find a equation of the plane that passes through the point(6,0,-2) and contains the linex= 4-2t,y=3+5t,z= 7+4t

Answer 1

Find the equation of the plane that passes through the point
P(6, 0, -2) and contains the line
x = 4 - 2t, y = 3 + 5t, z = 7 + 4t.

Write the equation of the line in vector form.

L(t) = Q + tv
L(t) = <4, 3, 7> + t<-2, 5, 4>
where t is a scalar ranging over the real numbers

We need two directional vectors and a point to define a plane. We already have a point and one directional vector. PQ will define a second directional vector.

u = PQ = = <4-6, 3-0, 7+2> = <-2, 3, 9>

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

n = u X v = <-2, 3, 9> X <-2, 5, 4> = <-37, -10, -4>

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

n = <33, 10, 4>

With the normal vector and a point in the plane we can write the equation of the plane. Let's choose P(6,0,-2).

33(x - 6) + 10(y - 0) + 4(z + 2) = 0
33x - 198 + 10y + 4z + 8 = 0
33x + 10y + 4z - 190 = 0

Answer 2

Hint:
Find two non-parallel vectors on the plane:
A = {-2, 5, 4}
B = {6-4, 0-3, -2-7} = {2, -3, -9}
Calculate r = A x B = {r1, r2, r3}
Then the equation of the plane is,
r1(x-6) + r2(y-0) + r3(z+2) = 0