Let $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}+\lambda \hat{k}$ be the co-terminal edges
of a parallelopiped whose volume is 5 units. Then the value of $\lambda$ is
🎥 Video solution / Text Solution of this question is given below:
Given
$\vec a = \langle 2,-3,4\rangle,\; \vec b = \langle 1,2,-1\rangle,\; \vec c = \langle 3,1,\lambda\rangle$
Volume condition:
$|\vec a \cdot (\vec b \times \vec c)| = 5$
Compute cross product using determinant:
\[
\vec b \times \vec c
=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & -1 \\
3 & 1 & \lambda
\end{vmatrix}
\]
\[
= (2\lambda + 1)\mathbf{i}
- (\lambda + 3)\mathbf{j}
- 5\mathbf{k}
\]
So
$\vec b \times \vec c = \langle 2\lambda + 1,\; -(\lambda + 3),\; -5\rangle$
Scalar triple product:
\[
\vec a \cdot (\vec b \times \vec c)
= 2(2\lambda + 1) + (-3)(-(\lambda + 3)) + 4(-5)
\]
Simplifying:
\[
= 4\lambda + 2 + 3(\lambda + 3) - 20
= 7\lambda - 9
\]
Volume equation:
\[
|7\lambda - 9| = 5
\]
Solve:
\[
7\lambda - 9 = 5 \Rightarrow \lambda = 2
\]
\[
7\lambda - 9 = -5 \Rightarrow \lambda = \frac{4}{7}
\]
Final answer:
\[
\boxed{\lambda = 2 \text{ or } \frac{4}{7}}
\]