# Thread: distance from the line to the plane (which one is correct?)

1. ## distance from the line to the plane (which one is correct?)

find the distance from the line x=2+t, y=1+t, z= -(1/2)-(1/2)t to the plane x+2y+6z =10

my first attempt ::

by setting t = 0, we get a point on the line, say P :<2,1,-1/2>. the point on the plane, say, S we get if we set y=0, z=0 ==> x= 0. thus the point S on the plane is <10,0,0>.

thus the vector PS = S - P =8i-j+1/2k

vector normal to the plane, n = i+2j+6k

so the distance point P to the plane = |PS.n|/|n| = 9/sqrt(41) unit

my second attempt ::

the vector normal to the line, n1 = i+j-1/2k

the vector normal to the plane, n2 = i+2j+6k

n1.n2 = 0 ==> the two normal are perpendicular ===> the line and the plane are parallel

thus the distance for the line to the plane = sqrt(2^2+1^2+(-1/2)^2) = sqrt(21)/2 unit

2. Originally Posted by bobey
find the distance from the line x=2+t, y=1+t, z= -(1/2)-(1/2)t to the plane x+2y+6z =10

my first attempt ::

by setting t = 0, we get a point on the line, say P :<2,1,-1/2>. the point on the plane, say, S we get if we set y=0, z=0 ==> x= 0. thus the point S on the plane is <10,0,0>.

thus the vector PS = S - P =8i-j+1/2k

vector normal to the plane, n = i+2j+6k

so the distance point P to the plane = |PS.n|/|n| = 9/sqrt(41) unit
This is incorrect. That gives |PS| which is the distance from P to S, not the distance from P to the nearest point on plane.

my second attempt ::

the vector normal to the line, n1 = i+j-1/2k

the vector normal to the plane, n2 = i+2j+6k

n1.n2 = 0 ==> the two normal are perpendicular ===> the line and the plane are parallel
Well, yes, of course. Otherwise, the two would intersect so the distance would be 0. There was really no reason to do this.

thus the distance for the line to the plane = sqrt(2^2+1^2+(-1/2)^2) = sqrt(21)/2 unit
What? This is simply the distance from the the given point to (0, 0, 0). What does that have to do with the plane?