# Thread: Cartesian equation of plane containing line

1. ## Cartesian equation of plane containing line

i) Find the distance from the origin to the line y = 1, z = 2. Answer: $\displaystyle \sqrt{5}$

Find a cartesian equation of the plane which contains the line in (i) and which is farthest from the origin.

Attempt: Plane = (0, 0, 0) + t(0, 1, 0) + s(0, 0, 2) t, s E R

The answer in the memorandum is y + 2z = 5. I am confused.

2. The answer in the memorandum seems to be the equaiton of the line. It is trivial after that. Since x is not significant, we can just work in the y-z plane and use 2D methods.

$\displaystyle \frac{|(0)+2(0)-5|}{\sqrt{1^{2}+2^{2}}}$

This should look familiar. I resisted simplifying so you can see all the pieces.

3. Originally Posted by SyNtHeSiS
i) Find the distance from the origin to the line y = 1, z = 2. Answer: $\displaystyle \sqrt{5}$

Find a cartesian equation of the plane which contains the line in (i) and which is farthest from the origin.

Attempt: Plane = (0, 0, 0) + t(0, 1, 0) + s(0, 0, 2) t, s E R
when s= t= 0, that gives the point (0, 0, 0). In other words, this is a plane that contains (0, 0, 0). Hardly the plane that is "farthest from the origin"!

Essentially, you found a plane that contains the given line and the origin. The plane that is "farthest from the origin" will be the plane, containing that line, that is perpendicular to the line from the plane to the origin. The point on the line y= 1, z= 2 that is closest to the origin is (0, 1, 2). The vector from the origin to that point is j+ 2k. What you want is the plane containing (0, 1, 2) that is perpendicular to that vector: 0(x- 0)+ 1(y- 1)+ 2(z- 2)= 0 or y+ 2z= 5.

The answer in the memorandum is y + 2z = 5. I am confused.