Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.7

Last Updated : 11 Feb, 2021

Question 1. Show that the points A, B, C with position vectors $\vec{a}-2\vec{b}+3\vec{c},\ 2\vec{a}+3\vec{b}-4\vec{c}$ and $-7\vec{b}+10\vec{c}$ are collinear.

Solution:

Given that,

Position vector of $A=\vec{a}-2\vec{b}+3\vec{c}$

Position vector of $B=2\vec{a}+3\vec{b}-4\vec{c}$

Position vector of $C=-7\vec{b}+10\vec{c}$

$\overrightarrow{AB}$ = Position vector B – Position vector of A

$= (2\vec{a}+3\vec{b}-4\vec{c})-(\vec{a}-2\vec{b}+3\vec{c})\\ =2\vec{a}+3\vec{b}-4\vec{c}-\vec{a}+2\vec{b}-3\vec{c}$

= $\vec{a}+5\vec{b}-7\vec{c}$

$\overrightarrow{BC}$ = Position vector of C – Position vector of B

$= (-7\vec{b}+10\vec{c})-(2\vec{a}+3\vec{b}-4\vec{c})\\ =-7\vec{b}+10\vec{c}-2\vec{a}-3\vec{b}+4\vec{c}$

= $-2\vec{a}-10\vec{b}+14\vec{c}$

Using $\overrightarrow{AB}$ and $\overrightarrow{BC}$ , we get,

$\overrightarrow{BC}=-2\overrightarrow{AB}$

So, $\overrightarrow{AB}$ || $\overrightarrow{BC}$ but $\vec{B}$ is a common vector.

Hence, proved that A, B, C are collinear.

Question 2 (i). If $\vec{a},\ \vec{b},\ \vec{c}$ are non-coplanar vectors, prove that the points having the position vectors $\vec{a},\ \vec{b},\ 3\vec{a}-2\vec{b}$ are collinear.

Solution:

Let us assume three points that are A, B, C

Position vector of A = $\vec{a}$

Position vector of B = $\vec{b}$

Position vector of C = $3\vec{a}-2\vec{b}$

$\overrightarrow{AB}$ = Position vector of B – Position vector of A

= $\vec{b}-\vec{a}$ -(1)

$\overrightarrow{BC}$ = Position vector of C – Position vector of B

$=3\vec{a}-2\vec{b}-\vec{b}\\ =3\vec{a}-3\vec{b}$ -(2)

Using equation(1) and (2), we get

$\overrightarrow{BC}$ = λ $(\overrightarrow{AB})$ -(where λ is a scalar)

$3\vec{a}-3\vec{b}=λ(\vec{b}-\vec{a})\\ 3\vec{a}-3\vec{b}=λ\vec{b}-λ\vec{a}\\ 3\vec{a}-3\vec{b}=λ\vec{a}+λ\vec{b}$

On comparing the coefficients of LHS and RHS,

-λ = 3

λ = 3

λ = -3

The value of λ is different

Therefore, points A, B, C are not collinear.

Question 2 (ii) If $\vec{a},\ \vec{b},\ \vec{c}$ are non-coplanar vectors, prove that the points having the position vectors $\vec{a}+\vec{b}+\vec{c},\ 4\vec{a}+3\vec{b},\ 10\vec{a}+7\vec{b}-2\vec{c}$ are collinear.

Solution:

Let us assume three points that are A, B, C

Position vector of A = $\vec{a}+\vec{b}+\vec{c}$

Position vector of B = $4\vec{a}+3\vec{b}$

Position vector of C = $10\vec{a}+7\vec{b}-2\vec{c}$

$\overrightarrow{AB}$ = Position vector of B – Position vector of A

= $=(4\vec{a}+3\vec{b})-(\vec{a}+\vec{b}+\vec{c})\\ =4\vec{a}+3\vec{b}-\vec{a}-\vec{b}-\vec{c}\\ \overrightarrow{AB}=3\vec{a}+2\vec{b}-\vec{c}$

$\overrightarrow{BC}$ = Position vector of C – Position vector of B

$=(10\vec{a}+7\vec{b}-2\vec{c})-(4\vec{a}+3\vec{b})\\ =10\vec{a}+7\vec{b}-2\vec{c}-4\vec{a}-3\vec{b}\\ \overrightarrow{BC}=6\vec{a}+4\vec{b}-2\vec{c}$

By using $\overrightarrow{AB}$ and $\overrightarrow{BC}$ , we get

$\overrightarrow{BC}$ = 2 $(\overrightarrow{AB})$

So, $(\overrightarrow{AB})$ || $(\overrightarrow{BC})$ but $\vec{B}$ is a common vector.

