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(B) The volume $V$ of a parallelepiped with coterminous edges $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the scalar trip

Vedclass · Accepted Answer

$\overrightarrow{b} \cdot \overrightarrow{c} = 10$

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(B) The volume $V$ of a parallelepiped with coterminous edges $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the scalar triple product $|[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]|$.$V = |\det(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c})| = 158$$\det(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}) = \begin{vmatrix} 1 & 1 & n \ 2 & 4 & -n \ 1 & n & 3 \end{vmatrix} = 1(12 + n^2) - 1(6 + n) + n(2n - 4)$$= 12 + n^2 - 6 - n + 2n^2 - 4n = 3n^2 - 5n + 6$Since $V = 158$,we have $|3n^2 - 5n + 6| = 158$. Given $n \geq 0$,$3n^2 - 5n + 6 = 158$ or $3n^2 - 5n + 6 = -158$.Case $1$: $3n^2 - 5n - 152 = 0$. Solving for $n$: $n = \frac{5 \pm \sqrt{25 - 4(3)(-152)}}{6} = \frac{5 \pm \sqrt{25 + 1824}}{6} = \frac{5 \pm \sqrt{1849}}{6} = \frac{5 \pm 43}{6}$.Since $n \geq 0$,$n = \frac{48}{6} = 8$.Case $2$: $3n^2 - 5n + 164 = 0$. The discriminant $D = 25 - 4(3)(164) Thus,$n = 8$.Now,check the options:$A$. $\overrightarrow{a} \cdot \overrightarrow{c} = (1)(1) + (1)(n) + (n)(3) = 1 + 4n = 1 + 4(8) = 33$.$B$. $\overrightarrow{b} \cdot \overrightarrow{c} = (2)(1) + (4)(n) + (-n)(3) = 2 + n = 2 + 8 = 10$.Therefore,option $B$ is correct.

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