrightarrow{OQ}[/Tex] is [Tex]2\hat{i}+\hat{j}[/Tex] and [Tex]-5\hat{i}+7\hat{j}[/Tex] respectively.

Therefore,

[Tex]\overrightarrow{PQ} = \overrightarrow{OQ}- \overrightarrow{OP}[/Tex]

i.e. [Tex]\overrightarrow{PQ} = (-5-2)\hat{i}+(7-1)\hat{j} = -7\hat{i}+6\hat{j}[/Tex]

So, the scalar components are -7 and 6 , and the vector components are [Tex]-7\hat{i}[/Tex] and [Tex]6\hat{j}[/Tex]

Q6. Find the sum of the vectors [Tex]\overrightarrow{a} = \hat{i}-2\hat{j}+\hat{k};[/Tex] [Tex]\overrightarrow{b} = -2\hat{i}-4\hat{j}+5\hat{k}[/Tex] and [Tex]\overrightarrow{c} = \hat{i}-6\hat{j}+7\hat{k}[/Tex]

Solution:

To find sum of vector we will add corresponding [Tex]\hat{i} , \hat{j} [/Tex] and [Tex]\hat{k}[/Tex] components

so,

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

[Tex]\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} = 0\hat{i}-4\hat{j}-\hat{k}[/Tex]

[Tex]\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} = -4\hat{j}-\hat{k}[/Tex]

Q7. Find the unit vector in the direction of the vector [Tex]\overrightarrow{a} = \hat{i}+\hat{j}+2\hat{k}[/Tex]

Solution:

Given that [Tex]\overrightarrow{a} = \hat{i}+\hat{j}+2\hat{k}[/Tex] and We know that,

[Tex]\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|}[/Tex]

So, [Tex]\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|} = \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{1^2+1^2+2^2}}= \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}[/Tex]

[Tex]\hat{a} = \frac{\hat{i}}{{\sqrt{6}}}+\frac{\hat{j}}{{\sqrt{6}}}+\frac{2\hat{k}}{{\sqrt{6}}}[/Tex]

Q8. Find the unit vector in the direction of vector [Tex]\overrightarrow{PQ}[/Tex] , where P and Q are the points (1 , 2 , 3) and (4 , 5 , 6) respectively.

Solution:

Given P and Q are the points (1 , 2 , 3) and (4 , 5 , 6).

Therefore, [Tex]\overrightarrow{PQ} = \overrightarrow{OQ}-\overrightarrow{OP}[/Tex]

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

[Tex]\overrightarrow{PQ}= 3\hat{i}+3\hat{j}+3\hat{k}[/Tex]

We know that [Tex]\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|} [/Tex]

Therefore,

[Tex]\hat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}= \frac{3\hat{i}+3\hat{j}+3\hat{k}}{\sqrt{3^2+3^2+3^2}} =\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}} [/Tex]

[Tex]\hat{PQ} = \frac{\hat{i}}{\sqrt{3}} +\frac{\hat{j}}{\sqrt{3}} +\frac{\hat{k}}{\sqrt{3}}[/Tex]

Q9. For given vectors [Tex]\overrightarrow{a} = 2\hat{i}-\hat{j}+2\hat{k}[/Tex], and [Tex]\overrightarrow{b} = -\hat{i}+\hat{j}-\hat{k}[/Tex], find the unit vector in the direction of the vector [Tex]\overrightarrow{a}+\overrightarrow{b}[/Tex].

Solution:

Given,

[Tex]\overrightarrow{a} = 2\hat{i}-\hat{j}+2\hat{k}[/Tex] and [Tex]\overrightarrow{b} = -\hat{i}+\hat{j}-\hat{k}[/Tex]

Therefore, [Tex]\overrightarrow{a}+\overrightarrow{b} = (2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}[/Tex]

[Tex]\overrightarrow{a}+\overrightarrow{b} = \hat{i}+\hat{k}[/Tex] and [Tex]|\overrightarrow{a}+\overrightarrow{b}| = \sqrt{1^2+1^2} = \sqrt{2}[/Tex]

We know that,

Unit vector = [Tex]\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}[/Tex]

[Tex]\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|} = \frac{\hat{i}+\hat{k}}{\sqrt{2}} = \frac{\hat{i}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}[/Tex]

Q10. Find a vector in the direction of vector [Tex]5\hat{i}-\hat{j}+2\hat{k}[/Tex] which has magnitude 8 units.

