RD Sharma Class 12 Exercise 22.8 Algebra of vectors Solutions Maths

RD Sharma Class 12 Exercise 22.8 Algebra of vectors Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 06:20 PM IST

The Class 12 RD Sharma chapter 22 exercise 22.8 solution covers the chapter 'Algebra of Vectors.' Though many students find it difficult to find the solutions to the problems provided in this chapter, the RD Sharma class 12th exercise 22.8 comes under consideration. RD Sharma solutions This solution provides the most basic ways to understand this chapter's theory, which helps them further solve questions much easier.

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RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise
Algebra of Vectors Excercise: 22.8
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise

Algebra of Vectors Excercise: 22.8

Algebra of vectors exercise 22.8 question 1(i)

Answer: Vectors are collinear
Hint: Prove one vector representing in the form of another vector
$\Rightarrow Given$ Let
$\begin{aligned} &A=2 \hat{i}+\hat{j}-\hat{k} \\ &B=3 \hat{i}-2 \hat{j}+\hat{k} \\ &C=\hat{i}+4 \hat{j}-3 \hat{k} \end{aligned}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{aligned} &=(3 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+\hat{j}-\hat{k}) \\ &=(3-2) \hat{i}+\{-(2+1)\} \hat{j}+(1+1) \hat{k} \\ &=\hat{i}-3 \hat{j}+2 \hat{k} \end{aligned}$
$\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{aligned} &=(\hat{i}+4 \hat{j}-3 \hat{k})-(3 \hat{i}-2 \hat{j}+\hat{k}) \\ &=(1-3) \hat{i}+(4+2) \hat{j}+(-3-1) \hat{k} \\ &=-2 \hat{i}+6 \hat{j}-4 \hat{k} \\ &=-2(\hat{i}-3 \hat{j}+2 \hat{k}) \end{aligned}$
From (1) we can say
$\overrightarrow{BC}=-2\overrightarrow{AB}$
Also, B is the common point and $\overrightarrow{BC}$ and $\overrightarrow{AB}$ are parallel
Hence, Vectors are collinear.

Algebra of vectors exercise 22.8 question 1(ii)

Answer: Vectors are collinear
Hint: Find PQ and QR
$\Rightarrow Given$
$\begin{aligned} &P=3 \hat{i}-2 \hat{j}+4 \hat{k} \\ &Q=\hat{i}+\hat{j}+\hat{k} \\ &R=-\hat{i}+4 \hat{j}-2 \hat{k} \end{aligned}$
Solution:
Let P,Q and R be the points represented and the factor vectors are
$\begin{aligned} &P=3 \hat{i}-2 \hat{j}+4 \hat{k} \\ &Q=\hat{i}+\hat{j}+\hat{k} \\ &R=-\hat{i}+4 \hat{j}-2 \hat{k} \end{aligned}$
$\overrightarrow{PQ}$ = Position of vector Q – Position of vector P
$\begin{aligned} &=(-\hat{i}+4 \hat{j}-2 \hat{k})-(3 \hat{i}-2 \hat{j}+4 \hat{k}) \\ &=-\hat{i}+\hat{j}+\hat{k}-3 \hat{i}+2 \hat{j}-4 \hat{k} \\ &=-2 \hat{i}+3 \hat{j}-3 \hat{k} \end{aligned}$
$\overrightarrow{QR}$ = Position of vector R – Position of vector Q
$\begin{aligned} &=(-\hat{i}+4 \hat{j}-2 \hat{k})-( \hat{i}+ \hat{j}+ \hat{k}) \\ &=-\hat{i}+4\hat{j}-2\hat{k}- \hat{i}- \hat{j}- \hat{k} \\ &=-2 \hat{i}+3 \hat{j}-3 \hat{k} \end{aligned}$
Here, $\overrightarrow{PQ}=\overrightarrow{QR}$
So, they are parallel to each other and B bring the common point implies that vectors are all collinear.

