Question

Find an equation of the plane orthogonal to the line

Answer 1

The given line is orthogonal to the plane you want to find, so the tangent vector of this line can be used as the normal vector for the plane.

The tangent vector for the line is

d/dt (⟨0, 9, 6⟩ + ⟨7, -7, -6⟩t ) = ⟨7, -7, -6⟩

Then the plane that passes through the origin with this as its normal vector has equation

⟨x, y, z⟩ • ⟨7, -7, -6⟩ = 0

We want the plane to pass through the point (9, 6, 0), so we just translate every vector pointing to the plane itself by adding ⟨9, 6, 0⟩,

(⟨x, y, z⟩ - ⟨9, 6, 0⟩) • ⟨7, -7, -6⟩ = 0

Simplifying this expression and writing it standard form gives

⟨x - 9, y - 6, z⟩ • ⟨7, -7, -6⟩ = 0

7 (x - 9) - 7 (y - 6) - 6z = 0

7x - 63 - 7y + 42 - 6z = 0

7x - 7y - 6z = 21

so that

a = 7, b = -7, c = -6, and d = 21