If `veca,vecb,vecc` be unit vectors such that `veca.vecb=veca.vecc=0` and the angle between `vecb` and `vecc` is `pi//6`. Prove that `veca=pm2(vecbxxvecc)`

Answer

1

Explanation

`vecA, vecB and vecC` are three unit vectors such that
`vecA. vecB = vecA. vecC=0`
and the angle between `vecB and vecC " is " pi//3`
Now Eq, (i) show that `vecA` is perpendicular to both `vecB and vecC`. Thus
` vecB xx vecC = lambda vecA` . where `lambda` is any scalar.
`or |vecB xx vecC|=|lamda vecA|`
` or sin pi//3 = +- lambda`
( as `pi//3` is the angle between ` vecB and vecC`)
`or lambda = +- sqrt3//2`
` Rightarrow vecB xx vecC = +- sqrt3/2 vecA`
` or vecA =+- 2/sqrt3 (vecB xx vecC)`
There, the given statement is false.