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  • Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 3 sub question 2 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 3 sub question 2 maths textbook solution

Answers (1)

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.

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