Intersection Line for Two Planes

[kieth89]

Homework Statement

Find an equation of the line where the planes Q and R intersect.
Q: -2x + 3y - z = 1; R: x + y + z = 0

Homework Equations

Equation of a Plane: ax + by + cz = d,where \vec{n} = <a, b, c>
Equation of a Line in R^{3}: \vec{r}(t)=<x_{0}, y_{0}, z_{0}> + t<x,y,z>

The Attempt at a Solution

First I find a point common to both planes, this will be P_{0}.
Set y = 0 and add the plane equations:
-2x + 0y - z - 1 = 0
1x + 0y + z - 0 = 0
Resulting in: -x - 1 = 0 so x = -1 , z = 1<br /> and P_{0} = (-1, 0, 1).

Now I find the direction vector for our line. This will just be the cross product of the normal vectors from the two plane equations:
&lt;-2, 3, -1&gt; X &lt;1, 1, 1&gt; = &lt;4, 1, -5&gt;

Now I just plug the obtained info into the equation for a line:
\vec{r}(t) = &lt;-1, 0, 1&gt; + t&lt;4, 1, -5&gt; -&gt; \vec{r}(t) = &lt;-1 + 4t, t, 1 - 5t&gt;

I felt pretty confident in this answer, but the answer key says it should be &lt;-\frac{1}{5} + 4t, \frac{1}{5} + t, -5t&gt;. I'm wondering if my answer is different due to using a different P_{0}, but I don't know...

 

[andrewkirk]

The answers are just different parametrisations of the line.
If we label your $t$ as $t$ and theirs as $t'$ then the conversion is $t'=t-\frac{1}{5}$.

This is hinted at by the question asking you to 'find an equation of...' rather than 'find the equation of...'

Strictly speaking though, neither answer is correct, as neither is an equation - where is the equals sign? But there is no single equation that denotes the line. Two are needed.

 

[ehild]

Now I just plug the obtained info into the equation for a line:
\vec{r}(t) = &lt;-1, 0, 1&gt; + t&lt;4, 1, -5&gt; -&gt; \vec{r}(t) = &lt;-1 + 4t, t, 1 - 5t&gt;

I felt pretty confident in this answer, but the answer key says it should be &lt;-\frac{1}{5} + 4t, \frac{1}{5} + t, -5t&gt;. I'm wondering if my answer is different due to using a different P_{0}, but I don't know...

Your equation is correct and also the given solution is really a parametric representation of the line of intersection. The difference is the starting point only. The equation of a line is given by $\vec r(t) = \vec a + \vec b t$

 

[ehild]

Strictly speaking though, neither answer is correct, as neither is an equation - where is the equals sign? But there is no single equation that denotes the line. Two are needed.

The answer of the OP is an equation. It has two sides and a "=" in between.

 

[kieth89]

Yay! So it is due to different starting points. Thanks for the help everyone.

 

[andrewkirk]

The answer of the OP is an equation. It has two sides and a "=" in between.

Oh my goodness, so it does! I don't know how I missed that. Either my mind's been playing tricks on me or my vision is deteriorating even faster than I feared.