The volume of a parallelepiped,whose cotermin | vedclass

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(B) The volume of a parallelepiped is given by the scalar triple product $|[\bar{u} \bar{v} \bar{w}]| = 1$. Calculating the determinant: $\begin{vmatr

Vedclass · Accepted Answer

$\frac{5}{6}$

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(B) The volume of a parallelepiped is given by the scalar triple product $|[\bar{u} \bar{v} \bar{w}]| = 1$. Calculating the determinant: $\begin{vmatrix} 1 & 1 & \lambda \ 1 & 1 & 3 \ 2 & 1 & 1 \end{vmatrix} = \pm 1$. Expanding along the first row: $1(1-3) - 1(1-6) + \lambda(1-2) = \pm 1$ $-2 + 5 - \lambda = \pm 1$ $3 - \lambda = \pm 1$. Case $1$: $3 - \lambda = 1 \Rightarrow \lambda = 2$. Case $2$: $3 - \lambda = -1 \Rightarrow \lambda = 4$. Given $\bar{u} = \hat{i} + \hat{j} + \lambda \hat{k}$ and $\bar{w} = 2\hat{i} + \hat{j} + \hat{k}$. For $\lambda = 2$: $\bar{u} = \hat{i} + \hat{j} + 2\hat{k}$. $\cos 	heta = \frac{\bar{u} \cdot \bar{w}}{|\bar{u}| |\bar{w}|} = \frac{(1)(2) + (1)(1) + (2)(1)}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+1^2+1^2}} = \frac{2+1+2}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6}$.

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

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