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).

[tex]Find an equation of the plane orthogonal to the line (x,y,z)=(0,9,6)+t(7,−7,−6) which passes thro[/tex]

fahrenheit

step-by-step explanation:

fahrenheit is more precise than celsius. the ambient temperature on most of the inhabited world ranges from -20 degrees fahrenheit to 110 degrees fahrenheit—a 130-degree range. on the celsius scale, that range is from -28.8 degrees to 43.3 degrees—a 72.1-degree range. this means that you can get a more exact measurement of the air temperature using fahrenheit because it uses almost twice the scale.

the two may have that, even a breakdown is probably offered

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