If the volume of the parallelepiped is 158 te | vedclass

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(A) The volume of a parallelepiped with coterminous edges $\bar{a}, \bar{b}, \bar{c}$ is given by the scalar triple product $|[\bar{a} \bar{b} \bar{c}

Vedclass · Accepted Answer

$8$

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(A) The volume of a parallelepiped with coterminous edges $\bar{a}, \bar{b}, \bar{c}$ is given by the scalar triple product $|[\bar{a} \bar{b} \bar{c}]|$.Given,volume $= 158$.$|\bar{a} \cdot (\bar{b} 	imes \bar{c})| = \left|\begin{vmatrix} 1 & 1 & n \ 2 & 4 & -n \ 1 & n & 3 \end{vmatrix}ight| = 158$.Expanding the determinant:$1(12 - (-n^2)) - 1(6 - (-n)) + n(2n - 4) = \pm 158$.$(12 + n^2) - (6 + n) + (2n^2 - 4n) = \pm 158$.$3n^2 - 5n + 6 = \pm 158$.Case $1$: $3n^2 - 5n + 6 = 158 \Rightarrow 3n^2 - 5n - 152 = 0$.Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{5 \pm \sqrt{25 - 4(3)(-152)}}{6} = \frac{5 \pm \sqrt{25 + 1824}}{6} = \frac{5 \pm \sqrt{1849}}{6} = \frac{5 \pm 43}{6}$.Since $n \geq 0$,$n = \frac{48}{6} = 8$.Case $2$: $3n^2 - 5n + 6 = -158 \Rightarrow 3n^2 - 5n + 164 = 0$.The discriminant $D = 25 - 4(3)(164) Thus,$n = 8$.

If the volume of the parallelepiped is $158 \text{ cubic units}$,whose coterminous edges are given by the vectors $\bar{a} = (\hat{i} + \hat{j} + n \hat{k})$,$\bar{b} = (2 \hat{i} + 4 \hat{j} - n \hat{k})$,and $\bar{c} = (\hat{i} + n \hat{j} + 3 \hat{k})$,where $n \geq 0$,then the value of $n$ is:

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