Algebra of vectors exercise 22.8 question 2(i)

Answer: A,B and C are collinear
Hint: Use formula $\overrightarrow{AB}=\lambda \overrightarrow{BC}$ i.e. representing one vector as a scalar product of another vector
$\Rightarrow Given$ , the points,
$\begin{aligned} &A=6 \hat{i}-7 \hat{j}-\hat{k} \\ &B=2 \hat{i}-3 \hat{j}+\hat{k} \\ &C=4 \hat{i}-5 \hat{j} \end{aligned}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{aligned} &=(2 \hat{i}-3 \hat{j}+\hat{k})-(6 \hat{i}-7 \hat{j}-\hat{k}) \\ &=(2-6) \hat{i}+(-3+7) \hat{j}+(1+1) \hat{k} \\ &=-4 \hat{i}+4 \hat{j}+2 \hat{k} \\ &=-2(2 \hat{i}-2 \hat{j}-\hat{k}) \end{aligned}$
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{aligned} &=(4 \hat{i}-5 \hat{j}+0 \hat{k})-(2 \hat{i}-3 \hat{j}+\hat{k}) \\ &=(4-2) \hat{i}+(-5+3) \hat{j}+(0-1) \hat{k} \\ &=2 \hat{i}-2 \hat{j}-\hat{k} \end{aligned}$
From $\left ( 1 \right )$
$\overrightarrow{AB}=-2\overrightarrow{BC}$
Thus, both vectors are parallel to each other and as B is the common point
$A\left ( 6,-7,-1 \right ),B\left ( 2,-3,1 \right )$ & $C\left ( 4,-5,0 \right )$ are collinear.

Algebra of vectors exercise 22.8 question 2(ii)

Answer: A,B and C are collinear
Hint: Try to form vectors such that one vector represents the another vector as a scalar product
$\Rightarrow Given$ the points,
$\begin{aligned} &A=2 \hat{i}-\hat{j}+3 \hat{k} \\ &B=4 \hat{i}+3 \hat{j}+\hat{k} \\ &C=3 \hat{i}+\hat{j}+2 \hat{k} \end{aligned}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{aligned} &=(4 \hat{i}+3 \hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k}) \\ &=(4-2) \hat{i}+(3+1) \hat{j}+(1-3) \hat{k} \\ &=2 \hat{i}+4 \hat{j}-2 \hat{k} \\ &=2(\hat{i}+2 \hat{j}-\hat{k}) \end{aligned}$ ............(1)
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{aligned} &=(3 \hat{i}+\hat{j}+2 \hat{k})-(4 \hat{i}+3 \hat{j}+\hat{k}) \\ &=(3-4) \hat{i}+(1-3) \hat{j}+(2-1) \hat{k} \\ &=-\hat{i}-2 \hat{j}+\hat{k} \\ &=-(\hat{i}+2 \hat{j}-\hat{k}) \end{aligned}$ ............(2)
From (1) & (2)
$\overrightarrow{AB}=-2\overrightarrow{BC}$
Thus, both vectors are parallel to each other and as B being the common point
Given, Points A, B and C are collinear.

Algebra of vectors exercise 22.8 question 2(iii)

Answer: A,B and C are collinear
Hint: Represent one vector as a scalar product of another.
$\Rightarrow Given$ the points,
$\begin{aligned} &A=\hat{i}+2 \hat{j}+7 \hat{k} \\ &B=2 \hat{i}+6 \hat{j}+3 \hat{k} \\ &C=3 \hat{i}+10 \hat{j}-\hat{k} \end{aligned}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{aligned} &=(2 \hat{i}+6 \hat{j}+3 \hat{k})-(\hat{i}+2 \hat{j}+7 \hat{k}) \\ &=(2-1) \hat{i}+(6-2) \hat{j}+(3-7) \hat{k} \\ &=\hat{i}+4 \hat{j}+(-4) \hat{k} \\ &=\hat{i}+4 \hat{j}-4 \hat{k} \end{aligned}$ ..........(1)
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{aligned} &=(3 \hat{i}+10 \hat{j}-\hat{k})-(2 \hat{i}+6 \hat{j}+3 \hat{k})\\ &=(3-2) \hat{i}+(10-6) \hat{j}+(-1-3) \hat{k}\\ &=\hat{i}+4 \hat{j}-4 \hat{k}\\ \end{aligned}$ .........(2)
$\text { From }(1) \&(2)\\$
$\overrightarrow{A B}=\overrightarrow{B C}$
So, vectors are parallel. But B is a point common to them
Hence, the given points A,B and C are collinear.