Hence, A, B, C are collinear

Question 3. Prove that the points having position vectors $\hat{i}+2\hat{j}+3\hat{k},\ 3\hat{i}+4\hat{j}+7\hat{k},\ -3\hat{i}-2\hat{j}-5\hat{k}$ are collinear.

Solution:

Let us considered points A, B, C

Position vector of A = $\hat{i}+2\hat{j}+3\hat{k}$

Position vector of B = $3\hat{i}+4\hat{j}+7\hat{k}$

Position vector of C = $-3\hat{i}-2\hat{j}-5\hat{k}$

$\overrightarrow{AB}$ = Position vector of B – Position vector of A

= $=(3\hat{i}+4\hat{j}+7\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})\\ =3\hat{i}+4\hat{j}+7\hat{k}-\hat{i}-2\hat{j}-3\hat{k}\\ \overrightarrow{AB}=2\hat{i}+2\hat{j}+4\hat{k}$

$\overrightarrow{BC}$ = Position vector of C – Position vector of B

$=(-3\hat{i}-2\hat{j}-5\hat{k})-(3\hat{i}+4\hat{j}+7\hat{k})\\ =-3\hat{i}-2\hat{j}-5\hat{k}-3\hat{i}-4\hat{j}-7\hat{k}\\ \overrightarrow{BC}=-6\hat{i}-6\hat{j}-12\hat{k}$

By using $\overrightarrow{AB}$ and $\overrightarrow{BC}$ , we get

$\overrightarrow{BC}$ = -3 $(\overrightarrow{AB})$

$(\overrightarrow{AB})$ || $(\overrightarrow{BC})$ but $\vec{B}$ is a common vector.

Hence, A, B, C are collinear

Question 4. If the points with position vectors $10\hat{i}+3\hat{j},\ 12\hat{i}-5\hat{j}\ and\ a\hat{i}+11\hat{j}$ are collinear, find the value of a.

Solution:

Let the points be A, B, C

Position vector of A = $10\hat{i}+3\hat{j}$

Position vector of B = $12\hat{i}-5\hat{j}$

Position vector of C = $a\hat{i}+11\hat{j}$

Given that, A, B, C are collinear

⇒ $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear

⇒ $\overrightarrow{AB}$ = λ $(\overrightarrow{BC})$ -(where λ is same scalar)

⇒ Position vector of B – Position vector of A = λ – (Position vector of C – Position vector of B)

$⇒ (12\hat{i}-5\hat{j})-(10\hat{i}+3\hat{j})=λ[(a\hat{i}+11\hat{j})-(12\hat{i}-5\hat{j})]\\ ⇒ 12\hat{i}-5\hat{j}-10\hat{i}-3\hat{j}=λ(a\hat{i}+11\hat{j}-12\hat{i}+5\hat{j})]\\ ⇒ 2\hat{i}-8\hat{j}=(λa-12λ)\hat{i}=(11λ+5λ)\hat{j}$

On comparing the coefficients of LHS and RHS, we get

λa – 12λ = 2 -(1)

-8 = 11λ + 5λ -(2)

-8 = 16λ

λ = -8/16

λ = -1/2

Now, Put the value of λ in eq(1), we get

λa – 12λ = 2

(-1/2)a – 12(-1/2) = 2

-a/2 +6 = 2

-a/2 = 2 – 6

-a/2 = -4

a = 8

So, the value of a is 8.

Question 5. If $\vec{a},\ \vec{b}$ are two non-collinear vectors, prove that the points with position vectors $\vec{a}+\vec{b},\ \vec{a}-\vec{b}\ and\ \vec{a}+λ\vec{b}$ are collinear for all real values of λ.

