If $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1 \hat{k})$ are coplanar vectors, then $\lambda$ is equal to

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Aicharya · Accepted Answer

Since $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1 \hat{k})$ are coplanar therefore $[\vec{a} \vec{b} \vec{c}]=0$ $ \begin{aligned} & 	ext { i.e., }\left|\begin{array}{ccc} 1 & 2 & \lambda \ -2 & 3 & 1 \ 3 & -1 & 2 \lambda-1 \end{array}ight|=0 \ & \Rightarrow 1(6 \lambda-2)-2(-4 \lambda-1)+\lambda(-7)=0 \ & \Rightarrow(6 \lambda-2)+8 \lambda+2+2+2 \lambda-9 \lambda=0 \ & \Rightarrow 7 \lambda=0 \Rightarrow \lambda=0 \end{aligned} $

If $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=\lambda…

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