Algebra of vectors exercise 22.8 question 2(iv)

Answer: Vectors are collinear
Hint: Use formula $\overrightarrow{AB}=\lambda \overrightarrow{BC}$ i.e. representing one vector as a scalar product of another vector
$\Rightarrow Given$ the points,
$\begin{aligned} &A=-3 \hat{i}-2 \hat{j}-5 \hat{k} \\ &B=\hat{i}+2 \hat{j}+3 \hat{k} \\ &C=3 \hat{i}+4 \hat{j}+7 \hat{k} \end{aligned}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{aligned} &=(\hat{i}+2 \hat{j}+3 \hat{k})-(-3 \hat{i}-2 \hat{j}-5 \hat{k})\\ &=(1+3) \hat{i}+(2+2) \hat{j}+(3+5) \hat{k}\\ &=4 \hat{i}+4 \hat{j}+8 \hat{k}\\ &=4(\hat{i}+\hat{j}+2 \hat{k}) \end{aligned}$ .....(1)
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{aligned} &=(3 \hat{i}+4 \hat{j}+7 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\\ &=(3-1) \hat{i}+(4-2) \hat{j}+(7-3) \hat{k}\\ &=2 \hat{i}+2 \hat{j}+4 \hat{k}\\ \end{aligned}$ .......(2)
$\text { From }(1) \&(2)\\$
$\overrightarrow{A B}=2 \overrightarrow{B C}$
Thus both vectors are parallel to each other and B being the common point
Hence, vectors are collinear.

Algebra of vectors exercise 22.8 question 3(i)

Answer: Vectors are co-planar
Hint: Vectors are coplanar if we express them as a linear combination of other two.
$\begin{aligned} &\Rightarrow 5 \vec{a}+6 \vec{b}+7 \vec{c}=x(7 \vec{a}-8 \vec{b}+9 \vec{c})+y(3 \vec{a}+20 \vec{b}+5 \vec{c}) \\ &\Rightarrow 5 \vec{a}+6 \vec{b}+7 \vec{c}=\vec{a}(7 x+3 y)+\vec{b}(-8 x+20 y)+\vec{c}(9 x+5 y) \end{aligned}$
Comparing the equations
$\begin{aligned} &5=7 x+3 y\\ &7 x+3 y=5\\ \end{aligned}$ .....(1)
$-8 x+20 y=6\\$ .......(2)
$9 x+5 y=7$ ........(3)
Solving equation (1) and (2) by elimination method
$\begin{aligned} &7 x+3 y=5\\ \end{aligned}$ ............ $\left ( 1 \right )\times \left ( -8 \right )$
$-8 x+20 y=6\\$ ............. $\left ( 2 \right )\times \left ( 7 \right )$
$\underline{\begin{gathered} \begin{array}{r} -56 x-24 y=-40 \\ -56 x+140 y=42 \\ \end{array} \end{gathered}}$
$-164 y=-82 \\ y=\frac{-82}{-164} \\ y=\frac{1}{2}$
Put value of y in (1)
$\begin{aligned} &7 x+3\left(\frac{1}{2}\right)=5 \\ &7 x+\frac{3}{2}=5 \\ &7 x=5-\frac{3}{2} \\ &7 x=\frac{10-3}{2} \\ &7 x=\frac{7}{2} \Rightarrow x=\frac{7}{7 \times 2} \\ \end{aligned}$
$\Rightarrow x=\frac{1}{2}$
Now $x=\frac{1}{2}$ & $y=\frac{1}{2}$ satisfies the third equation
Hence given vectors are coplanar.

Algebra of vectors exercise 22.8 question 3(ii)

Answer: Vectors are not co-planar
Hint: Place the three vectors in the form of linear combination of the other two
Let,
$\begin{aligned} &\Rightarrow \vec{a}-2 \vec{b}+3 \vec{c}=x(-3 \vec{b}+5 \vec{c})+y(-2 \vec{a}+3 \vec{b}-4 \vec{c}) \\ &\Rightarrow \vec{a}-2 \vec{b}+3 \vec{c}=\vec{a}(-2 y)+\vec{b}(-3 x+3 y)+\vec{c}(5 x-4 y) \end{aligned}$
Comparing the terms we get,
$\begin{aligned} &1=-2 y \Rightarrow y=\frac{-1}{2} \\ &-2=-3 x+3 y \Rightarrow-2=-3 x+3\left(\frac{-1}{2}\right) \\ &-2=-3 x-\frac{3}{2} \\ &3 x=-\frac{3}{2}+2 \\ &3 x=\frac{-3+4}{2} \\ \end{aligned}$
$x=\frac{1}{6} \\$
$\text { Similarly }, 3=5 x-4 y \\ \text { Put } \\ x=\frac{1}{6} \& y=\left(-\frac{1}{2}\right) \\ \text { R.H.S }=5 x-4 y$
$\begin{aligned} &=5\left(\frac{1}{6}\right)-4\left(-\frac{1}{2}\right) \\ &=\frac{5}{6}+2 \\ &=\frac{5+12}{6} \\ &=\frac{17}{6} \neq 3 \end{aligned}$
Thus, here value of x and y do not satisfy the third equation.
Hence given vectors are not coplanar.

