Distance from the line to the plane (which one is correct?)

[bobey]

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

Homework Statement

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

Homework Equations

The Attempt at a Solution

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) unitmy 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
WHICH ATTEMPT IS CORRECT SINCE I GOT TWO DIFFERENT ANSWERS? PLEASE HELP ME...

 

[lanedance]

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

this is not correct n1 = i+j-1/2k is parallel to your line

 

[HallsofIvy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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. You are trying to use the fact that |PS.n|= |PS||n| cos(theta) but in this case what you get is |PS|, the straight line distance from the point <2, 1, -1/2> to the point <10, 0, 0> While one is a point on the line and the other is a point on the plane, it doesn't follow that this is the shortest possible distance between a point on the line and a point on the plane, which is, by definition, the 'distance between line and 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

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

This is just the distance from (2, 1, -1/2) to (0, 0, 0). It has nothing to do with the plane!

WHICH ATTEMPT IS CORRECT SINCE I GOT TWO DIFFERENT ANSWERS? PLEASE HELP ME...

It looks to me like neither is correct. I simply don't follow your "logic".