Question

Find an equation of the plane orthogonal to the line
(x,y,z)=(0,9,6)+t(7,−7,−6)

which passes through the point (9, 6, 0).

Give your answer in the form ax+by+cz=d (with a=7).

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

Answer 2

An equation of the plane orthogonal to the line 7x - 7y - 6z = 21.

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,

What is the tangent vector?

A tangent vector is a vector that is tangent to a curve or surface at a given point.

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 the 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 in 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.

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