Thread: Lines and Planes

Consider the plane: -5x+8y-7z=-8. and the line:[x,y,z]^T=[-7,6,1]+s[6,7,4]^T. Find the smallest distance between the two.

I just have no idea how to do this question. I can find the normal vector to be the coefficients of the plane which is (-5,8,-7). P can be said as an arbitrary point when t=0. so, P=(-7,6-1). I can also find Po on the plane which can be (0,-1,0). I tried to do this question in terms of a projection by finding the vector V1. The answer i got is 84sqrt(138)/113. I dont believe this is correct at all. Any help is appreciated!

Originally Posted by Belowzero78

Consider the plane: -5x+8y-7z=-8. and the line:[x,y,z]^T=[-7,6,1]+s[6,7,4]^T. Find the smallest distance between the two.

I just have no idea how to do this question. I can find the normal vector to be the coefficients of the plane which is (-5,8,-7). P can be said as an arbitrary point when t=0. so, P=(-7,6-1). I can also find Po on the plane which can be (0,-1,0). I tried to do this question in terms of a projection by finding the vector V1. The answer i got is 84sqrt(138)/113. I dont believe this is correct at all. Any help is appreciated!

First check to see if the line intersects the plane! If it does, then the "shortest distance between the two" is 0. If not, if the line and plane are parallel, then the "shortest distance" is along the normal to the plane. (Strictly speaking, there is no "shortest" distance, the distance between parallel line and plane is constant.)

So go ahead and pick a point on the line line as you did: (-7, 6, -1) works fine. But don't just pick a point on the plane at random; find the equation of the line through that point perpendicular to the plane. Since <-5, 8, -7> is the normal vector, x= -7- 5t, y= 6+ 8t, z= -1- 7t is the equation of that line. Find the point where that line intersects the plane. The distance between that point of intersection and (-7, 6, -1) is the distance between the line and plane.

the answer i get is 84/sqrt(138) is this correct?