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

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
$\begin{array}{rl}& A=2\hat{i}+\hat{j}-\hat{k}\\ & B=3\hat{i}-2\hat{j}+\hat{k}\\ & C=\hat{i}+4\hat{j}-3\hat{k}\end{array}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{array}{rl}& =(3\hat{i}-2\hat{j}+\hat{k})-(2\hat{i}+\hat{j}-\hat{k})\\ & =(3-2)\hat{i}+\{-(2+1)\}\hat{j}+(1+1)\hat{k}\\ & =\hat{i}-3\hat{j}+2\hat{k}\end{array}$
$\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{array}{rl}& =(\hat{i}+4\hat{j}-3\hat{k})-(3\hat{i}-2\hat{j}+\hat{k})\\ & =(1-3)\hat{i}+(4+2)\hat{j}+(-3-1)\hat{k}\\ & =-2\hat{i}+6\hat{j}-4\hat{k}\\ & =-2(\hat{i}-3\hat{j}+2\hat{k})\end{array}$
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
$\Rightarrow Given$
$\begin{array}{rl}& P=3\hat{i}-2\hat{j}+4\hat{k}\\ & Q=\hat{i}+\hat{j}+\hat{k}\\ & R=-\hat{i}+4\hat{j}-2\hat{k}\end{array}$
Solution:
Let P,Q and R be the points represented and the factor vectors are
$\begin{array}{rl}& P=3\hat{i}-2\hat{j}+4\hat{k}\\ & Q=\hat{i}+\hat{j}+\hat{k}\\ & R=-\hat{i}+4\hat{j}-2\hat{k}\end{array}$
$\overrightarrow{PQ}$ = Position of vector Q – Position of vector P
$\begin{array}{rl}& =(-\hat{i}+4\hat{j}-2\hat{k})-(3\hat{i}-2\hat{j}+4\hat{k})\\ & =-\hat{i}+\hat{j}+\hat{k}-3\hat{i}+2\hat{j}-4\hat{k}\\ & =-2\hat{i}+3\hat{j}-3\hat{k}\end{array}$
$\overrightarrow{QR}$ = Position of vector R – Position of vector Q
$\begin{array}{rl}& =(-\hat{i}+4\hat{j}-2\hat{k})-(\hat{i}+\hat{j}+\hat{k})\\ & =-\hat{i}+4\hat{j}-2\hat{k}-\hat{i}-\hat{j}-\hat{k}\\ & =-2\hat{i}+3\hat{j}-3\hat{k}\end{array}$
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,
$\begin{array}{rl}& A=6\hat{i}-7\hat{j}-\hat{k}\\ & B=2\hat{i}-3\hat{j}+\hat{k}\\ & C=4\hat{i}-5\hat{j}\end{array}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{array}{rl}& =(2\hat{i}-3\hat{j}+\hat{k})-(6\hat{i}-7\hat{j}-\hat{k})\\ & =(2-6)\hat{i}+(-3+7)\hat{j}+(1+1)\hat{k}\\ & =-4\hat{i}+4\hat{j}+2\hat{k}\\ & =-2(2\hat{i}-2\hat{j}-\hat{k})\end{array}$
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{array}{rl}& =(4\hat{i}-5\hat{j}+0\hat{k})-(2\hat{i}-3\hat{j}+\hat{k})\\ & =(4-2)\hat{i}+(-5+3)\hat{j}+(0-1)\hat{k}\\ & =2\hat{i}-2\hat{j}-\hat{k}\end{array}$
From $\left(1\right)$
$\overrightarrow{AB}=-2\overrightarrow{BC}$
Thus, both vectors are parallel to each other and as B is the common point
$A(6,-7,-1),B(2,-3,1)$&$C(4,-5,0)$ 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,
$\begin{array}{rl}& A=2\hat{i}-\hat{j}+3\hat{k}\\ & B=4\hat{i}+3\hat{j}+\hat{k}\\ & C=3\hat{i}+\hat{j}+2\hat{k}\end{array}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{array}{rl}& =(4\hat{i}+3\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})\\ & =(4-2)\hat{i}+(3+1)\hat{j}+(1-3)\hat{k}\\ & =2\hat{i}+4\hat{j}-2\hat{k}\\ & =2(\hat{i}+2\hat{j}-\hat{k})\end{array}$ ............(1)
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{array}{rl}& =(3\hat{i}+\hat{j}+2\hat{k})-(4\hat{i}+3\hat{j}+\hat{k})\\ & =(3-4)\hat{i}+(1-3)\hat{j}+(2-1)\hat{k}\\ & =-\hat{i}-2\hat{j}+\hat{k}\\ & =-(\hat{i}+2\hat{j}-\hat{k})\end{array}$ ............(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,
