Question

A non zero vector `veca` is parallel to the line of intersection of the plane determined by the vectors `hati,hati+hatj`and the plane determined by the vectors `hati-hatj , hati + hatk`. The angle between `veca` and `hati-2hatj + 2hatk` can be

Answer 1

Correct Answer - `(pi)/(4) " or " (3pi)/(4)`
Equation of the plane containing `hat(i) " and " hat(j)` is
`[( vec(r ) - hat(i)) hat(i) (hat(i) + hat(j)) ]=0`
`rArr (vec( r) -hat(i)) .[(hat(i) xx hat(j))]=0`
`rArr {(x hat(i) + y hat(j) + zhat(k)) - hat(i)} . [hat(i) xx hat(i) + hat(i) xx hat(j)] =0`
`rArr {(x-1) hat(i) +y hat(j) + xhat(k)}. [hat(k)] =0`
`rArr (x-1) hat(i) . hat(k) + y hat(j) . hat(k) + zhat(k) . hat(k) =0`
Equations of the plane containing `hat(i) - hat(j) " and " hat(i) + hat(k)` is
`[(vec(r )- (hat(i) - hat(j)) (hat(i) - hat(j)) (hat(i) + hat(k))]=0`
`rArr (vec(r ) - hat(i) + hat(j)) . [(hat(i) - hat(j)) xx ( hat(i) + hat(k)) ]=0`
`rArr {(xhat(i) +y hat(j) + zhat(k)) - (hat(i) - hat(j))} . [ hat(i) xx hat(i) + hat(i) xx hat(k) - hat(j) xx hat(i) - hat(j) xx hat(k))]=0`
`rArr {(x -1) hat(i) + (y +1) hat(j) + zhat(k) }. [-hat(j) + hat(k) - hat(i) ]=0`
`rArr -(x -1) -( y+1) +z=0 .....(ii)`
Let `vec(a) = a_(1) hat(i) + a_(2) hat(j) + a_(3) hat(k)`
Since `vec(a) ` is parallel to Eqs . (i) and (ii) we obtain
`a_(3) =0`
and ` a_(1) + a_(2) -a_(3) =0 rArr a_(1) =a_(1) =-a_(2), a_(3)=0`
Thus a vectors in the direction `vec(a) " is " hat(i) - hat(j)`
If 0 is the angle between `vec(a) " and " hat(i) - 2 hat(j) + 2hat(k)` then
` cos theta = +- ((1)(1) + (-1)(-2))/(sqrt(1+1sqrt(1+4+4)) = +- (3)/(sqrt(2.3))`
`rArr cos theta = +- (1)/(sqrt(2)) rArr theta =(pi)/(4) " or " (3pi)/(4)`