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Question: Answered & Verified by Expert
If $2 \mathbf{i}+3 \mathbf{j}, \mathbf{i}+\mathbf{j}+\mathbf{k}$ and $\lambda \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$ taken in an order are coterminous edges of a parallelopiped of volume $2 \mathrm{cu}$ units, then value of $\lambda$ is
MathematicsVector AlgebraKCETKCET 2011
Options:
  • A $-4$
  • B 2
  • C 3
  • D 4
Solution:
2887 Upvotes Verified Answer
The correct answer is: 4
Given, let $A=2 \mathbf{i}+3 \mathbf{j}+0 \mathbf{k}$
$B=\mathbf{i}+\mathbf{j}+\mathbf{k}$
$C=\lambda i+4 j+2 k$
If $A, B, C$ are the coterminous edges of a parallelopiped then its volume is
$$
=(A \times B) \cdot C
$$
$$
=\{(2 \mathbf{i}+3 \mathbf{j}+0 \mathbf{k}) \times(\mathbf{i}+\mathbf{j}+\mathbf{k})\}
$$
$(\lambda \mathbf{i}+4 \mathbf{j}+2 \mathbf{k})$
$=(3 \mathbf{i}-2 \mathbf{j}-\mathbf{k}) \cdot(\lambda \mathbf{i}+4 \mathbf{j}+2 \mathbf{k})$
$=3 \lambda-8-2$
$=3 \lambda-10$
Given volume $=2$
$$
\begin{aligned}
& \Rightarrow & 3 \lambda-10 &=2 \\
\Rightarrow & & 3 \lambda &=12 \\
\Rightarrow & \lambda &=4
\end{aligned}
$$

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