Equation of a place from a point and parametrics

[memish]

Homework Statement

Find the equation of the plane that passes through the point (1,−2,−1) and contains the line x(t)=−1−3ty(t)=−3−2tz(t)=4+4t.

Homework Equations

none

The Attempt at a Solution

I used vector <-3,-2,4> from the parametrics and crossed it with <2,1,-5> which i got by doing given point (1-2,-1)-(-1,-3,4). I got the 2nd point by puting t=0 into the parametric. When I crossed those i got <6, -7, -7> which is the normal I think? Then I just did 6(x-1) - 7(y+2) - 7 (z+1) =0

so that seems ok tome... but i checked and that's not correct! where am i going wrong?

 

[Office_Shredder]

I think you did the cross product wrong.

(-3,-2,4) dotted with (6,-7,-7) is:

-3*6 + (-2)(-7) + 4(-7) = -18 + 14 - 28

which is not 0. So the vector you got is not normal to your first vector

 

[memish]

ok thanks! i got it :)