Equation of a place from a point and parametrics

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SUMMARY

The discussion centers on finding the equation of a plane that passes through the point (1,−2,−1) and contains the line defined by the parametric equations x(t) = −1 − 3t, y(t) = −3 − 2t, z(t) = 4 + 4t. The user initially calculated the normal vector by crossing the direction vector from the parametric line, <-3, -2, 4>, with the vector from the point to a point on the line, <2, 1, -5>. However, the resulting normal vector <6, -7, -7> was incorrect, as it did not satisfy the condition of being orthogonal to the direction vector, leading to the conclusion that the cross product was performed incorrectly.

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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.


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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?
 
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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
 
ok thanks! i got it :)
 

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