Question

Write the normal form of the equation of the plane

2x–3y + 6z + 14 = 0.

Answer 1

The given equation of the plane

2x–3y + 6z + 14 = 0

2x–3y + 6z = – 14 ……(i)

Now,

 \(\sqrt{2^2+(-3)^2+(-6)^2}\)

= \(\sqrt{4+9+36}\)

= \(\sqrt{49}\) = 7

Dividing (i) by 7, we get

Multiplying both sides by – 1, we get

This is the normal form of the given equation of the plane.