1.

Three point A, B and C with position vectors a_(1)=3hati-2hatj-hatk, a_(2)=hati+3hatj+4hatk and a_(3)=2hati+hatj-2hatk relative to an origin O. The distance of A from the plane OBC is (magnitude)

Answer»

5
`sqrt3`
3
`2sqrt3`

Solution :Equation of PLANE `OBC= a_(1) = ALPHA a_(2) +beta a_(3)`,
i.e., `a_(1)*(a_(2)xxa_(3))=0`
The distanceof A from the plane `OBC = |(a_(1)*(a_(2)xxa_(3)))/(|a_(2)xxa_(3)|)|`
`a_(2)xxa_(3) = -10hati + 10 hatj - 5hatk`
`rArr| a_(2) xx a_(3)|=15`
`THEREFORE ` DISTANCE ` = (1)/(15)||(3,-2,-1),(1,3,4),(2,1,-2)||=3`


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