Question

Find an equation of the plane. the plane through the point (2, 0, 5) and perpendicular to the line x = 7t, y = 6 − t, z = 4 + 5t

Answer 1

Given:
P: (2,0,5)
L: (0,6,4)+t(7,-1,5)
and required plane, Π , passes through P and perpendicular to L.

The direction vector of L is V=<7,-1,5>.
For &Pi; to be perpendicular to V, &Pi; has V as the normal vector.

The equation of a plane with normal vector <7,-1,5> passing through a given point P(xp,yp,zp) is
7(x-xp)-1(y-yp)+5(z-zp)=0

Thus the equation of plane &Pi; passing through P(2,0,5) is
7(x-2)-y+5(z-5)=0
or alternatively,
7x-y+5z = 14+25 
7x-y+5z = 39