Question 259977: let y=(-7,-6,-2), u1=(6,-6,1), u2=(-4,2,36). Compute the distance d from y to the plane in R3, spanned by u1 and u2. The distance d is supposed to be the shortest distance and I understand that to be the length from y to where it is perpendicular to the plane. The confusion is with a method that uses a normal vector to the plane and is that now considered the projection of y onto the normal vector such that I am now looking for the parallel component of y on a? Can you provide a visual model perhaps?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Here's the basic outline to solving this problem.
1) First find the equation of the plane. To do this, compute the cross product of u1 and u2 to find a third vector (say u3). This vector is the normal which will help you find the equation of the plane.
2) This normal will essentially be a line. So you can find the equation of that line using that normal vector and the vector (-7, -6, -2) (since you want the line to go through this point)
3) Use both the line and the plane to find the intersection between the two figures. This point will be closest to the given point (-7, -6, -2)
4) Finally, find the distance (using the distance formula) from the point found in step 3 and (-7, -6, -2) to find the distance from (-7, -6, -2) to the plane.
Let me know if this helps. If not, then repost or ask me.
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