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). 

Answer 1

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.