Solution:

Given, [Tex]\overrightarrow{a} = 5\hat{i}-\hat{j}+2\hat{k}[/Tex]

and [Tex]|\overrightarrow{a}| = \sqrt{5^2+(-1)^2+(2)^2} = \sqrt{25+1+4} [/Tex]

[Tex]|\overrightarrow{a}| = \sqrt{30}[/Tex]

Therefore,

[Tex]\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|} = \frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}[/Tex]

Thus, a vector parallel to [Tex]5\hat{i}-\hat{j}+2\hat{k}[/Tex] with magnitude 8 units is

So,

[Tex]8\hat{a} = 8(\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}})[/Tex]

[Tex]= \frac{40\hat{i}}{\sqrt{30}} -\frac{8\hat{j}}{\sqrt{30}} +\frac{16\hat{k}}{\sqrt{30}}[/Tex]

Q11. Show that the vectors [Tex] 2\hat{i}-3\hat{j}+4\hat{k}[/Tex] and [Tex] -4\hat{i}+6\hat{j}-8\hat{k}[/Tex] are collinear.

Solution:

Given that,

[Tex]\overrightarrow{b} = -4\hat{i}+6\hat{j}-8\hat{k}[/Tex]

[Tex]\overrightarrow{b} = -2(2\hat{i}-3\hat{j}+4\hat{k})[/Tex]

[Tex]\overrightarrow{b} = -2\overrightarrow{a}[/Tex]

and we know that,

When [Tex]\overrightarrow{b} = \lambda\overrightarrow{a}[/Tex] for two Collinear.

Where [Tex]\lambda = -2[/Tex]

So, the vectors [Tex] \overrightarrow{a}[/Tex] and [Tex]\overrightarrow{b}[/Tex] are Collinear.

Q12. Find the direction cosines of the vector [Tex] \hat{i}+2\hat{j}+3\hat{k}[/Tex].

Solution:

Let [Tex]\overrightarrow{a} = \hat{i}+2\hat{j}+3\hat{k}[/Tex]

and using magnitude concept of vector

[Tex]|\overrightarrow{a}| = \sqrt{1^2+2^2+3^2} = \sqrt{14}[/Tex]

So, Direction Cosine (D.C) of vector a will be

[Tex]\ell = \frac{x}{|\overrightarrow{a}|} = \frac{1}{\sqrt{14}}[/Tex]

[Tex]m = \frac{y}{|\overrightarrow{a}|} = \frac{2}{\sqrt{14}}[/Tex]

[Tex]n = \frac{z}{|\overrightarrow{a}|} = \frac{3}{\sqrt{14}}[/Tex]

The D.C s of vector a are [Tex](\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}})[/Tex]

Q13. Find the direction of the cosines of the vectors joining the points A(1 , 2 ,-3)and B(-1 , -2 , 1) directions from A to B.

Solution:

Let position vectors of point A and B are [Tex]\overrightarrow{OA}[/Tex] and [Tex]\overrightarrow{OB}[/Tex] respectively.

So, [Tex]\overrightarrow{AB} = \overrightarrow{OB}- \overrightarrow{OA}[/Tex]

i.e. [Tex]\overrightarrow{AB} = (-1-1)\hat{i}+(-2-2)\hat{j}+(1-(-3))\hat{k}[/Tex]

[Tex]\overrightarrow{AB} = -2\hat{i}-4\hat{j}+4\hat{k}[/Tex]

and [Tex]|\overrightarrow{AB}| = \sqrt{(-2)^2+(-4)^2+4^2} = \sqrt{4+16+16}[/Tex]

[Tex]|\overrightarrow{AB}| = \sqrt{36} = 6[/Tex]

Therefore, The Direction Cosine of vector AB [Tex](-\frac{2}{6},-\frac{4}{6},\frac{4}{6}) = (-\frac{1}{3},-\frac{2}{3},\frac{2}{3})[/Tex]

Q14. Show that the vector [Tex]\hat{i}+\hat{j}+\hat{k}[/Tex] is equally inclined to the axis OX, OY and OZ.

