Find the equation of the plane that passes through (-1,2,0), (2,0,1) and (-5,3,1)?

Question

Find the equation of the plane that passes through (-1,2,0), (2,0,1) and (-5,3,1)

Answer 1

Find the equation of the plane that passes through
P(-1,2,0), Q(2,0,1) and R(-5,3,1).

First create the directional vectors PQ and PR, for the plane .

PQ = = <2+1, 0-2, 1-0> = <3, -2, 1>
PR = = <-5+1, 3-2, 1-0> = <-4, 1, 1>

The normal vector n, of the plane is perpendicular to both of them. Take the cross product.

n = PQ X PR = <3, -2, 1> X <-4, 1, 1> = <-3, -7, -5>

Any non-zero multiple of n is also a cross product. Multiply by -1.

n = <3, 7, 5>

With a point in the plane and the normal vector to the plane we can write the equation of the plane. Let's use Q(2, 0, 1).

3(x - 2) + 7(y - 0) + 5(z - 1) = 0
3x - 6 + 7y + 5z - 5 = 0
3x + 7y + 5z - 11 = 0

Answer 2

this can be the good way around... the line given is parallel to <3, a million, -a million> and passes by way of (0, a million, 2). for this reason, the plan contains the factors (a million, 2, 3) and (0, a million, 2), meaning it contains the line by way of those factors as properly. That line is parallel to . Taking the pass made of those vectors supplies a vector prevalent to the airplane. N = <3, a million, -a million> X = < a million - -a million, -a million - 3, 3 -a million> = <2, -4, 2>. So, the equation of the airplane has the form 2x -4y + 2z + C = 0. utilizing between the factors to locate C, 2*0 - 4*a million + 2*2 + C = 0 C = 0 And the airplane is in simple terms 2x - 4y + 2z = 0.