Solution:

Let us considered points A, B, C

Position vector of A = $\vec{a}+\vec{b}$

Position vector of B = $\vec{a}-\vec{b}$

Position vector of C = $\vec{a}+λ\vec{b}$

$\overrightarrow{AB}$ = Position vector of B – Position vector of A

$=(\vec{a}-\vec{b})-(\vec{a}+\vec{b})\\ =\vec{a}-\vec{b}-\vec{a}-\vec{b}\\ \overrightarrow{AB}=-2\vec{b}$

$\overrightarrow{BC}$ = Position vector of C – Position vector of B

$=(\vec{a}+λ\vec{b})-(\vec{a}-\vec{b})\\ =\vec{a}+λ\vec{b}-\vec{a}+\vec{b}\\ =λ\vec{b}+\vec{b}\\ \overrightarrow{BC}=(λ+1)\vec{b}$

Using $\overrightarrow{AB}$ and $\overrightarrow{BC}$

$\overrightarrow{AB}$ = $\left[\frac{(λ+1)}{-2}\right](\overrightarrow{BC})$

Let $\left(\frac{λ+1}{-2}\right)$ = μ

Since λ is a real number. So, μ is also a real number.

$(\overrightarrow{AB})$ || $(\overrightarrow{BC})$ but $\vec{B}$ is a common vector.

Hence, A, B, C are collinear.

Question 6. If $\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OB}+\overrightarrow{OC}$ , prove that A, B, C are collinear points

Solution:

According to the question

$\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OB}+\overrightarrow{OC}\\ \overrightarrow{OA}+\overrightarrow{BO}=\overrightarrow{BO}-\overrightarrow{CO}\\ \overrightarrow{AB}=\overrightarrow{BC}$

So, $(\overrightarrow{AB})$ || $(\overrightarrow{BC})$ but $\vec{B}$ is a common vector.

Hence, A, B, C are collinear.

Question 7. 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.

Solution:

Le us considered, the position vector A = $2\hat{i}-3\hat{j}+4\hat{k}$

Position vector B = $-4\hat{i}+6\hat{j}-8\hat{k}$

Let us assume O be the initial point having position vector

$0\times\hat{i}+0\times\hat{j}+0\times\hat{k}$

$\overrightarrow{OA}$ = Position vector of A – Position vector of O

$=(2\hat{i}-3\hat{j}+4\hat{k})-(0\times\hat{i}+0\times\hat{j}+0\times\hat{k})\\ =2\hat{i}-3\hat{j}+4\hat{k}$

$\overrightarrow{OB}$ = Position vector of B – Position vector of O

$=(-4\hat{i}+6\hat{j}-8\hat{k})-(0\times\hat{i}+0\times\hat{j}+0\times\hat{k})\\ =-4\hat{i}+6\hat{j}-8\hat{k}$

By u8sing OA and OB, we get

$\overrightarrow{OB}=-2(\overrightarrow{OA})$

Therefore, $\overrightarrow{OA}$ || $\overrightarrow{OB}$ but O is the common point to them.

Hence, A and B are collinear.

Question 8. If the points A(m, -1), B(2, 1), C(4, 5) are collinear, find the value of m.

Solution:

Let us considered

A = (m, -1)

B = (2, 1)

C = (4, 5)

$\overrightarrow{AB}$ = Position vector of B – Position vector of A

$=(2\hat{i}+\hat{j})-(m\hat{i}-\hat{j})\\ =2\hat{i}+\hat{j}-m\hat{i}+\hat{j}\\ =(2-m)\hat{i}+2\hat{j}$

$\overrightarrow{BC}$ = Position vector of C – Position vector of B

$=(4\hat{i}+5\hat{j})-(2\hat{i}+\hat{j})\\ =4\hat{i}+5\hat{j}-2\hat{i}-\hat{j}\\ =2\hat{i}+4\hat{j}$

A, B, C are collinear.

So, $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear.

So, $\overrightarrow{AB}$ = λ $(\overrightarrow{BC})$

$(2-m)\hat{i}+2\hat{j}=λ(2\hat{i}+4\hat{j}),\ \ for\ λ\ scalar\\ (2-m)\hat{i}+2\hat{j}=2λ\hat{i}+4λ\hat{j}$

On comparing the coefficient of LHS and RHS, we get

2 – m = 2λ

$\frac{2-m}{2}=λ$ -(1)

2 = 4λ

$\frac{2}{4}$ = λ

$\frac{1}{2}$ = λ -(2)

From eq(1) and (2), we get

$\frac{2-m}{2}=\frac{1}{2}$

4 – 2m = 2

-2m = 2 – 4

m = $\frac{-2}{-2}$

m = 1

So, the value of m is 1

Question 9. Show that the points (3, 4), (-5, 16), (5, 1) are collinear.

