Can I get some clarification on this Calc III problem?

Posted on 2017-03-24 18:40:00 in Math by LUIS

The question states "Find an equation for the plane containing the line r(t) = <3t-1, 
-t+4, 5> and the vector v = <1, 1, -2>. 

The final answer I got was -2x-6y-4z=-42 but the TA said that this was incorrect. My steps: 

1. Write r(t) in the form of l = u + tv such that r(t) = (-1, 4, 5) + t<3, -1 0>. 

2. Find the cross product of <1, 1, -2> and <3, -1, 0> to give the vector perpendicular to both, denoted ; it is found to be <-2, -6, -4>. 

3. Plug into equation of the plane, 
• (x-x0, y-y0, z-z0) = 0, yielding 
<-2, -6, -4> • (x-(-1), y-4, z-5) = 0. 

4. After simplifying, I ended up with 
-2x-6y-4z=-42, but this was apparently incorrect. 

Can someone show me where I went wrong?

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