Fine the scalar equation of the plane that is perpendicular to the plane with equation.....?

Question

Fine the scalar equation of the plane that is perpendicular to the plane with equation r = [-2,3,1] + s[5,-2,-2] + t[-1,0,1], and intersects it along the line with equation r = [9, -1, -5] + p[2,-2,2].


please show all logical work and reasoning.

Answer 1

First take the cross product of the vectors
[5,-2,-2] and [-1,0,1]. This gives the vector [2,3,2] which will be perpendicular to the first two vectors. The new plane will be r=[9,-1,-5] +p[2,-2,2]+ t[2,3,2].

Answer 2

The first plane has the vector equation:
r = [-2,3,1] + s[5,-2,-2] + t[-1,0,1]

Its normal vector n1, can be found by taking the cross product of the two directional vectors.

n1 = <5, -2, -2> X <-1, 0, 1> = <-2, -3, -2>

Any non-zero multiple of the vector will also be a normal vector to the plane. Multiply by -1.

n1 = <2, 3, 2>

The first and second planes intersect along the line given by the equation:

r = [9, -1, -5] + p[2,-2,2]

The normal vector n2, of the second plane is normal to the normal vector of the first plane and the directional vector of the line of intersection of the two planes. Therefore n2 can be found by taking the cross product of those two vectors.

n2 = <2, 3, 2> X <2, -2, 2> = <10, 0, -10>

Any non-zero multiple of the vector will also be a normal vector to the plane. Divide by 10.

n2 = <1, 0, -1>

We need to find a point in the second plane. Set p = 0 for the line of intersection and we have the point (9, -1, -5). Now we can write the equation of the second plane.

1(x - 9) + 0(y + 1) - 1(z + 5) = 0
x - 9 - z - 5 = 0
x - z - 14 = 0