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Question
Find the volume of the parallelopiped whose vertices are A (3, 2, −1), B (−2, 2, −3) C (3, 5, −2) and D (−2, 5, 4).
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Solution
A = (3, 2, −1),
B (−2, 2, −3),
C (3, 5, −2),
D (−2, 5, 4)
Let `bar a, bar b, bar c, bar d` be the position vectors of points A, B, C and D respectively.
`bar(AB) = bar b - bar a = -5hati - 2k`
`bar(AC) = bar c - bar a = 3hat j - hat k`
`bar(AD)= bar d - bar a = -5hat i + 3hat j + 5hat k`
Volume of parallelopiped = `[bar (AB) bar(AC) bar(AD)]`
= `[(-5, 0, -2), (0, 3, -1), (-5, 3, 5)]`
= −5(15 + 3) −2(0 + 15)
= −5(18) −2(15)
= −90 − 30
= −120
∴ Volume of parallelopiped = 120 cu. units.
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