Question

Find an equation of the plane consisting of all points that are equidistant from P=(-1, -3, 5) and Q=(5, 2, 0), and having 6 as the coefficient of z

= 0
Hint: The midpoint between P and Q is a point on the plane and the vector pointing from P to Q (or vice versa) is a normal vector for the plane

Answer 1

6x + 5y - 5z = -17

Explanation:

Find the midpoint between P and Q.

Midpoint = (-1 + 5)/2, (-3 + 2)/2, (5 + 0)/2) = 2, -1/2, 2.5)

Find the vector pointing from P to Q.

Vector PQ = (5 - (-1), 2 - (-3), 0 - 5) = 6, 5, -5)

The normal vector of the plane is perpendicular to the vector pointing from P to Q.

Normal Vector = (6, 5, -5)

The equation of the plane can be written in the form of ax + by + cz + d = 0, where (a, b, c) is the normal vector and d is the distance between the plane and the origin.

(6x + 5y - 5z + d) = 0

We know that the point (2, -1/2, 2.5) lies on the plane

(6 * 2 + 5 * (-1/2) - 5 * 2.5 + d) = 0

-17/2 + d = 0

d = 17/2

(6x + 5y - 5z + 17/2) = 0

12x + 10y - 10z + 17 = 0

6x + 5y - 5z = -17

This is the equation of the plane consisting of all points that are equidistant from P=(-1, -3, 5) and Q=(5, 2, 0), and having 6 as the coefficient of z.