$\begin{array}{rl}& A=\hat{i}+2\hat{j}+7\hat{k}\\ & B=2\hat{i}+6\hat{j}+3\hat{k}\\ & C=3\hat{i}+10\hat{j}-\hat{k}\end{array}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{array}{rl}& =(2\hat{i}+6\hat{j}+3\hat{k})-(\hat{i}+2\hat{j}+7\hat{k})\\ & =(2-1)\hat{i}+(6-2)\hat{j}+(3-7)\hat{k}\\ & =\hat{i}+4\hat{j}+(-4)\hat{k}\\ & =\hat{i}+4\hat{j}-4\hat{k}\end{array}$ ..........(1)
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{array}{rl}& =(3\hat{i}+10\hat{j}-\hat{k})-(2\hat{i}+6\hat{j}+3\hat{k})\\ & =(3-2)\hat{i}+(10-6)\hat{j}+(-1-3)\hat{k}\\ & =\hat{i}+4\hat{j}-4\hat{k}\end{array}$ .........(2)
$\text{ From }(1)\mathrm{\&}(2)$
$\overrightarrow{AB}=\overrightarrow{BC}$
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,
$\begin{array}{rl}& A=-3\hat{i}-2\hat{j}-5\hat{k}\\ & B=\hat{i}+2\hat{j}+3\hat{k}\\ & C=3\hat{i}+4\hat{j}+7\hat{k}\end{array}$
$\overrightarrow{AB}$ = Position of vector B – Position of vector A
$\begin{array}{rl}& =(\hat{i}+2\hat{j}+3\hat{k})-(-3\hat{i}-2\hat{j}-5\hat{k})\\ & =(1+3)\hat{i}+(2+2)\hat{j}+(3+5)\hat{k}\\ & =4\hat{i}+4\hat{j}+8\hat{k}\\ & =4(\hat{i}+\hat{j}+2\hat{k})\end{array}$ .....(1)
Similarly, $\overrightarrow{BC}$ = Position of vector C – Position of vector B
$\begin{array}{rl}& =(3\hat{i}+4\hat{j}+7\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})\\ & =(3-1)\hat{i}+(4-2)\hat{j}+(7-3)\hat{k}\\ & =2\hat{i}+2\hat{j}+4\hat{k}\end{array}$ .......(2)
$\text{ From }(1)\mathrm{\&}(2)$
$\overrightarrow{AB}=2\overrightarrow{BC}$
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.
$\begin{array}{rl}& \Rightarrow 5\overrightarrow{a}+6\overrightarrow{b}+7\overrightarrow{c}=x(7\overrightarrow{a}-8\overrightarrow{b}+9\overrightarrow{c})+y(3\overrightarrow{a}+20\overrightarrow{b}+5\overrightarrow{c})\\ & \Rightarrow 5\overrightarrow{a}+6\overrightarrow{b}+7\overrightarrow{c}=\overrightarrow{a}(7x+3y)+\overrightarrow{b}(-8x+20y)+\overrightarrow{c}(9x+5y)\end{array}$
Comparing the equations
$\begin{array}{rl}& 5=7x+3y\\ & 7x+3y=5\end{array}$ .....(1)
$-8x+20y=6$ .......(2)
$9x+5y=7$ ........(3)
Solving equation (1) and (2) by elimination method
$\begin{array}{rl}& 7x+3y=5\end{array}$ ............$\left(1\right)\times (-8)$
$-8x+20y=6$ .............$\left(2\right)\times \left(7\right)$
$\underset{―}{\begin{array}{c}\begin{array}{r}-56x-24y=-40\\ -56x+140y=42\end{array}\end{array}}$
$$\begin{gathered} -164y=-82 \\ y=\frac{-82}{-164} \\ y=\frac{1}{2} \end{gathered}$$
Put value of y in (1)
$\begin{array}{rl}& 7x+3\left(\frac{1}{2}\right)=5\\ & 7x+\frac{3}{2}=5\\ & 7x=5-\frac{3}{2}\\ & 7x=\frac{10-3}{2}\\ & 7x=\frac{7}{2}\Rightarrow x=\frac{7}{7\times 2}\end{array}$
$\Rightarrow x=\frac{1}{2}$
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,
