Where does the line through A(1,0,1) and B(4,-2,2) intersect the plane x+y+z=6?

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SUMMARY

The intersection of the line through points A(1,0,1) and B(4,-2,2) with the plane defined by the equation x+y+z=6 can be determined using parametric equations. The direction vector of the line, calculated as B-A, is <3,-2,1>. By substituting the parametric equations x=1+3t, y=0-2t, and z=1+t into the plane equation, one can solve for the parameter t to find the intersection point.

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Homework Statement


Where does the line through A(1,0,1) and B(4,-2,2) intersect the plane x+y+z=6?


Homework Equations


r= r_0 + tv


The Attempt at a Solution



The line through A and B is the vector AB and can be found by subtracting the A components from the B components. So B-A= <3,-2,1>.

They give x+y+z=6 so the components of that vector is <1,1,1>.

Not sure where to go from here...
 
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Put the components of your parametric equation for the line, e.g x=1+3t into the equation for the plane and solve for t, please? The normal vector to the plane doesn't have much to do with it.
 

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