Let a,b,c be three vectors such that `a ne 0` and `a xx b = 2a xx c,|a| = |c| = 1, |b| = 4 and |b xx c| = sqrt(15)`. If `b - 2 c = lambda a,` then `lambda` is equal ot

Options

A. 1
B. `+4`
C. 3
D. `-2`

Answer

B

Explanation

Given that `|a| = |c| = 1` and `|b| = 4`
Let angle between b and c is `alpha` then
`|b xx c| = sqrt(15)`
`rArr " "|b||c| sin alpha = sqrt(15)`
`rArr sin alpha (sqrt(15))/(4 xx1) = (sqrt(15))/(4)`
`therefore cos alpha = sqrt(1-sin^(2) alpha) =(1)/(4)`
We have , `b - 2c = lambda a`
On squaring both sides, we get
`(b-2c)^(2) = lambda^(2)(a)^(2)`
`rArr " " b^(2) + 4c^(2) - 4bc = lambda^(2)a^(2)`
`rArr 16+4 - 4 |b||c| cos alpha = lambda^(2)`
`rArr 16 + 4-4 xx 4 xx 1 xx (1)/(4) = lambda^(2)`
`rArr" "lambda^(2) = 16`
`rArr" "lambda = pm 4`