Solution:

Let [Tex]\overrightarrow{a} = \hat{i}+\hat{j}+\hat{k}[/Tex]

So, [Tex]|\overrightarrow{a}| = \sqrt{1^2+1^2+1^2} = \sqrt{3} [/Tex]

Therefore Direction Cosine will be [Tex](\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})[/Tex]

Let [Tex]\alpha[/Tex], [Tex]\beta[/Tex] and [Tex]\gamma[/Tex] be the angles formed by [Tex]\overrightarrow{a} [/Tex] with the positive directions of x , y and z axes respectively.

So, [Tex]\cos\alpha = \frac{1}{\sqrt{3}}[/Tex] , [Tex]\cos\beta = \frac{1}{\sqrt{3}}[/Tex] , [Tex]\cos\gamma = \frac{1}{\sqrt{3}}[/Tex]

All angles [Tex]\alpha, \beta [/Tex]and [Tex]\gamma[/Tex] are equal.

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

Q15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are [Tex] \hat{i}+2\hat{j}-\hat{k}[/Tex]and [Tex] -\hat{i}+\hat{j}+\hat{k}[/Tex] respectively, in the ratio 2:1

(i) Internally

(ii) Externally

Solution:

Let the position vectors of P and Q are [Tex]\overrightarrow{OP}[/Tex] and [Tex]\overrightarrow{OQ}[/Tex]

[Tex]\overrightarrow{OP} = \hat{i}+2\hat{j}-\hat{k}[/Tex]

[Tex]\overrightarrow{OQ} = -\hat{i}+\hat{j}+\hat{k}[/Tex]

(i) The position vector of R which divides the line joining two points P and Q internally in the ratio 2:1 is

using Section Formula(internally)

[Tex]\overrightarrow{OR} = \frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2\hat{j}-\hat{k})}{2+1}[/Tex]

[Tex]\overrightarrow{OR} = \frac{(-2\hat{i}+2\hat{j}+2\hat{k})+(\hat{i}+2\hat{j}-\hat{k})}{3}[/Tex]

[Tex]\overrightarrow{OR} = \frac{-\hat{i}+4\hat{j}+\hat{k}}{3}[/Tex]

[Tex]\overrightarrow{OR} = -\frac{\hat{i}}{3}+\frac{4\hat{j}}{3}+\frac{\hat{k}}{3}[/Tex]

(ii) The position vector of R which divides the line joining two points P and Q externally in the ratio 2:1 is

using Section Formula(externally)

[Tex]\overrightarrow{OR} = \frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2\hat{j}-\hat{k})}{2-1}[/Tex]

[Tex]\overrightarrow{OR} = \frac{(-2\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-\hat{k})}{1}[/Tex]

[Tex]\overrightarrow{OR} = -3\hat{i}+3\hat{k}[/Tex]

Q16. Find the position vector of the mid-point of the vector joining the points (2 , 3 , 4) and (4 , 1 , -2).

Solution:

Mid point divide the vector joining the points (2 , 3 , 4) and (4 , 1 , -2) in the ratio 1:1 internally.

The position vector of the mid-point R will be

[Tex]\overrightarrow{OR} = \frac{(2\hat{i}+3\hat{j}+4\hat{k})+1(4\hat{i}+1\hat{j}-2\hat{k})}{1+1}[/Tex]

[Tex]\overrightarrow{OR} = \frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}[/Tex]

[Tex]\overrightarrow{OR} = 3\hat{i}+2\hat{j}+\hat{k}[/Tex]

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

Solution:

Position vectors of points A, B, and C are respectively given as:

[Tex] \overrightarrow{a} = 3\hat{i}-4\hat{j}-4\hat{k}[/Tex] , [Tex] \overrightarrow{b} = 2\hat{i}-\hat{j}+\hat{k}[/Tex] and [Tex] \overrightarrow{c} = \hat{i}-3\hat{j}-5\hat{k}[/Tex]

So,[Tex] \overrightarrow{AB} = \overrightarrow{b}- \overrightarrow{a} = (2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}[/Tex]

[Tex]\overrightarrow{AB} = -\hat{i}+3\hat{j}+5\hat{k}[/Tex]

[Tex] \overrightarrow{BC} = \overrightarrow{c}- \overrightarrow{b} = (1-2)\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}[/Tex]

