RD Sharma Class 12 Ex 23.7 Solutions Chapter 23 Algebra of Vectors

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Here we provide RD Sharma Class 12 Ex 23.7 Solutions Chapter 23 Algebra of Vectors for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 23.7 Solutions Chapter 23 Algebra of Vectors book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter23
Exercise23.7
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 23.7 Solutions Chapter 23 Algebra of Vectors

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

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