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Lines whose equation are `(x-3)/2=(y-2)/3=(z-1)/(lamda)` and `(x-2)/3=(y-3)/2=(z-2)/3` lie in same plane, then.
Angle between the plane containing both lines and the plane `4x+y+2z=0` is

A

`(pI)/(3)`

B

`(pi)/(2)`

C

`(pi)/(6)`

D

`cos^(-1)((1)/(sqrt(186)))`

Text Solution

Verified by Experts

The correct Answer is:
B
normal vector to plane containing lines is
`vecn=|{:(hati,hatj,hatk),(2,3,-1),(3,2,3):}|=hati+6hatj+5hatk`
So angle between planes
`costheta=(4+6-10)/(sqrt(21)sqrt(62))=0impliestheta=(pi)/(2)`
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