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(A) The volume of a parallelepiped is given by the absolute value of the scalar triple product $|[\overrightarrow{u}, \overrightarrow{v}, \overrightar

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The volume of a parallelepiped is given by the absolute value of the scalar triple product $|[\overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w}]| = 1$.$|\overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w}| = \begin{vmatrix} 1 & 1 & \lambda \ 1 & 1 & 3 \ 2 & 1 & 1 \end{vmatrix} = 1(1-3) - 1(1-6) + \lambda(1-2) = -2 + 5 - \lambda = 3 - \lambda$.Since the volume is $1$,$|3 - \lambda| = 1$,which gives $3 - \lambda = 1$ or $3 - \lambda = -1$.Thus,$\lambda = 2$ or $\lambda = 4$.Case $1$: If $\lambda = 2$,then $\overrightarrow{u} = \hat{i} + \hat{j} + 2\hat{k}$ and $\overrightarrow{w} = 2\hat{i} + \hat{j} + \hat{k}$.$\overrightarrow{u} \cdot \overrightarrow{w} = (1)(2) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5$.$|\overrightarrow{u}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$ and $|\overrightarrow{w}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$.$\cos 	heta = \frac{\overrightarrow{u} \cdot \overrightarrow{w}}{|\overrightarrow{u}||\overrightarrow{w}|} = \frac{5}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6}$.Case $2$: If $\lambda = 4$,then $\overrightarrow{u} = \hat{i} + \hat{j} + 4\hat{k}$ and $\overrightarrow{w} = 2\hat{i} + \hat{j} + \hat{k}$.$\overrightarrow{u} \cdot \overrightarrow{w} = (1)(2) + (1)(1) + (4)(1) = 2 + 1 + 4 = 7$.$|\overrightarrow{u}| = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{18} = 3\sqrt{2}$ and $|\overrightarrow{w}| = \sqrt{6}$.$\cos 	heta = \frac{7}{\sqrt{18} \cdot \sqrt{6}} = \frac{7}{3\sqrt{2} \cdot \sqrt{6}} = \frac{7}{3\sqrt{12}} = \frac{7}{3 \cdot 2\sqrt{3}} = \frac{7}{6\sqrt{3}}$.Comparing with the options,$\frac{7}{6\sqrt{3}}$ is present.

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