Solution:

Let su considered

A = (3, 4)

B = (-5, 16)

C = (5, 1)

$\overrightarrow{AB}$ = Position vector of B – Position vector of A

$=(-5\hat{i}+16\hat{j})-(3\hat{i}+4\hat{j})\\ =-5\hat{i}+16\hat{j}-3\hat{i}-4\hat{j}\\ \overrightarrow{AB}=-8\hat{i}+12\hat{j}$

$\overrightarrow{BC}$ = Position vector of C – Position vector of B

$=(5\hat{i}+\hat{j})-(-5\hat{i}+16\hat{j})\\ =5\hat{i}+\hat{j}+5\hat{i}-16\hat{j}\\ \overrightarrow{BC}=10\hat{i}-15\hat{j}$

So, $4(\overrightarrow{AB})=-5(\overrightarrow{BC})$

$\overrightarrow{AB}$ || $\overrightarrow{BC}$ but B is a common point.

Hence, A, B, C are collinear.

Question 10. If the vectors $\vec{a}=2\hat{i}-3\hat{j}\ and\ \vec{b}=-6\hat{i}+m\hat{j}$ are collinear, find the value of m.

Solution:

Given: $a=2\hat{i}-3\hat{j}$ and $b=-6\hat{i}+m\hat{j}$ are collinear.

So, a = λb

$2\hat{i}-3\hat{j}=λ(-6\hat{i}+m\hat{j})\\ 2\hat{i}-3\hat{j}=-6λ\hat{i}+mλ\hat{j}$

On comparing the coefficients of LHS and RHS, we get

2 = -6λ

λ = 2/(-6)

λ = -1/3 -(1)

-3 = λm

λ = -3/m -(2)

From eq(1) and (2), we have

-1/3 = -3/m

m = 3 × 3

m = 9

So, the value of m is 9

Question 11. Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Solution:

Given: A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7).

$\overrightarrow{AB}=(5-1)\hat{i}+(0+2)\hat{j}+(-2+8)\hat{k}=4\hat{i}+2\hat{j}+6\hat{k}\\ \overrightarrow{BC}=(11-5)\hat{i}+(3-0)\hat{j}+(7+2)\hat{k}=6\hat{i}+3\hat{j}+9\hat{k}\\ \overrightarrow{AC}=(11-1)\hat{i}+(3+2)\hat{j}+(7+8)\hat{k}=10\hat{i}+5\hat{j}+15\hat{k}\\ \left|\overrightarrow{AB}\right|=\sqrt{4^2+2^2+6^2}=\sqrt{16+4+36}=\sqrt{56}=2\sqrt{14}\\ \left|\overrightarrow{BC}\right|=\sqrt{6^2+3^2+9^2}=\sqrt{36+9+81}=\sqrt{126}=3\sqrt{14}\\ \left|\overrightarrow{AC}\right|=\sqrt{10^2+5^2+15^2}=\sqrt{100+25+225}=\sqrt{350}=5\sqrt{14}\\ \left|\overrightarrow{AC}\right|=\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|$

So the given points are collinear

Now, let us considered point B divide AC in the ratio λ : 1.

Then

$\overrightarrow{OB}=\frac{λ\overrightarrow{OC}+\overrightarrow{OA}}{(λ+1)}$

$⇒5\hat{i}-2\hat{k}=\frac{λ(11\hat{i}+3\hat{j}+7\hat{k})+(\hat{i}-2\hat{j}-8\hat{k})}{λ+1}\\ ⇒(λ+1)(5\hat{i}-2\hat{k})=11λ\hat{i}+3λ\hat{j}+7λ\hat{k}+\hat{i}-2\hat{j}-8\hat{k}\\ ⇒5(λ+1)\hat{i}-2(λ+1)\hat{k}=(11λ+1)\hat{i}+(3λ-2)\hat{j}+(7λ-8)\hat{k}$

5(λ + 1) = 11λ + 1

⇒ 5λ + 5 = 11λ + 1

⇒ 6λ = 4

⇒ λ = $\frac{4}{6}=\frac{2}{3}$

So, the point B divides AC in the ratio 2 : 3.

Question 12. Using vector show that the points A(-2, 3, 5), B(7, 0, -1) and C(-3, -2, -5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3).

Solution:

According to the question, we have,

$\overrightarrow{AP}$ = Position vector of P – Position vector of A

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

$\overrightarrow{PB}$ = Position vector of B – Position vector of P

$\overrightarrow{PB}=7\hat{i}-\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})=6\hat{i}-2\hat{j}-4\hat{k}$

Hence, $\overrightarrow{PB}=2\overrightarrow{AP}$ . So, the vectors are collinear.

