Find the equation of a line passing through the point `(2hati-3hatj-5hatk)` and perpendicular to the plane`vecr.(6hati-3hatj+5hatk) +2=0`. Also find the point of intersection of this line and the plane.

Answer

`vecr.(2hati-3hatj-5hatk)+lambda(6hati-3hatj+5hatk), (76/35,-108/35,-34/4)`

Explanation

Clearly, the required line passes through the point `(2hati-3hatj-5hatk)` and is parallel to the normal of the given plane, which is `(6ahti-3hatj+5hatk)`.
So, the rquired vector equation is
`vecr=(2hati-3hatj-5hatk)+lambda(6hati-3hatj+5hatk)`.
The Cartesian equation of the line is `(x-2)/6=(y+3)/-3=(z+5)/5=k`.
A general point on this line cut the plane `6x-3y+5z+2=0`. ltrbgt Then, `6(6k+2)-3(3k-3)+5(5k-2)=0`.
`rArr (36k+9k+25k)=2 rArr 70k=2`
`rArr k-1/35`.
`therefore` required point of intersection of the line and plane is
`P(6/35+2,-3/35-3,1/7-5)`, i.e., `P(76/35,-108/35,-34/7)`.