If two diagonals of one of its faces are ` 6hati + 6 hatk and 4 hatj + 2hatk` and of the edges not containing the given diagonals is `vecc= 4 hatj - 8 hatk` , then the volume of a parallelpiped is

Options

A. 60
B. 80
C. 100
D. 120

Answer

d

Explanation

Let `veca = 6hati + 6 hatk, vecb = 4 hatj + 2hatk ,vecc = 4hatj - 8hatk`
then `veca xx vecb = -24hati -12 hatj + 24hatk`
` 12 ( - 2hati - hatj + 2hatk)`
Area of the base of the parallelepiped
` = 1/2 |veca xx vecb|`
` 1/2 (12xx 3)`
18
Height of the parallelepiped
= Length of projection of `vecc on veca xx vecb`
` (|vecc.vecaxxvecb|)/(|veca xx vecb|)`
`(|12(-4-16)|)/36`
` 20/3`
Volume of the parallelpiped =` 18xx 20/3 = 120`