1.

Vectors a and b are such that |a|=1,|b|=4 and a.b=2." If " c=2axxb-3b, then the angle between b and c is

Answer»

`(pi)/(6)`
`(5PI)/(6)`
`(pi)/(3)`
`(2pi)/(3)`

Solution :Now , `|axxb|^(2)=|a|^(2)|b|^(2)-(a*b)^(2)=16-4=12`
` and |c|^(2)=(2axxb-3b)*(2axxb-3b)`
`=4|axxb|^(2)+9|b|^(2)=4*12+9*16=192`
`rArr |c|=8sqrt(3)`
Again now `b*c = b*(2axxb-3b)2(b*axxb)-3|b|^(2)`
`=-3|b|^(2)=-48`
`therefore cos theta=(b*c)/(|b||c|)=-(48)/(4*8sqrt(3))=-(SQRT(3))/(2)`
`therefore theta = (5pi)/(6)`


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