# Find the shortest distance between the plane with equation x-y+2z = 2 and a line with parametric equations:

Find the shortest distance between the plane P with equation x-y+2z = 2 and a line L with parametric equations: $$L:\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-3}{-1}=t$$

L: { x = 1 + t } { y = 1 - t } { z = 3 - t }

From the equation of the plane, I found the normal vector to be <1, -1, 2>. From the parametric equations I identified a point on the line, B (1,1,3). I also found a point on the plane, A (2,0,0). I found the vector AB (going from point A to B) as <-1,1,3>.

I know in order to find the shortest distance between the plane and the line I need to do a projection of AB onto the normal vector.

I know from doing the dot product of AB and the normal vector, that the numerator for the projection will be 4 (-1 + -1 + 6).

I keep coming up with the wrong answer and am looking for some help in figuring out the equation for this projection, as I simply cannot figure out how the solution manual got to that answer.

In your question line is $$\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-3}{-1}=t$$ Hence one of Direction Ratio of line is (1,-1,-1) and direction ratio of normal to plane is (1,-1,2) Hence dot product of line and plane is zero. Line is parallel to plane . Shortest distance will be distance of any one point on line from plane . One of the point is (1,1,3). $$Shortest \,distance=\vert \frac{1-1+2×3-2}{\sqrt{1+1+2}} \vert$$ $$=\frac{4}{\sqrt{6}}$$