Question

If the volume of parallelepiped whose coterminous edges are given by $\overrightarrow{u}=\hat{i}+\hat{j}+\lambda \hat{k},\overrightarrow{v}=\hat{i}+\hat{j}+3\hat{k},\overrightarrow{w}=2\hat{i}+\hat{j}+\hat{k}$ be $1$ cubic unit. If $\theta$ be the angle between the edges $\overrightarrow{u},\overrightarrow{w}$ then, $\cos \theta$ can be:

• A. $\frac{7}{6\sqrt{6}}$
• B. $\frac{5}{7}$
• C. $\frac{7}{6\sqrt{3}}$
• D. $\frac{5}{3\sqrt{3}}$

Answer 1

The correct option is C

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

Explanation for the correct answer:

Find the value of $\cos \theta$:

Given, that the volume of the parallelepiped is $1$ cubic unit.

We know, that the volume of the parallelepiped is given by $\left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]$

So,

$\begin{array}{rcl}\left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]& =& \pm 1\\ & \Rightarrow & \left|\begin{array}{ccc}1& 1& \lambda \\ 1& 1& 3\\ 2& 1& 1\end{array}\right|=\pm 1\\ & \Rightarrow & 1\left(1-3\right)-1\left(1-6\right)+\lambda \left(1-2\right)=\pm 1\\ & \Rightarrow & -2+5-\lambda =\pm 1\\ & \Rightarrow & \lambda =2,4\end{array}$

at $\lambda =4$

$\begin{array}{rcl}\cos \theta & =& \frac{\overrightarrow{u}.\overrightarrow{w}}{\left|u\right|\left|w\right|}\\ & =& \frac{2+1+4}{\left[\sqrt{1+1+16}\right]\left[\sqrt{4+1+1}\right]}\\ & =& \frac{7}{\sqrt{18}\sqrt{6}}\\ & =& \frac{7}{6\sqrt{3}}\end{array}$

Hence, the correct option is C.