Let `vec a ` be vector parallel to line of intersection of planes `P_1 and P_2` through origin. If `P_1`is parallel to the vectors `2 bar j + 3 bar k and 4 bar j - 3 bar k` and `P_2` is parallel to `bar j - bar k` and ` 3 bar I + 3 bar j `, then the angle between `vec a` and `2 bar i +bar j - 2 bar k` is :

Options

A. `pi//2`
B. `pi//4`
C. `pi//6`
D. `3pi//4`

Answer

b,d

Explanation

Normal to plane `P_(1)` is
`vecn_(1)= (2hatj+3hatk)xx)(4hatj-3hatk)=-18hati`
Normal to plane `P_(2)` is
Therefore, `vecn_(2)= (hatj-hatk)xx (3hati +3hatj)=3hati -3hatj-3hatk`
`vecA` is parallel to `+-(vecn_(1)xx vecn_(2))=+- (-54hatj+54hatk)`
Now , the angle between `vecA nad 2hati +hatj - 2hatk` is given by
`cos thet=+-((-54hatj+54hatk).(2hati+hatj-2hatk))/(54sqrt2 .3)`
`=+-1/sqrt2`
`theta= pi//4 or 3pi//4`