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Intersection and distance in three-dimensional space

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1. Find the intersection point of the line (x-1)/2=(y+1)/3=z-2 and the plane 2x+y-z=17.

2. Find the distance from point Q(1,-2,3) to the plane 2x-y-z=6.

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Solution Summary

This shows how to find an intersection of a line and plane and the distance from a point to a plane.

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1. Let's first parameterize the line so t=(x-1)/2=(y+1)/3=z-2, so then a point on the line is

u = (2t+1,3t-1,t+2) and if the equation of the plane is 2x+y-z=17 then their intersection occurs when

2(2t+1)+(3t-1)-(t+2)=17 that is t=3, substituting this back into the general form of the line we ...

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