Plane and Line Question?

Question

A: (1,3,1) B:(1,2,4) C:(2,3,6) D:(5,-2,1)
The plane containing the points A, B, C is denoted by P and the line passing through D perpendicular to P is L. The point of intersection of P and L is denoted by M.

a. find the cartesian equation of P and L
b. determine the coordinates of M
c. Find the perpendicular distance of D from P

Answer 1

a.
Find the equation of the plane P defined by the points
A(1,3,1), B(1,2,4), and C(2,3,6).

Define the vectors
AB = <1-1,2-3,4-1> = <0,-1,3>
AC = <2-1,3-3,6-1> = <1,0,5>

The cross product of AB and AC will give the normal vector to the plane.

AB X AC = -5i + 3j + k

The equation of the plane can be obtained from the cross product of the two vectors and one of the points in the plane. Let's select (1,3,1).

P: -5(x - 1) + 3(y - 3) + (z - 1) = 0
P: -5x + 5 + 3y - 9 + z - 1 = 0
P: -5x + 3y + z - 5 = 0

The normal vector to plane P is also the directional vector of the line L that is perpendicular to the plane. The line L is defined by the directional vector and the point D(5,-2,1).

L = <5,-2,1> + t<-5,3,1>
where t varies over the interval (-∞, ∞)
__________________________

b.
Determine the coordinates of M, the point of intersection of plane P and line L.

P: -5x + 3y + z - 5 = 0
L = <5,-2,1> + t<-5,3,1>

Parametrically L can be expressed as:
x = 5 - 5t
y = -2 + 3t
z = 1 + t

Rewrite the equation for P in terms of t.

P: -5(5 - 5t) + 3(-2 + 3t) + (1 + t) - 5 = 0
P: -25 + 25t - 6 + 9t + 1 + t - 5 = 0
P: 35t - 35 = 0
t = 1

The point of intersection can be found when t = 1.

x = 5 - 5t = 5 - 5 = 0
y = -2 + 3t = -2 + 3 = 1
z = 1 + t = 1 + 1 = 2

The point of intersection of plane P and line L is M(0,1,2).
__________________________

c.
Find the perpendicular distance of D from P.

Since we already have the point of intersection M(0,1,2) this problem reduces to finding the distance between the two points
D(5,-2,1) and M(0,1,2).

Distance = √[(5 - 0)² + (-2 - 1)² + (1 - 2)²] = √(25 + 9 + 1) = √35

However we could still find the perpendicular distance from point D(5,-2,1) to plane P even if we didn't have the point of intersection. We would use the distance formula from a point to a plane.

P: -5x + 3y + z - 5 = 0
D(5,-2,1)

Distance = |(-5)(5) + (3)(-2) + (1)(1) - 5| / √(5² + 3² + 1²)
Distance = |-25 - 6 + 1 - 5| / √(25 + 9 + 1)
Distance = 35 / √35 = √35

As you can see, both calculations of distance agree as they should.