Algebra of vectors exercise 22.8 question 4

Answer: Vectors are co-planar
Hint: Express four points as the linear combination of one vector into two vectors
Given: Let,
$\begin{aligned} &P=6 \hat{i}-7 \hat{j} \\ &Q=16 \hat{i}-19 \hat{j}-4 \hat{k} \\ &R=3 \hat{j}-6 \hat{k} \\ &S=2 \hat{i}-5 \hat{j}+10 \hat{k} \end{aligned}$
$\Rightarrow$ Now, Expressing one vector as a linear combination of other two.
$\begin{aligned} &\overrightarrow{P Q}=(16 \hat{i}-19 \hat{j}-4 \hat{k})-(6 \hat{i}-7 \hat{j}+0 \hat{k})\\ &=(16-6) \hat{i}+(-19+7) \hat{j}+(-4-0) \hat{k}\\ &=10 \hat{i}-12 \hat{j}-4 \hat{k}\\ &\overrightarrow{P R}=(0 \hat{i}+3 \hat{j}-6 \hat{k})-(6 \hat{i}-7 \hat{j}+0 \hat{k})\\ &=(0-6) \hat{i}+(3+7) \hat{j}+(-6-0) \hat{k}\\ &=-6 \hat{i}+10 \hat{j}-6 \hat{k}\\ &\overrightarrow{P S}=(2 \hat{i}-5 \hat{j}+10 \hat{k})-(6 \hat{i}-7 \hat{j}+0 \hat{k})\\ &=(2-6) \hat{i}+(-5+7) \hat{j}+(0-0) \hat{k} \end{aligned}$
$\begin{aligned} &=-4 \hat{i}+2 \hat{j}+10 \hat{k} \\ &\overrightarrow{P Q}=x \overrightarrow{P R}+y \overrightarrow{P S} \\ &10 \hat{i}-12 \hat{j}-4 \hat{k}=x(-6 \hat{i}+10 \hat{j}-6 \hat{k})+y(-4 \hat{i}+2 \hat{j}+10 \hat{k}) \\ &10 \hat{i}-12 \hat{j}-4 \hat{k}=\hat{i}(-6 x-4 y)+\hat{j}(10 x+2 y)+\hat{k}(-6 x+10) \end{aligned}$
Comparing coefficients of i, j and k on both sides
$\begin{aligned} &-6 x-4 y=10 \Rightarrow-2(3 x+2 y)=10 \Rightarrow 3 x+2 y=-5 \ldots \ldots .(1)\\ &10 x+2 y=-12 \Rightarrow 2(5 x+y)=-12 \Rightarrow 5 x+y=-6 \ldots \ldots .\left ( 2 \right )\\ &-6 x+10 y=-4 \Rightarrow-2(3 x-5 y)=-4 \Rightarrow 3 x-5 y=2 \ldots \ldots .(3) \end{aligned}$
Solving (1) and (3) by elimination
$\begin{aligned} &\begin{array}{c} 3 x+2 y=-5 \\ 3 x-5 y=2 \\ \frac{7 y}{y}=-7 \\ y=\frac{-7}{7} \Rightarrow y=-1 \\ 3 x-5(-1)=2 \\ 3 x+5=2 \\ 3 x=2-5 \\ x=\frac{-3}{3}=-1 \end{array} \end{aligned}$
Now, Put $x=-1$ & $y=-1$ in equation (2)
$\begin{aligned} &5 x+y=-6 \\ &\text { L.H } S=5(-1)+(-1) \\ &=-5-1 \\ &=-6 \\ &=R \cdot H . S \end{aligned}$
As $x=-1$ & $y=-1$ satisfies all the equation we can say that
All four points are coplanar.

