# Let the volume of a parallelopiped

Question:

Let the volume of a parallelopiped whose coterminous edges are given by $\vec{u}=\hat{i}+\hat{j}+\lambda \hat{k}, \vec{v}=\hat{i}+\hat{j}+3 \hat{k}$ and

$\vec{w}=2 \hat{i}+\hat{j}+\hat{k}$ be 1 cu. unit. If $\theta$ be the angle between the edges $\vec{u}$ and $\vec{w}$, then $\operatorname{cost} \theta$ can be:

1. $\frac{7}{6 \sqrt{6}}$

2. $\frac{7}{6 \sqrt{3}}$

3. $\frac{5}{7}$

4. $\frac{5}{3 \sqrt{3}}$

Correct Option: , 2

Solution:

It is given that $\vec{u}=\hat{i}+\hat{j}+\lambda \hat{k}, \quad \vec{v}=\hat{i}+\hat{j}+3 \hat{k}$ and

$w=2 \hat{i}+\hat{j}+\hat{k}$

Volume of parallelopiped $=[\vec{u} \cdot \vec{v} \cdot \vec{w}]$

$\Rightarrow \pm 1=\left|\begin{array}{lll}1 & 1 & \lambda \\ 1 & 1 & 3 \\ 2 & 1 & 1\end{array}\right| \Rightarrow-\lambda+3=\pm 1 \Rightarrow \lambda=2$ or $\lambda=4$

For $\lambda=2$

$\cos \theta=\frac{2+1+2}{\sqrt{6} \sqrt{6}}=\frac{5}{6}$

For $\lambda=4$

$\cos \theta=\frac{2+1+4}{\sqrt{6} \sqrt{18}}=\frac{7}{6 \sqrt{3}}$