Finding the equation of a plane

[navalava]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Here's what I did:
I substituted t=2 to get:
x=0, y=17, z= 15
So in total, we have three points that the plane contains: P1(6,0,-2), P2(4,3,7), P3(0,17,15)
Vector P1P2=<-2,3,9>
Vector P1P3=<-6,17,17>
I took a cross product of these vectors to get <-102,-20,-16>
I used the point P1(6,0,-2) to obtain:

-102(x-6)-20(y-0)-16(z+2)=0

But the solution is 33(x-6)-10(y-0)-4(z+2)=0
I'm unable to understand what I did wrong. I'd really appreciate any help!

 

[Dick]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Here's what I did:
I substituted t=2 to get:
x=0, y=17, z= 15
So in total, we have three points that the plane contains: P1(6,0,-2), P2(4,3,7), P3(0,17,15)
Vector P1P2=<-2,3,9>
Vector P1P3=<-6,17,17>
I took a cross product of these vectors to get <-102,-20,-16>
I used the point P1(6,0,-2) to obtain:

-102(x-6)-20(y-0)-16(z+2)=0

But the solution is 33(x-6)-10(y-0)-4(z+2)=0
I'm unable to understand what I did wrong. I'd really appreciate any help!

Substituting t=2 gives me x=0, y=13, z= 15. Check the y value again.

 

[dynamicsolo]

In fact, why bother finding another point on the line for which you have the parametric equation? You already have its direction vector and you know a point on that line by using t = 0 .

Find the vector from that t = 0 point to (6, 0, -2). You now have two vectors in the plane that you can find the cross product for. That would be less work than dealing with a third point.

 

[navalava]

Thank you! I see what I did wrong in substitution now!