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Let $\vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}$ and
$\vec{c}=2 \hat{i}+4 \hat{j}+\left(\lambda^{2}-1\right) \hat{k}$ be coplanar vectors. Then the
non-zero vector $\vec{a} \times \vec{c}$ is:
Options:
$\vec{c}=2 \hat{i}+4 \hat{j}+\left(\lambda^{2}-1\right) \hat{k}$ be coplanar vectors. Then the
non-zero vector $\vec{a} \times \vec{c}$ is:
Solution:
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Verified Answer
The correct answer is:
$-10 \hat{i}+5 \hat{j}$
$\because \bar{a}, \bar{b}$ and $\bar{c}$ are coplanar
$\therefore\left|\begin{array}{ccc}1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & \left(\lambda^{2}-1\right)\end{array}\right|=0$
$\Rightarrow \quad \lambda^{3}-\lambda-16+2\left(8-\lambda^{2}+1\right)+4(4-2 \lambda)=0$
$\Rightarrow \quad \lambda^{3}-2 \lambda^{2}-9 \lambda+18=0$
i.e. $\quad(\lambda-2)(\lambda-3)(\lambda+3)=0$
For $\lambda=2, \vec{c}=2 \hat{i}+4 \hat{j}+3 \hat{k}$
$\therefore \quad \vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3\end{array}\right|=-10 \hat{i}+5 \hat{j}$
For $\lambda=3$ or $-3, \bar{c}=2 \bar{a} \Rightarrow \bar{a} \times \bar{c}=0$ (Rejected)
$\therefore\left|\begin{array}{ccc}1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & \left(\lambda^{2}-1\right)\end{array}\right|=0$
$\Rightarrow \quad \lambda^{3}-\lambda-16+2\left(8-\lambda^{2}+1\right)+4(4-2 \lambda)=0$
$\Rightarrow \quad \lambda^{3}-2 \lambda^{2}-9 \lambda+18=0$
i.e. $\quad(\lambda-2)(\lambda-3)(\lambda+3)=0$
For $\lambda=2, \vec{c}=2 \hat{i}+4 \hat{j}+3 \hat{k}$
$\therefore \quad \vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3\end{array}\right|=-10 \hat{i}+5 \hat{j}$
For $\lambda=3$ or $-3, \bar{c}=2 \bar{a} \Rightarrow \bar{a} \times \bar{c}=0$ (Rejected)
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