Find the distance from the origin to the line on the intersection of the planes 2x+y-z=4 and x-y+z=-3?

Question

Find the distance from the origin to the line on the intersection of the planes 2x+y-z=4 and x-y+z=-3

I found the line of intersection is (x-1/3)/2=-y/3=(z+10/3)/(-3)
I dont know what to do next

Answer 1

Find the distance from the origin to the line on the intersection of the planes

2x + y - z = 4
x - y + z = -3

First, find the line of intersection of the two planes.

The directional vector v for the line will be normal to the normal vectors of both planes. So take the cross product of the normal vectors.

v = <2, 1, -1> X <1, -1, 1> = 0i - 3j - 3k
Any non-zero multiple will do as well. Divide by -3.
v = j + k = <0, 1, 1>

Find a point on the line.

2x + y - z = 4
x - y + z = -3

Add the two equations together.

3x = 1
x = 1/3

Plug in for x.

2(1/3) + y - z = 4
(1/3) - y + z = -3

y - z = 10/3
-y + z = -10/3

Select a value for y. Let y = 3.

Then

y - z = 10/3
3 - z = 10/3
z = -1/3

A point on the line of intersection is Q(1/3, 3, -1/3).

The line of intersection is:

L = Q + tv = <1/3, 3, -1/3> + t<0, 1, 1>
where t is a scalar ranging over the real numbers
_______________________

The problem has now been reduced to finding the distance from the origin to the line L.

Create a vector u from the origin to any point on the line L. Select point Q.

u = <0 - 1/3, 0 - 3, 0 - (-1/3)> = <-1/3, -3, 1/3>
v = <0, 1, 1>
|| v || = √(0² + 1² + 1²) = √(0 + 1 + 1) = √2

The distance d from the origin to the line L is given by:

d = || u X v || / || v || = || -10i/3 + j/3 - k/3 || / √2

d = √[(-10/3)² + (1/3)² + (1/3)²] / √2 = √(102/9) / √2

d = √51 / 3 ≈ 2.3804761

Answer 2

Let's see. It's been awhile since Calc III. Assuming your line of intersection is correct.

P1=(1/3,1,-10/3)
PoP1=((1/3)i+j-(10/3)k)

Cross Product=

i j k
1/3 1 -10/3
2 3 -3

=7i-(17/3)j-k

Use formula for distance between point and line.

sqrt(7^2+(17/3)^2+1^2)/sqrt(2^2+3^2+3^2)=

sqrt(16258)/66 = 1.932....

See if anyone concurs.