Find an Equation to a plane that passes through the point (6, 0, -2) and contains the lines ...?

Question

contains the lines:

x = 4-2t

y = 3+5t

z = 7+4t

Answer 1

The directional vector of the given line is:

v = <-2, 5, 4>

A point P, on the line can be found by letting parameter t = 0.

P(4, 3, 7)

A second directional vector v, of the plane can be found from the two identified points in the plane P(4,3,7) and Q(6,0,-2).

PQ = <Q - P> = <6-4, 0-3, -2-7> = <2, -3, -9>

The normal vector n, of the plane is orthogonal to both directional vectors.  Take the cross product.

n = v X PQ = <-2, 5, 4> X <2, -3, -9> = <-33, -10, -4>

Any non-zero multiple of n is also a normal vector of the plane.  Multiply by -1.

n = <33, 10, 4>

With the normal vector of the plane and a point in the plane we can write the equation of the plane.  Let's choose point P(4, 3, 7).  Remember, the normal vector of a plane is orthogonal to any vector that lies in the plane.  And the dot product of orthogonal vectors is zero.  Define R(x,y,z) to be an arbitrary point in the plane.  Then vector PR lies in the plane.

n • PR = 0

n • <R - P> = 0

<33, 10, 4> • <x - 4, y - 3, z - 7> = 0

33(x - 4) + 10(y - 3) + 4(z - 7) = 0

33x - 132 + 10y - 30 + 4z - 28 = 0

33x + 10y + 4z - 190 = 0