- #1
MartinBob
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Hi, I have the following question:
Find the perpendicular distance from the origin to the plane x + 2y + 2z = 6.
My solution is incorrect but I can't find what's wrong. I did this:
I know that the distance will be a vector from the the origin to the plane, and it will be along a normal vector. Normal vector must be parallel to (1,2,2), so I solve O + (1,2,2)t satisfies plane equation to find the point where a normal to the plane will hit the origin. So 1t + 2t + 2t = 6, t = 6/5. Then the point (6/5, 12/5, 12/5) is on the plane, and it is on the normal vector from the plane towards (0,0,0) because (6/5, 12/5, 12/5) - (6/5)(1, 2, 2) = (0, 0, 0). But when I take the distance of this point, I get sqrt((36/25)+(144/25)+(144/25)) = 18/5, and this answer is not correct.
The solution in the book used projections and I understand why it worked (the answer is 2), but I can't find what went wrong in my solution. Thanks for helping.
Find the perpendicular distance from the origin to the plane x + 2y + 2z = 6.
My solution is incorrect but I can't find what's wrong. I did this:
I know that the distance will be a vector from the the origin to the plane, and it will be along a normal vector. Normal vector must be parallel to (1,2,2), so I solve O + (1,2,2)t satisfies plane equation to find the point where a normal to the plane will hit the origin. So 1t + 2t + 2t = 6, t = 6/5. Then the point (6/5, 12/5, 12/5) is on the plane, and it is on the normal vector from the plane towards (0,0,0) because (6/5, 12/5, 12/5) - (6/5)(1, 2, 2) = (0, 0, 0). But when I take the distance of this point, I get sqrt((36/25)+(144/25)+(144/25)) = 18/5, and this answer is not correct.
The solution in the book used projections and I understand why it worked (the answer is 2), but I can't find what went wrong in my solution. Thanks for helping.