Let $\vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}$ and $\vec{c}=2 \hat{i}+4 \hat{j}+\left(\lambda^{2}-1 ight) \hat{k}$ be coplanar vectors. Then the non-zero vector $\vec{a} imes \vec{c}$ is:

Question

Let $\vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}$ and $\vec{c}=2 \hat{i}+4 \hat{j}+\left(\lambda^{2}-1ight) \hat{k}$ be coplanar vectors. Then the non-zero vector $\vec{a} 	imes \vec{c}$ is:

Aicharya · Accepted Answer

$\because \bar{a}, \bar{b}$ and $\bar{c}$ are coplanar $	herefore\left|\begin{array}{ccc}1 & 2 & 4 \ 1 & \lambda & 4 \ 2 & 4 & \left(\lambda^{2}-1ight)\end{array}ight|=0$ $\Rightarrow \quad \lambda^{3}-\lambda-16+2\left(8-\lambda^{2}+1ight)+4(4-2 \lambda)=0$ $\Rightarrow \quad \lambda^{3}-2 \lambda^{2}-9 \lambda+18=0$ i.e. $\quad(\lambda-2)(\lambda-3)(\lambda+3)=0$ For $\lambda=2, \vec{c}=2 \hat{i}+4 \hat{j}+3 \hat{k}$ $	herefore \quad \vec{a} 	imes \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 4 \ 2 & 4 & 3\end{array}ight|=-10 \hat{i}+5 \hat{j}$ For $\lambda=3$ or $-3, \bar{c}=2 \bar{a} \Rightarrow \bar{a} 	imes \bar{c}=0$ (Rejected)

Let $\vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}$ and $\vec{c}=2…

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