Let the volume of a parallelepiped with coterminous edges u= i+ j+ k, v= i+ j+3 k, w=2 i+ …

Question

Let the volume of a parallelepiped with coterminous edges $\vec u=\hat i+\hat j+\lambda\hat k$, $\vec v=\hat i+\hat j+3\hat k$, $\vec w=2\hat i+\hat j+\hat k$ be 1 cu. unit. If $	heta$ is the angle between $\vec u,\vec w$, $\cos	heta$ can be:

Accepted Answer

$$[\vec u\ \vec v\ \vec w] = \begin{vmatrix}1&1&\lambda\1&1&3\2&1&1\end{vmatrix} = \pm1 \implies -\lambda+3=\pm1 \implies \lambda=2\text{ or }4$$

For $\lambda=4$: $|\vec u|=\sqrt{18}$, $|\vec w|=\sqrt6$, $\vec u\cdot\vec w=2+1+4=7$:
$$\cos\theta = \frac7{\sqrt6\sqrt{18}} = \frac7{6\sqrt3}$$

(For $\lambda=2$, $\cos\theta=5/6$, not among the options --- the $\lambda=4$ branch matches option B.)

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