Calculus Scalar

Question

Find an equation of the plane through the point (-5, 3, -3) and perpendicular to the vector (-4, -2, 0).

Answer 1

First we write down the point-normal equation of a plane: a (x - x0) + b (y-y0) + c (z-z0) = 0

The given point is: (-5, 3, -3) therefore x0 = -5, y0 = 3, z0 = -3

The given vector is : [-4, -2, 0] therefore a = -4, b = -2, c = 0

Plug in variables into the above equation to get: -4 (x + 5) + (-2) (y - 3) + 0 (z + 3) = 0

**Note: 0 (z + 3) goes to zero.

Simplify the equation to get: -4x - 20 + (-2y + 6) + 0 = 0

Further simplify to get: -4x - 20 - 2y + 6 = 0

Add -20 + 6 to get -14 and the final answer is: -4x - 2y - 14 = 0

Hence, the equation of a line passing through given point and perpendicular to given vector is:

-4x - 2y = 14