$\begin{array}{rl}& \Rightarrow \overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}=x(-3\overrightarrow{b}+5\overrightarrow{c})+y(-2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c})\\ & \Rightarrow \overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{a}(-2y)+\overrightarrow{b}(-3x+3y)+\overrightarrow{c}(5x-4y)\end{array}$
Comparing the terms we get,
$\begin{array}{rl}& 1=-2y\Rightarrow y=\frac{-1}{2}\\ & -2=-3x+3y\Rightarrow -2=-3x+3\left(\frac{-1}{2}\right)\\ & -2=-3x-\frac{3}{2}\\ & 3x=-\frac{3}{2}+2\\ & 3x=\frac{-3+4}{2}\end{array}$
$x=\frac{1}{6}$
$$\begin{gathered} \text{ Similarly },3=5x-4y \\ \text{ Put } \\ x=\frac{1}{6}\mathrm{\&}y=(-\frac{1}{2}) \\ \text{ R.H.S }=5x-4y \end{gathered}$$
$\begin{array}{rl}& =5\left(\frac{1}{6}\right)-4(-\frac{1}{2})\\ & =\frac{5}{6}+2\\ & =\frac{5+12}{6}\\ & =\frac{17}{6}\ne 3\end{array}$
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,
$\begin{array}{rl}& P=6\hat{i}-7\hat{j}\\ & Q=16\hat{i}-19\hat{j}-4\hat{k}\\ & R=3\hat{j}-6\hat{k}\\ & S=2\hat{i}-5\hat{j}+10\hat{k}\end{array}$
$\Rightarrow$Now, Expressing one vector as a linear combination of other two.
$\begin{array}{rl}& \overrightarrow{PQ}=(16\hat{i}-19\hat{j}-4\hat{k})-(6\hat{i}-7\hat{j}+0\hat{k})\\ & =(16-6)\hat{i}+(-19+7)\hat{j}+(-4-0)\hat{k}\\ & =10\hat{i}-12\hat{j}-4\hat{k}\\ & \overrightarrow{PR}=(0\hat{i}+3\hat{j}-6\hat{k})-(6\hat{i}-7\hat{j}+0\hat{k})\\ & =(0-6)\hat{i}+(3+7)\hat{j}+(-6-0)\hat{k}\\ & =-6\hat{i}+10\hat{j}-6\hat{k}\\ & \overrightarrow{PS}=(2\hat{i}-5\hat{j}+10\hat{k})-(6\hat{i}-7\hat{j}+0\hat{k})\\ & =(2-6)\hat{i}+(-5+7)\hat{j}+(0-0)\hat{k}\end{array}$
$\begin{array}{rl}& =-4\hat{i}+2\hat{j}+10\hat{k}\\ & \overrightarrow{PQ}=x\overrightarrow{PR}+y\overrightarrow{PS}\\ & 10\hat{i}-12\hat{j}-4\hat{k}=x(-6\hat{i}+10\hat{j}-6\hat{k})+y(-4\hat{i}+2\hat{j}+10\hat{k})\\ & 10\hat{i}-12\hat{j}-4\hat{k}=\hat{i}(-6x-4y)+\hat{j}(10x+2y)+\hat{k}(-6x+10)\end{array}$
Comparing coefficients of i, j and k on both sides
$\begin{array}{rl}& -6x-4y=10\Rightarrow -2(3x+2y)=10\Rightarrow 3x+2y=-5\dots \dots .(1)\\ & 10x+2y=-12\Rightarrow 2(5x+y)=-12\Rightarrow 5x+y=-6\dots \dots .\left(2\right)\\ & -6x+10y=-4\Rightarrow -2(3x-5y)=-4\Rightarrow 3x-5y=2\dots \dots .(3)\end{array}$
Solving (1) and (3) by elimination
$\begin{array}{rl}& \begin{array}{c}3x+2y=-5\\ 3x-5y=2\\ \frac{7y}{y}=-7\\ y=\frac{-7}{7}\Rightarrow y=-1\\ 3x-5(-1)=2\\ 3x+5=2\\ 3x=2-5\\ x=\frac{-3}{3}=-1\end{array}\end{array}$
Now, Put $x=-1$ & $y=-1$ in equation (2)
$\begin{array}{rl}& 5x+y=-6\\ & \text{ L.H }S=5(-1)+(-1)\\ & =-5-1\\ & =-6\\ & =R\cdot H.S\end{array}$
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
$\begin{array}{rl}& A=2\hat{i}-\hat{j}+\hat{k}\\ & B=\hat{i}-3\hat{j}-5\hat{k}\\ & C=3\hat{i}-4\hat{j}-4\hat{k}\\ & \text{ then, }\\ & A=xB+yC\\ & 2\hat{i}-\hat{j}+\hat{k}=x(\hat{i}-3\hat{j}-5\hat{k})+y(3\hat{i}-4\hat{j}-4\hat{k})\\ & 2\hat{i}-\hat{j}+\hat{k}=\hat{i}(x+3y)+\hat{j}(-3x-4y)+\hat{k}(-5x-4y)\end{array}$
Comparing coefficients of $\hat{i},\hat{j}$& $\hat{k}$ on both sides
$\begin{array}{rl}& x+3y=2\end{array}$ .....(1)
$-3x-4y=-1$ ....(2)
$-5x-4y=1$ ....(3)
Subtracting (2) and (3)
