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Lines whose equation are (x-3)/2=(y-2)/3...

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
`vec(n)=|(hati, hatj, hatk),(2, 3, -1),(3, 2, 3)|=1hat(i)+6hatj-5hat(k)`
So angle between planes `costheta=(4+6-10)/(sqrt(21)sqrt(62))=0impliestheta=(pi)/2`
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