Algebra of vectors exercise 22.8 question 5(i)

Answer: Vectors are co-planar
Hint: We know that vectors are coplanar if one can be expressed as a linear combination into other two.
Given:
So, if
$\begin{aligned} &A=2 \hat{i}-\hat{j}+\hat{k}\\ &B=\hat{i}-3 \hat{j}-5 \hat{k}\\ &C=3 \hat{i}-4 \hat{j}-4 \hat{k}\\ &\text { then, }\\ &A=x B+y C\\ &2 \hat{i}-\hat{j}+\hat{k}=x(\hat{i}-3 \hat{j}-5 \hat{k})+y(3 \hat{i}-4 \hat{j}-4 \hat{k})\\ &2 \hat{i}-\hat{j}+\hat{k}=\hat{i}(x+3 y)+\hat{j}(-3 x-4 y)+\hat{k}(-5 x-4 y) \end{aligned}$
Comparing coefficients of $\hat{i},\hat{j}$ & $\hat{k}$ on both sides
$\begin{aligned} &x+3 y=2\\ \end{aligned}$ .....(1)
$-3 x-4 y=-1\\$ ....(2)
$-5 x-4 y=1$ ....(3)
Subtracting (2) and (3)
$-3 x-4 y=-1$
$\underline{\\ -5 x-4 y=1}$
$2x$ $=-2$
$x=-1$
Put, $x=-1$ in (2)
$\begin{aligned} &-3(-1)-4 y=-1 \\ &3-4 y=-1 \\ &-4 y=-4 \\ &y=1 \end{aligned}$
Now, put x = -1 and y = 1 in (1)
$\begin{aligned} &\text { L.H.S }=x+3 y \\ &=(-1)+3(1) \\ &=-1+3 \\ &=2 \\ &=R \cdot H . S \end{aligned}$
Thus, the values satisfy all the three we can say that vectors are coplanar.

Algebra of vectors exercise 22.8 question 5(ii)

Answer: vectors are co-planar
Hint: If vectors are given express as linear combination
$\Rightarrow$ Given vectors as follows:
Let,
$\begin{aligned} &P=\hat{i}+\hat{j}+\hat{k} \\ &Q=2 \hat{i}+3 \hat{j}-\hat{k} \\ &R=-\hat{i}-2 \hat{j}+2 \hat{k} \\ &P=x Q+y R \\ &\hat{i}+\hat{j}+\hat{k}=x(2 \hat{i}+3 \hat{j}-\hat{k})+y(-\hat{i}-2 \hat{j}+2 \hat{k}) \\ &\hat{i}+\hat{j}+\hat{k}=\hat{i}(2 x-y)+\hat{j}(3 x-2 y)+\hat{k}(-x+2 y) \end{aligned}$
Comparing coefficient of $\hat{i},\hat{j}$ & $\hat{k}$ on both sides
$2x-y=1$ ........(1)
$3x-2y=1$ .........(2)
$-x+2y=1$ ..........(3)
Add eqn (2) and (3)
$3x-2y=1$
$\underline{-x+2y=1}$
$2x$ $=2$
$x=1$
Put, x = 1 in (3)
$\begin{aligned} &-1+2 y=1 \\ &2 y=1+1 \\ &y=\frac{2}{2}=1 \\ &\text { Put, } x=1 \& y=1 in\left ( 1 \right ) \end{aligned}$
$L.H.S=2x-y$
$=2\left ( 1 \right )-\left ( 1 \right )$
$=2-1$
$=1$
$=R.H.S$
As values satisfies the third equation we can say that the given vectors are coplanar.

