Find an equation of the plane.

Question

Find an equation of the plane.
The plane that passes through the point (2, 4, 5) and contains the line x = 5t, y = 2 + t, z = 4 − t

Answer 1

We seek a normal n of the plane, and for this we would like to find two vectors parallel to the plane so that we can take their cross product to obtain a normal (a vector perpendicular to the plane). We can then use the equation

 

n • <x-2,y-4,z-5> = 0  as the equation for the plane. (**)

 

To find these two vectors, it is enough to find three points that lie on the plane. We have one, (2,4,5). We have one from setting t=0, and one from setting t=1: (0,2,4) and (5,3,3). Now we form the vector from (2,4,5) to (0,2,4), a = <-2,-2,-1>, and from (2,4,5) to (5,3,3), b = <3,-1,-2>.

 

We take the cross product a x b:

|   i   j   k  |

| -2 -2 -1  |

|  3  -1 -2 |

 

= 3i - 7j + 8k , call it n

 

Then plugging into (**) above and computing the dot product we have

 

0 = 3(x-2)-7(y-4) + 8(z-5)

  = 3x - 6 - 7y + 28 + 8z - 40

18 = 3x - 7y + 8z

 

We verify that the three points we found above are on this plane.

 

Alternately, we could have used the two vectors a = <-2,-2,-1> and b = <5,1,-1>, in other words, getting b directly from the coefficients of t in the equation of the line. But we still would have had to find one point on the line, by plugging in t=0.

 

I hope that helps!