Calculus 3 Proof

Question

Prove that the vector n=ai+bj+ck is perpendicular to the plane A: ax+by+cz=d by showing that n is perpendicular to PQ for any two points P(x1,y1,z1) and Q(x2,y2,z2) in A. 

Answer 1

Given P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) where P and Q are in A.

Since P is in A, then ax₁ + by₁ + cz₁ = d

Since Q is in A, then ax₂ + by₂ + cz₂ = d

 

Then vector PQ is (x₂ - x₁)i + (y₂ - y₁)j + (z₂ - z₁)k

 

Now n⋅PQ = (ai + bj + ck)⋅((x₂ - x₁)i + (y₂ - y₁)j + (z₂ - z₁)k)

               = a(x₂ - x₁₎ + b(y₂ - y₁) + c (z₂ - z₁)

               = (ax₂ + by₂ + cz₂) - (ax₁ + by₁ + cz₁)

               = d  d = 0

 

Since n⋅PQ =0, the vectors n and PQ are perpendicular.

Since PQ is in plane A, then n is perpendicular to plane A.

 

 

Answer 2

Let n = ai + bj + ck

 

If P = (x₁, y₁, z₁) and Q = (x₂, y₂, z₂) are points in the plane, then the vector v = PQ =

 

(x₂-x₁)i + (y₂-y₁)j + (z₂-z₁)k is parallel to the plane.  

 

v • n = a(x₂-x₁) + b(y₂-y₁) + c(z₂-z₁)

 

        = (ax₂+by₂+cz₂) - (ax₁+by₁+cz₁)

 

        = d - d  (since P and Q are both points in the plane)

 

        = 0 

 

Since v • n = 0, v ⊥ n.