Algebra of vectors exercise 22.8 question 6(i)

Answer: vectors are not co-planar
Hint: If in given vectors we can’t express one vector in linear combination of other two, then they are non-coplanar
$\Rightarrow$ Given,
$\begin{aligned} &P=3 \hat{i}+\hat{j}-\hat{k}\\ &Q=2 \hat{i}-\hat{j}+7 \hat{k}\\ &R=7 \hat{i}-\hat{j}+23 \hat{k}\\ &\text { then, } \end{aligned}$
$P\neq xQ+yR$ then they are non-coplanar
$\begin{aligned} &\text { Taking } \\ &P=x Q+y R \\ &3 \hat{i}+\hat{j}-\hat{k}=x(2 \hat{i}-\hat{j}+7 \hat{k})+y(7 \hat{i}-\hat{j}+23 \hat{k}) \\ &3 \hat{i}+\hat{j}-\hat{k}=\hat{i}(2 x+7 y)+\hat{j}(-x-y)+\hat{k}(7 x+23 y) \end{aligned}$
Comparing coefficient of $\hat{i},\hat{j}$ & $\hat{k}$ on both sides
$2x+7y=3$ .......(1)
$-x-y=1$ .........(2)
$7x+23y=-1$ ..........(3)
To eliminate x from (1) and (2) multiply equation (1) with (-1) and equation (2) with (2) and subtract
$-2x-7y=-3$
$\underline{-2x-2y=2}$
$-5y=-5$
$y=1$
Put value of y = 1 in equation (2)
$\begin{aligned} &-x-1=1 \\ &\begin{array}{l} -x=2 \\ x=-2 \end{array} \\ &\text { Put } x=-2 \& y=\operatorname{lin}(3) \\ &7 x+23 y=-1 \\ &=7(-2)+23 \\ &=-14+23 \\ &=9 \\ &\neq R \cdot H \cdot S \end{aligned}$
As values of x and y do not satisf the third equation
$P\neq xQ+yR$
Thus, they are non-coplanar.

Algebra of vectors exercise 22.8 question 6(ii)

Answer: vectors are not co-planar
Hint: Form vectors in linear combination of other two vectors
$\Rightarrow$ Given, three vectors
Let,
$\begin{aligned} &A=\hat{i}+2 \hat{j}+3 \hat{k} \\ &B=2 \hat{i}+\hat{j}+3 \hat{k} \\ &C=\hat{i}+\hat{j}+\hat{k} \\ &A=x B+y C \\ &\hat{i}+2 \hat{j}+3 \hat{k}=x(2 \hat{i}+\hat{j}+3 \hat{k})+y(\hat{i}+\hat{j}+\hat{k}) \\ &\hat{i}+2 \hat{j}+3 \hat{k}=\hat{i}(2 x+y)+\hat{j}(x+y)+\hat{k}(3 x+y) \end{aligned}$
Comparing coefficient of $\hat{i},\hat{j}$ & $\hat{k}$
$2x+y=1$ .............(1)
$x+y=2$ ..............(2)
$3x+y=3$ .............(3)
Subtracting (1) and (2)
$\begin{aligned} &2 x+y=1\\ &\begin{aligned} &x+y=2 \\ &x=-1 \end{aligned}\\ &x+y=2\\ &-1+y=2\\ &y=3\\ &\text { Put } x=-1 \quad \& \quad y=3 \quad \text { in }\\ &3 x+y=3\\ &\text { Substitute } x=-1 \text { and } y=3 in \left ( 3 \right ) \end{aligned}$
$3x+y=3$
Substitute x=-1 and y=3
$3\left ( -1 \right )+3=3$
$= -3+3$
$=0$
$\neq 3$
Thus, as x and y do not satisfy the third equation
Hence, given vectors are non-coplanar.

Algebra of vectors exercise 22.8 question 7(i)

Answer: vectors are not co-planar
Hint: If vectors are in linear combination then they are coplanar and if they can’t be represented then they are not.
$\Rightarrow$ Given the vectors as follows:
$\begin{aligned} &A=2 \vec{a}-\vec{b}+3 \vec{c} \\ &B=\vec{a}+\vec{b}-2 \vec{c} \\ &C=\vec{a}+\vec{b}-3 \vec{c} \\ &A=x B+y C \\ &2 \vec{a}-\vec{b}+3 \vec{c}=x(\vec{a}+\vec{b}-2 \vec{c})+y(\vec{a}+\vec{b}-3 \vec{c}) \\ &2 \vec{a}-\vec{b}+3 \vec{c}=\vec{a}(x+y)+\vec{b}(x+y)+\vec{c}(-2 x-3 y) \end{aligned}$
Comparing coefficient of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$
$x+y=2$ .............(1)
$x+y= -1$ ..............(2)
$-2x-3y= 3$ ...............(3)
Subtracting equation (1) and (2)
$x+y=2$
$\underline{x+y=-1}$
No solution can be obtained
Hence given vectors are non-coplanar.

