Question

Let P(3,2,6) be point in the space and Q be a point on the line r=(i-j+2k)+const(-3i+j+5k), then find the value of const for which the vector PQ is parallel to the plane x-4y+3z=1.
thank you

Answer 1

let the const be a

so line pq becomes = q-p = -(2+3a)i + (a-3) j +(5a-6) k

so now we need the const such that it's parallel to the plane => it's perpendicular to the normal of the plane.

so the dot product of the line pq with the normal of the plane must result in 0.

(1,-4,3) directions of the normal of the plane

=> [-(2+3a)i + (a-3) j +(5a-6) k]  . [i-4j+3k] = 0
=> -2-3a-4a+12+152-18 = 0
=> a = 1

so the const value is 1