$-3x-4y=-1$
$$\begin{gathered} \underset{―}{ \\ -5x-4y=1} \end{gathered}$$
$2x$ $=-2$
$x=-1$
Put,$x=-1$ in (2)
$\begin{array}{rl}& -3(-1)-4y=-1\\ & 3-4y=-1\\ & -4y=-4\\ & y=1\end{array}$
Now, put x = -1 and y = 1 in (1)
$\begin{array}{rl}& \text{ L.H.S }=x+3y\\ & =(-1)+3(1)\\ & =-1+3\\ & =2\\ & =R\cdot H.S\end{array}$
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,
$\begin{array}{rl}& P=\hat{i}+\hat{j}+\hat{k}\\ & Q=2\hat{i}+3\hat{j}-\hat{k}\\ & R=-\hat{i}-2\hat{j}+2\hat{k}\\ & P=xQ+yR\\ & \hat{i}+\hat{j}+\hat{k}=x(2\hat{i}+3\hat{j}-\hat{k})+y(-\hat{i}-2\hat{j}+2\hat{k})\\ & \hat{i}+\hat{j}+\hat{k}=\hat{i}(2x-y)+\hat{j}(3x-2y)+\hat{k}(-x+2y)\end{array}$
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)
$3x-2y=1$
$\underset{―}{-x+2y=1}$
$2x$ $=2$
$x=1$
Put, x = 1 in (3)
$\begin{array}{rl}& -1+2y=1\\ & 2y=1+1\\ & y=\frac{2}{2}=1\\ & \text{ Put, }x=1\mathrm{\&}y=1in\left(1\right)\end{array}$
$L.H.S=2x-y$
$=2\left(1\right)-\left(1\right)$
$=2-1$
$=1$
$=R.H.S$
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,
$\begin{array}{rl}& P=3\hat{i}+\hat{j}-\hat{k}\\ & Q=2\hat{i}-\hat{j}+7\hat{k}\\ & R=7\hat{i}-\hat{j}+23\hat{k}\\ & \text{ then, }\end{array}$
$P\ne xQ+yR$ then they are non-coplanar
$\begin{array}{rl}& \text{ Taking }\\ & P=xQ+yR\\ & 3\hat{i}+\hat{j}-\hat{k}=x(2\hat{i}-\hat{j}+7\hat{k})+y(7\hat{i}-\hat{j}+23\hat{k})\\ & 3\hat{i}+\hat{j}-\hat{k}=\hat{i}(2x+7y)+\hat{j}(-x-y)+\hat{k}(7x+23y)\end{array}$
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
$-2x-7y=-3$
$\underset{―}{-2x-2y=2}$
$-5y=-5$
$y=1$
Put value of y = 1 in equation (2)
$\begin{array}{rl}& -x-1=1\\ & \begin{array}{c}-x=2\\ x=-2\end{array}\\ & \text{ Put }x=-2\mathrm{\&}y=\mathrm{lin}(3)\\ & 7x+23y=-1\\ & =7(-2)+23\\ & =-14+23\\ & =9\\ & \ne R\cdot H\cdot S\end{array}$
As values of x and y do not satisf the third equation
$P\ne xQ+yR$
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,
$\begin{array}{rl}& A=\hat{i}+2\hat{j}+3\hat{k}\\ & B=2\hat{i}+\hat{j}+3\hat{k}\\ & C=\hat{i}+\hat{j}+\hat{k}\\ & A=xB+yC\\ & \hat{i}+2\hat{j}+3\hat{k}=x(2\hat{i}+\hat{j}+3\hat{k})+y(\hat{i}+\hat{j}+\hat{k})\\ & \hat{i}+2\hat{j}+3\hat{k}=\hat{i}(2x+y)+\hat{j}(x+y)+\hat{k}(3x+y)\end{array}$
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)
$\begin{array}{rl}& 2x+y=1\\ & \begin{array}{rl}& x+y=2\\ & x=-1\end{array}\\ & x+y=2\\ & -1+y=2\\ & y=3\\ & \text{ Put }x=-1\mathrm{\&}y=3\text{ in }\\ & 3x+y=3\\ & \text{ Substitute }x=-1\text{ and }y=3in\left(3\right)\end{array}$
$3x+y=3$
Substitute x=-1 and y=3
$3(-1)+3=3$
$=-3+3$
$=0$
$\ne 3$
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:
$\begin{array}{rl}& A=2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}\\ & B=\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}\\ & C=\overrightarrow{a}+\overrightarrow{b}-3\overrightarrow{c}\\ & A=xB+yC\\ & 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}=x(\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c})+y(\overrightarrow{a}+\overrightarrow{b}-3\overrightarrow{c})\\ & 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{a}(x+y)+\overrightarrow{b}(x+y)+\overrightarrow{c}(-2x-3y)\end{array}$
