Find the rectangular equation of the plane that contains the line r=(1+4t)i+(-3+4t)j+(-1+8t)k?

Question

and is perpendicular to the plane 3x-5y+z=17. I get how to find a plane containing the line but I don't get how to get the plane containing the line that is perpendicular to the plane 3x-5y+z=17.

*** r=(1+4t)i+ (-3+4t)j +(-1+8t)k

Answer 1

Let r_0 = r at t = 0 so r_0 = (1, -3, -1).
Let r_1 = r at t =1 so r_1 = (5, 1, 7).

Let Plane 1 have the plane equation 3x -5y + z = 17.

Substitute r_0 and r_1 into this plane equation to confirm that they lie in the plane; (otherwise, the parametric line does not lie in the plane as claimed).

Let dr = r_1 - r_0 = (4, 4, 8).
dr is a direction vector parallel to the plane because r_0 and r_1 each lie in the plane. Any non-zero scalar multiple points in the same direction so we'll take dr = (1, 1, 2) to simplify the following math.

The vector normal to the first plane is n1 = (3, -5, 1). I'll leave this part of the proof for you since it's trivial. At this point, you should check that dr and n1 are orthogonal by taking the dot product.

A second plane (to be found) is orthogonal to both the first plane and the line. Its normal vector is the cross product of n1 with dr.

Let n2 = n1 x dr = (-11, -5, 8).

The plane equation of the second plane is
-11x - 5y + 8z = w, where w is an unknown to be found by substituting a point in this plane. We could use either r_0 or r_1. Substituting r_0, we find w = -4.

The equation of the second plane is

-11x - 5y + 8z = -4.

Answer 2

Find the rectangular equation of the plane that contains the line r = (1+4t)i + (-3+4t)j + (-1+8t)k and is perpendicular to the plane 3x - 5y + z = 17.

Let's rewrite the equation for the line r.

r(t) = <1, -3, -1> + t<4, 4, 8>

The directional vector can be multiplied or divided by any non-zero scalar since that doesn't change its direction. Divide by 4 to simplify the math.

r(t) = <1, -3, -1> + t<1, 1, 2>

The problem doesn't require it but I noticed something. The given line is required to be in the plane to be found but not necessarily in the given plane. Let's see if it is in the given plane also. Two points define a line. So let's find two points on the line.

r(0) = P(1, -3, -1)
r(1) = <1, -3, -1> + <1, 1, 2> = Q(2, -2, 1)

Here is the equation for the given plane.

3x - 5y + z = 17

Let's plug in the points P and Q to see if they are in the plane.

P: 3*1 - 5(-3) + 1(-1) = 3 + 15 - 1 = 17 ||| So P is in the plane.
Q: 3*2 - 5(-2) + 1*1 = 6 + 10 + 1 = 17 ||| So Q is in the plane.

Since both P and Q, two points on the line, are in the given plane, the whole line is in the given plane. Therefore the line r, is in both planes. That allows us to take a shortcut.

The normal vector of the plane to be found must be perpendicular to both the given plane and any vector in the given plane. So it is perpendicular to the directional vector of the given line since that is in the given plane. Now let's calculate the normal vector n for the plane to be found.

n = X

n = <1, 1, 2> X < 3, -5, 1> = <11, 5, -8>

To write the equation of the plane to be found we need its normal vector n, and a point on the plane. Let's use
point Q(2, -2, 1).

11(x - 2) + 5(y + 2) - 8(z - 1) = 0
11x - 22 + 5y + 10 - 8z + 8 = 0
11x + 5y - 8z - 4 = 0