Algebra of vectors exercise 22.8 question 7(ii)

Answer: vectors are not co-planar
Hint: The coefficients of a,b and c
$\Rightarrow$ Given: $\begin{aligned} &\vec{a}+ \vec{b}+3 \vec{c}, \end{aligned}$ $\begin{aligned} &2\vec{a}+ \vec{b}+3 \vec{c} \end{aligned}$ and $\begin{aligned} &\vec{a}+ \vec{b}+ \vec{c} \end{aligned}$
Solution : We know that
Three vectors are coplanar if any one of them can be expressed sa the linear combination of other two vectors.
Let
$\begin{aligned} &\vec{a}+2 \vec{b}+3 \vec{c}=x(2 \vec{a}+\vec{b}+3 \vec{c})+y(\vec{a}+\vec{b}+\vec{c}) \\ &\vec{a}+2 \vec{b}+3 \vec{c}=\vec{a}(2 x+y)+\vec{b}(x+y)+\vec{c}(3 x+y) \end{aligned}$
Comparing the coefficients of L.H.S and R.H.S.
$2x+y=1$ ....(1)
$x+y=2$ ....(2)
$3x+y=3$ ....(3)
Solving (1) and (2)
$2x+y=1$
$\underline{x+y=2}$
$x$ $=-1$
And from (i)
$2\times \left ( -1 \right )+y=1$
$y=1+2$
$y=3$
There is no value of x and y that can satisfy the equation (3)
Thus, vectors are not co-planar.

Algebra of vectors exercise 22.8 question 8

Answer: $\overrightarrow{d}$ is expressible as linear combination of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$
Hint: Express one vector in linear combination of the other two
$\Rightarrow$ Given,
$\begin{aligned} &\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\ &\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k} \\ &\vec{c}=\hat{i}+\hat{j}+\hat{k} \\ &\Rightarrow \vec{a}=x \vec{b}+y \vec{c} \\ &\Rightarrow \hat{i}+2 \hat{j}+3 \hat{k}=x(2 \hat{i}+\hat{j}+3 \hat{k})+y(\hat{i}+\hat{j}+\hat{k}) \\ &\Rightarrow \hat{i}+2 \hat{j}+3 \hat{k}=\hat{i}(2 x+y)+\hat{j}(x+y)+\hat{k}(3 x+y) \end{aligned}$
Comparing coefficients of $\hat{i} ,\hat{j}$ & $\hat{k}$
$2x+y=1$ ...(1)
$x+y=2$ ....(2)
$3x+y=3$ ....(3)
Subtracting (1) and (2)
$2x+y=1$
$\underline{x+y=2}$
$x$ $=-1$
$x+y=2$
$-1+y=2$
$y=3$
Put, x=-1 and y=3
$3x+y=3$
$L.H.S=3x+y$
$=3\left ( -1 \right )+3$
$=-3+3$
$=0$
$\neq R.H.S$
Thus, $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ are non-coplanar
Now, expressing $\overrightarrow{a}$ as a linear combination of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$
$\begin{aligned} &\vec{d}=\vec{a} x+\vec{b} y+\vec{c} z \\ &\Rightarrow 2 \hat{i}-\hat{j}-3 \hat{k}=x(\hat{i}+2 \hat{j}+3 \hat{k})+y(2 \hat{i}+\hat{j}+3 \hat{k})+z(\hat{i}+\hat{j}+\hat{k}) \\ &\Rightarrow 2 \hat{i}-\hat{j}-3 \hat{k}=\hat{i}(x+2 y+z)+\hat{j}(2 x+y+z)+\hat{k}(3 x+3 y+z) \end{aligned}$
Comparing co-efficient of $\hat{i} ,\hat{j}$ & $\hat{k}$
$x+2y+z=2$ ...(1)
$2x+y+z=-1$ ....(2)
$3x+3y+z=-3$ ....(3)
Subtracting (1) and (2)
$x+2y+z=2$
$\underline{2x+y+z=-1}$
$-x+y$ $=3$
Subtracting (2) and (3)
$2x+y+z=-1$
$\underline{3x+3y+z=-3}$
$-x-2y$ $=2$
$x+2y=-2$ .......(5)
Subtracting (4) and (5)
$x-y=-3$
$\underline{x+2y=-2}$
$-3y=-1$
$y=\frac{1}{3}$
Put $y=\frac{1}{3}$ in (4)
$\begin{aligned} &x-\left(\frac{1}{3}\right)=-3 \\ &x=-3+\frac{1}{3} \\ &x=\frac{-8}{3} \end{aligned}$
Put $x=\frac{-8}{3}$ & $y=\frac{1}{3}$ in eq (1)
$\begin{aligned} &x+2 y+z=2 \\ &\frac{-8}{3}+2\left(\frac{1}{3}\right)+z=2 \\ &\frac{-6}{3}+z=2 \\ &z=2+2 \\ &z=4 \end{aligned}$
Hence, $\overrightarrow{d}$ is expressible as linear combination of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ .