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)
$x+y=2$
$\underset{―}{x+y=-1}$
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: $\begin{array}{rl}& \overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c},\end{array}$ $\begin{array}{rl}& 2\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c}\end{array}$ and $\begin{array}{rl}& \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\end{array}$
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
$\begin{array}{rl}& \overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=x(2\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c})+y(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\\ & \overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{a}(2x+y)+\overrightarrow{b}(x+y)+\overrightarrow{c}(3x+y)\end{array}$
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)
$2x+y=1$
$\underset{―}{x+y=2}$
$x$ $=-1$
And from (i)
$2\times (-1)+y=1$
$y=1+2$
$y=3$
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,
$\begin{array}{rl}& \overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}\\ & \overrightarrow{b}=2\hat{i}+\hat{j}+3\hat{k}\\ & \overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}\\ & \Rightarrow \overrightarrow{a}=x\overrightarrow{b}+y\overrightarrow{c}\\ & \Rightarrow \hat{i}+2\hat{j}+3\hat{k}=x(2\hat{i}+\hat{j}+3\hat{k})+y(\hat{i}+\hat{j}+\hat{k})\\ & \Rightarrow \hat{i}+2\hat{j}+3\hat{k}=\hat{i}(2x+y)+\hat{j}(x+y)+\hat{k}(3x+y)\end{array}$
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)
$2x+y=1$
$\underset{―}{x+y=2}$
$x$ $=-1$
$x+y=2$
$-1+y=2$
$y=3$
Put, x=-1 and y=3
$3x+y=3$
$L.H.S=3x+y$
$=3(-1)+3$
$=-3+3$
$=0$
$\ne R.H.S$
Thus, $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ are non-coplanar
Now, expressing $\overrightarrow{a}$ as a linear combination of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$
$\begin{array}{rl}& \overrightarrow{d}=\overrightarrow{a}x+\overrightarrow{b}y+\overrightarrow{c}z\\ & \Rightarrow 2\hat{i}-\hat{j}-3\hat{k}=x(\hat{i}+2\hat{j}+3\hat{k})+y(2\hat{i}+\hat{j}+3\hat{k})+z(\hat{i}+\hat{j}+\hat{k})\\ & \Rightarrow 2\hat{i}-\hat{j}-3\hat{k}=\hat{i}(x+2y+z)+\hat{j}(2x+y+z)+\hat{k}(3x+3y+z)\end{array}$
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)
$x+2y+z=2$
$\underset{―}{2x+y+z=-1}$
$-x+y$ $=3$
Subtracting (2) and (3)
$2x+y+z=-1$
$\underset{―}{3x+3y+z=-3}$
$-x-2y$ $=2$
$x+2y=-2$ .......(5)
Subtracting (4) and (5)
$x-y=-3$
$\underset{―}{x+2y=-2}$
$-3y=-1$
$y=\frac{1}{3}$
Put $y=\frac{1}{3}$ in (4)
$\begin{array}{rl}& x-\left(\frac{1}{3}\right)=-3\\ & x=-3+\frac{1}{3}\\ & x=\frac{-8}{3}\end{array}$
Put $x=\frac{-8}{3}$ & $y=\frac{1}{3}$ in eq (1)
$\begin{array}{rl}& x+2y+z=2\\ & \frac{-8}{3}+2\left(\frac{1}{3}\right)+z=2\\ & \frac{-6}{3}+z=2\\ & z=2+2\\ & z=4\end{array}$
Hence, $\overrightarrow{d}$ is expressible as linear combination of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$.