First slide

Scalar triple product of vectors

Question

If the two diagonals of one of its faces are $6 \hat{i} + 6 \hat{k} and 4 \hat{j} + 2 \hat{k}$ and of the edges not containing the given diagonals $\vec{c} = 4 \hat{j} - 8 \hat{k}$ then the volume of a parallelepiped is

Moderate

A

60
B

80
C

100
D

120

Solution

$Let \vec{a} = 6 \hat{i} + 6 \hat{k}, \vec{b} = 4 \hat{j} + 2 \hat{k}, \vec{c} = 4 \hat{j} - 8 \hat{k}$
$Then \vec{a} \times \vec{b} = - 24 \hat{i} - 12 \hat{j} + 24 \hat{k}$
$= 12 (- 2 \hat{i} - \hat{j} + 2 \hat{k})$
Area of the base of the parallelepiped
$\begin{matrix} = \frac{1}{2} | \vec{a} \times \vec{b} | \\ = \frac{1}{2} (12 \times 3) \\ = 18 \end{matrix}$
Height of the parallelepiped = Length of projection of $\vec{c} on \vec{a} \times \vec{b}$
$\begin{array}{l} = \frac{| \vec{c} \cdot \vec{a} \times \vec{b} |}{| \vec{a} \times \vec{b} |} \\ = \frac{| 12 (- 4 - 16) |}{36} \\ = \frac{20}{3} \end{array}$
$∴ Volume of the parallelepiped = 18 \times \frac{20}{3} = 120$

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