[Tex]\overrightarrow{BC} = -\hat{i}+2\hat{j}-6\hat{k}[/Tex]

[Tex] \overrightarrow{CA} = \overrightarrow{a}- \overrightarrow{c} = (3-1)\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}[/Tex]

[Tex]\overrightarrow{CA} = 2\hat{i}-\hat{j}+\hat{k}[/Tex]

Using magnitude of a vector formula,

[Tex]|\overrightarrow{AB}|^2 = {(-1)^2+3^2+5^2} = 35[/Tex]

[Tex]|\overrightarrow{BC}|^2 = {(-1)^2+(-2)^2+(-6)^2} = 41[/Tex]

[Tex]|\overrightarrow{CA}|^2 = {2^2+(-1)^2+1^2} = 6[/Tex]

As [Tex]|\overrightarrow{AB}|^2+|\overrightarrow{CA}| = 35+6 = 41 = |\overrightarrow{BC}|[/Tex]

Therefore, ABC is a right-angled triangle.

Q18. In triangle ABC (Fig 10.18), which of the following is not true:

(A) [Tex] \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{0}[/Tex]

(B) [Tex] \overrightarrow{AB} + \overrightarrow{BC} – \overrightarrow{AC} = \overrightarrow{0}[/Tex]

(C) [Tex] \overrightarrow{AB} + \overrightarrow{BC} – \overrightarrow{CA} = \overrightarrow{0}[/Tex]

(D) [Tex] \overrightarrow{AB} – \overrightarrow{CB} + \overrightarrow{CA} = \overrightarrow{0}[/Tex]

Solution: (C)

Using Triangle law of addition,

[Tex] \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} [/Tex]

[Tex] \overrightarrow{AB} + \overrightarrow{BC} = -\overrightarrow{CA} [/Tex]

[Tex] \overrightarrow{AB} + \overrightarrow{BC} +\overrightarrow{CA} = \overrightarrow{0}[/Tex]

So, the equation given in option A is true.

[Tex] \overrightarrow{AB} + \overrightarrow{BC} +\overrightarrow{CA} = \overrightarrow{0}[/Tex]

[Tex] \overrightarrow{AB} + \overrightarrow{BC} -\overrightarrow{AC} = \overrightarrow{0}[/Tex]

So, the equation given in option B is true.

and [Tex] \overrightarrow{AB} – \overrightarrow{CB} +\overrightarrow{CA} = \overrightarrow{0}[/Tex]

Hence, the equation given in option D is true.

Thus, the correct option will be (C).

Q19. If a and b are two collinear vectors, then which of the following are incorrect?

(A) [Tex]b = \lambda a[/Tex] , for some scalar [Tex]\lambda[/Tex]

(B) [Tex]a = \pm b[/Tex]

(C) the respective components of a and b are proportional.

(D) both the vectors a and b have same direction, but different magnitudes.

Solution: (D)

If a and b are collinear vectors, they are parallel.

Therefore, for some scalar [Tex]\lambda[/Tex]

[Tex]b = \lambda a [/Tex]

i.e. option A is true.

If [Tex]\lambda = \pm 1[/Tex], then [Tex]a = \pm b[/Tex]

option B is true.

[Tex]\overrightarrow{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k};[/Tex] and [Tex]\overrightarrow{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}[/Tex]

Then, [Tex]b = \lambda a [/Tex]

[Tex] b_1\hat{i}+b_2\hat{j}+b_3\hat{k} = \lambda( a_1\hat{i}+a_2\hat{j}+a_3\hat{k})[/Tex]

[Tex] b_1\hat{i}+b_2\hat{j}+b_3\hat{k} = \lambda a_1\hat{i}+\lambda a_2\hat{j}+\lambda a_3\hat{k}[/Tex]

Comparing the components of both the sides

we have [Tex]b_1 = \lambda a_1 , b_2 = \lambda a_2 , b_3 = \lambda a_3 [/Tex]

[Tex]\frac{b_1} {a_1}= \frac{b_2} {a_2}= \frac{b_3} {a_3} = \lambda[/Tex]

So, the respective components of vector a and b are proportional.

Thus, the correct option is D

Also, Check

Class 12 NCERT Solutions- Mathematics Part ii – Chapter 10 – Vector Algebra Exercise 10.1