Let the volume of a parallelepiped whose cote | vedclass

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Question

(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.

Similar Questions

If the vectors $(1 - x)\hat i + \hat j + \hat k$,$\hat i + (1 - y)\hat j + \hat k$,and $\hat i + \hat j + (1 - z)\hat k$ are coplanar,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is $(x, y, z \neq 0)$.

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

If the volume of a parallelepiped with coterminous edges $4 \hat{i} + 5 \hat{j} + \hat{k}$, $-\hat{j} + \hat{k}$, and $3 \hat{i} + 9 \hat{j} + p \hat{k}$ is $34$ cubic units, then $p$ is equal to:

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