The area of the triangle having vertices as `hati-2hatj+3hatk,-2hati+3hatj-hatk,4hati-7hatj+7hatk` is

Options

A. 36 sq unit
B. 0 sq unit
C. 39 sq unit
D. 11 sq unit

Answer

B

Explanation

Let `A=hati-2hatj+3hatk,B=-2hati+3hatj-hatk`
and `C=4hati-7hatj+7hatk`
`thereforeAB=-2hati+3hatj-hatk-hati+2hatj-3hatk`
`=-3hati+5hatj-4hatk`
and `AC=4hati-7hatj+7hatk-hati+2hatj-3hatk` ,brgt `=3hati-5hatj+4hatk`
Hence, area of
`DeltaABC=(1)/(2)|ABxxAC|`
`=(1)/(2)|(hati,hatj,hatk),(-3,5,-4),(3,-5,4)|`
`=-(1)/(2)|(hati,hatj,hatk),(3,-5,4),(3,-5,4)|`
`=0` (`because` two rows are identical)