But P is a point common so, P, A, B are collinear points.

Similarly, $\overrightarrow{CP}=\hat{i}+2\hat{j}+3\hat{k}-(-3\hat{i}-2\hat{j}-5\hat{k})=4\hat{i}+4\hat{j}+8\hat{k}\\ \overrightarrow{PD}=3\hat{i}+4\hat{j}+7\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})=2\hat{i}+2\hat{j}+4\hat{k}$

$\overrightarrow{CP}\ and\ \overrightarrow{PD}$ are collinear,

But P is a common point to $\overrightarrow{CP}\ and\ \overrightarrow{CD}$ . So, C, P, D are collinear points.

Hence, AB and CD intersect at the point P.

Question 13. Using vectors, find the value of I such that the points (I, -10, 3), (1, -1, 3), and (3, 5, 3) are collinear.

Solution:

Given: Points (I, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.

Therefore, (I, -10, 3) = x(1, -1, 3) + y(3, 5, 3) for some scalars x and y

I = x + 3y -(1)

-10 = -x + 5y -(2)

3 = 3x + 3y -(3)

On solving eq(2) and (3), we get,

x = 5/2 and y = -3/2

Now, put the value of x and y in eq(1), we get

I = (5/2) + 3(-3/2)

I = -2

So the value of I is -2

Suggest improvement

Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.6 | Set 2

Class 12 RD Sharma- Chapter 23 Algebra of Vectors - Exercise 23.8

Share your thoughts in the comments

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.7

Question 1. Show that the points A, B, C with position vectors and are collinear.

Question 2 (i). If are non-coplanar vectors, prove that the points having the position vectors are collinear.

Question 2 (ii) If are non-coplanar vectors, prove that the points having the position vectors are collinear.

Question 3. Prove that the points having position vectors are collinear.

Question 4. If the points with position vectors are collinear, find the value of a.

Question 5. If are two non-collinear vectors, prove that the points with position vectors are collinear for all real values of λ.

Question 6. If , prove that A, B, C are collinear points

Question 7. Show that the vectors are collinear.

Question 8. If the points A(m, -1), B(2, 1), C(4, 5) are collinear, find the value of m.

Question 9. Show that the points (3, 4), (-5, 16), (5, 1) are collinear.

Question 10. If the vectors are collinear, find the value of m.

Question 11. Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Question 12. Using vector show that the points A(-2, 3, 5), B(7, 0, -1) and C(-3, -2, -5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3).

Question 1. Show that the points A, B, C with position vectors $\vec{a}-2\vec{b}+3\vec{c},\ 2\vec{a}+3\vec{b}-4\vec{c}$ and $-7\vec{b}+10\vec{c}$ are collinear.

Question 2 (i). If $\vec{a},\ \vec{b},\ \vec{c}$ are non-coplanar vectors, prove that the points having the position vectors $\vec{a},\ \vec{b},\ 3\vec{a}-2\vec{b}$ are collinear.

Question 2 (ii) If $\vec{a},\ \vec{b},\ \vec{c}$ are non-coplanar vectors, prove that the points having the position vectors $\vec{a}+\vec{b}+\vec{c},\ 4\vec{a}+3\vec{b},\ 10\vec{a}+7\vec{b}-2\vec{c}$ are collinear.

Question 3. Prove that the points having position vectors $\hat{i}+2\hat{j}+3\hat{k},\ 3\hat{i}+4\hat{j}+7\hat{k},\ -3\hat{i}-2\hat{j}-5\hat{k}$ are collinear.

Question 4. If the points with position vectors $10\hat{i}+3\hat{j},\ 12\hat{i}-5\hat{j}\ and\ a\hat{i}+11\hat{j}$ are collinear, find the value of a.

Question 5. If $\vec{a},\ \vec{b}$ are two non-collinear vectors, prove that the points with position vectors $\vec{a}+\vec{b},\ \vec{a}-\vec{b}\ and\ \vec{a}+λ\vec{b}$ are collinear for all real values of λ.

Question 6. If $\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OB}+\overrightarrow{OC}$ , prove that A, B, C are collinear points

Question 7. 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.

Question 10. If the vectors $\vec{a}=2\hat{i}-3\hat{j}\ and\ \vec{b}=-6\hat{i}+m\hat{j}$ are collinear, find the value of m.