Algebra of vectors exercise 22.8 question 9

Answer: Vectors are coplanar
Hint: Prove and express that vectors can be coplanar if we can express them as the linear combination.
$\Rightarrow$ Given, three vectors $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ to be proved as coplanar
Now “Necessary condition”
To be coplanar vectors must be expressed as linear combination of other two.
If $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ has to be coplanar.
There exists, $\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}$ .....(1)
Where x and y be some scalars.
Hence, $l\overrightarrow{c}+m\overrightarrow{b}+n\overrightarrow{c}=0$ ....(2)
Comparing (1) and (2)
We can say, l=x
m = y
c = -1
For $l\overrightarrow{c}+m\overrightarrow{b}+n\overrightarrow{c}=0$ to be coplanar they must satisfy this.Wherel,m and n are non-zero simultaneously.
“Sufficient Condition”
If we suppose that $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ be three vectors satisfying $l\overrightarrow{c}+m\overrightarrow{b}+n\overrightarrow{c}=0$
Where l,m and n not all zero simultaneously as scalars.
So from, $l\overrightarrow{c}+m\overrightarrow{b}+n\overrightarrow{c}=0$
$\begin{aligned} &n \vec{c}=-l \vec{a}-m \vec{b} \\ &\vec{c}=\left(\frac{-l}{n}\right) \vec{a}+\left(\frac{-m}{n}\right) \vec{b} \end{aligned}$
Thus, $\overrightarrow{c}$ can be written as linear combination of $\overrightarrow{a}$ & $\overrightarrow{b}$ where $\left ( \frac{-l}{n} \right )$ and $\left ( \frac{-m}{n} \right )$ be some scalars.
Hence, $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ are coplanar vectors.

Algebra of vectors exercise 22.8 question 10

Answer: Vectors are coplanar
Hint: From vectors as a linear combination to prove they are coplanar
$\Rightarrow$ Given $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ & $\overrightarrow{d}$ be four position vectors.
If these vectors are coplanar then, $l\overrightarrow{d}+m\overrightarrow{b}+n\overrightarrow{c}+w\overrightarrow{d}=0$ where, l,m,n and w be some scalars
As per given condition
$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ & $\overrightarrow{d}$ are coplanar if they are in the form of
$3\overrightarrow{a}-2\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{d}=0$
Comparing,
l = 3
m = -2
n = 1
w = -2
Now, $l+m+n+w=3-2+1-2=0$
Thus, vectors are coplanar. Where l, m, n and w are non-zero scalar simultaneously.

The RD Sharma class 12 solution of Algebra of vectors exercise 22.8 is divided into two levels- Level 1 and Level 2. The level 1 questions are quite the basic ones to clear up the introductory concepts of the chapter, and the level 2 questions are the ones that ask for brief solutions so that the students can prepare themselves for any questions that can be asked in the exam. The RD Sharma class 12 solutions chapter 22 exercise 22.8 consist a total of 18 questions that gives a brief of all the essential concepts of this chapter mentioned below-

Using vector method, prove the points are Collinear or Non-collinear vectors
Prove the vectors are Coplanar or Non-coplanar
Linear combination of vectors

Hence, we can find more benefits of using the RD Sharma class 12 solution chapter 22 exercise 22.7 in the list given below:-

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3. What is a coplanar vector?

Coplanar vectors are defined as the vectors which are lying on the same in a three-dimensional plane. The vectors are parallel to the same plan. The Co-planarity of two lines lies in a three-dimensional space, which is represented in vector form.

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