Chad W. answered • 09/14/16

Experienced and Professional Tutor on a Bicycle

Fun problem!

Let's first determine the intersection of the planes 2x-y=0 and y+z=-1. We determine normal vectors from those two planes (why this works is outside the scope of this answer).

**n1**= [2,1,0]

**n2**= [0,1,1]

where

**n1**and**n2**represent normal vectors of the planes. A line that exists as the intersection of two planes must be perpendicular to both planes. Thus, we can determine a vector (**v1**) parallel to the intersection of the planes by taking a cross product of**n1**and**n2**.**v1**=

**n1**x

**n2**= [1,-2,2]

The final plane must be perpendicular to both

**v1**and a vector parallel to the line x=3t, y=2+t, z=-2t, which will be called**v2**. We first determine a simple choice of**v2**.**v2**= [3,1,-2]

So, now we can find a possible

**n3**, a normal of the final plane, by taking a cross product of**v1**and**v2**.**n3**=

**v1**x

**v2**= [2,8,7]

We will also need a point in the plane. A simple choice is picked from t=0, giving p=(0,2,0). Thus, we can say that the equation of the plane is:

2x + 8(y-2) + 7z = 0

or

2x + 8y + 7z = 16