$\bar{a}=\hat{i}+j+\hat{k}, \bar{b}=\hat{i}-j+2 \hat{k}$ and $\bar{c}=x|+(x-1)\rangle-\hat{k}$. If the vector $\bar{c}$ lies in the plane of $\bar{u}$ and $\bar{b}$, then $x=$

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Aicharya · Accepted Answer

Given vectors are coplanar, we write : $\bar{a} \cdot(\bar{b} 	imes \bar{c})=0$ $\begin{aligned} &\left|\begin{array}{ccc} 1 & 1 & 1 \ 1 & -1 & 2 \ x & x-1 & -1 \end{array}ight|=0 \ 	herefore & 1(1-2 x+2)-1(-1-2 x)+1(x-1+x)=0 \ & 3-2 x+1+2 x+2 x-1=0 \Rightarrow 2 x+3=0 \Rightarrow x=\frac{-3}{2} \end{aligned}$

$\bar{a}=\hat{i}+j+\hat{k}, \bar{b}=\hat{i}-j+2 \hat{k}$ and $\bar{c}=x|+(x-1)\rangle-\hat{k}$. If the…

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