Question

Find the equation of the plane that is orthogonal to the plane 3 2 4 x y z + − = and goes
through the points P(1,2,4) and Q(−1,3,2).

Answer 1

Answer:

Let’s assume this is asking for the plane through P(1,2,4) and Q(−1,3,2) that also includes the direction of the normal vector of the given plane, d=(3,2,4) .

Another direction in the plane is e=QP=P−Q=(2,−1,2)

The normal vector of the plane we seek is normal to both of these,

n=d×e=(3,2,4)×(2,−1,2)=(8,2,−7)

For every point (x,y,z) in the plane the vector (x,y,z)−P is orthogonal to n .

n⋅((x,y,z)−P)=0

n⋅(x,y,z)=n⋅P

(8,2,−7)⋅(x,y,z)=(8,2,−7)⋅(1,2,4)

Answer: 8x+2y−7z=−16

Check:

P: 8(1)+2(2)−7(4)=−16✓

That’s the same calculation we just did.

Q: 8(−1)+2(3)−7(2)=−16✓

Also something in the direction of D from P, say

P+D=(4,4,8), 8(4)+2(4)−7(8)=−16✓