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  • Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 5 sub question 1 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22 point 8 question 5 sub question 1 maths textbook solution

Answers (1)

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

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