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Distance Formula

Three vectors...

Question

Three vectors $a = 12, 4, 3; b = 8, - 12, - 9; c = 33, - 4, - 24$ define a parallelopiped. Evaluate the area of its faces and its volume respectively

A

209, 3675

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B

214, 3788

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C

220, 3696

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D

342, 4022

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Solution

The correct option is D $220, 3696$
Let the three coterminous edges OA, OB, OC be represented by the vectorsa, b and crespectively. Then
$\to O A = 12 i + 4 j + 3 k, \to O B = 8 i - 12 j - 9 k$ and $\to O C = 33 i - 4 j - 24 k .$
$∴ O A = \sqrt{12^{2} + 4^{2} + 3^{2}} = 13,$ $O B = \sqrt{8^{2} + {(- 12)}^{2} + {(- 9)}^{2}} = 17$ and
$O C = \sqrt{33^{2} + (- 4^{2}) + {(- 24)}^{2}} = 41.$
Hence lengths of the edges of the parallelopiped
are 13, 17 and 41.Now vectors area of the face determined by OA and OB
is given by
$∣ ∣ ∣ ∣ \begin{matrix} i & j & k 12 & 4 & 3 8 & - 12 & - 9 \end{matrix} ∣ ∣ ∣ ∣ = 0 i + 132 j - 176 k .$ Hence area of the face determined by OA and
OB. $= \sqrt{132^{2} + 176^{2}} = 44 \sqrt{3^{2} + 4^{2}} = 220.$
Similarly areas of other two faces can be found. This is left for the reader.
Volume
$= | [a b c] | = ∣ ∣ ∣ ∣ \begin{matrix} 12 & 4 & 3 8 & - 12 & - 9 33 & - 4 & - 24 \end{matrix} ∣ ∣ ∣ ∣ = 12 ∣ ∣ ∣ ∣ \begin{matrix} 12 & 1 & 1 8 & - 3 & - 3 33 & - 1 & - 8 \end{matrix} ∣ ∣ ∣ ∣$
$= 12 ∣ ∣ ∣ ∣ \begin{matrix} 12 & 1 & 1 44 & 0 & 0 33 & - 1 & - 8 \end{matrix} ∣ ∣ ∣ ∣ = - 12 \times 44 (- 8 + 1)$
$= 12 \times 44 \times 7 = 3696.$

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