The volume of the tetrahedron having the edge | vedclass

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(A) Let the coterminous edges be $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, and $\vec{c} = \hat{i} - \hat{j} +

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$1$

Vedclass · Answer

(A) Let the coterminous edges be $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, and $\vec{c} = \hat{i} - \hat{j} + \lambda\hat{k}$.The volume of a tetrahedron is given by $V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Given $V = \frac{2}{3}$, we have $\frac{2}{3} = \frac{1}{6} |\vec{a} \cdot (\vec{b} 	imes \vec{c})|$.Calculating the scalar triple product:$|\vec{a} \vec{b} \vec{c}| = \begin{vmatrix} 1 & 2 & -1 \ 1 & 1 & 1 \ 1 & -1 & \lambda \end{vmatrix} = 1(\lambda + 1) - 2(\lambda - 1) - 1(-1 - 1) = \lambda + 1 - 2\lambda + 2 + 2 = 5 - \lambda$.Thus, $\frac{2}{3} = \frac{1}{6} |5 - \lambda|$.$4 = |5 - \lambda|$.This implies $5 - \lambda = 4$ or $5 - \lambda = -4$.If $5 - \lambda = 4$, then $\lambda = 1$.If $5 - \lambda = -4$, then $\lambda = 9$.Since the options provided only include $1$, the correct value is $\lambda = 1$.

The volume of the tetrahedron having the edges $\hat{i}+2\hat{j}-\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{i}-\hat{j}+\lambda\hat{k}$ as coterminous edges is $\frac{2}{3}$ cubic units. Then $\lambda$ equals:

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