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

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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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