- #1

kieth89

- 31

- 0

## Homework Statement

Find an equation of the line where the planes Q and R intersect.

[itex]Q: -2x + 3y - z = 1; R: x + y + z = 0[/itex]

## Homework Equations

Equation of a Plane: [itex]ax + by + cz = d, [/itex]where [itex]\vec{n} = <a, b, c>[/itex]

Equation of a Line in [itex]R^{3}[/itex]: [itex]\vec{r}(t)=<x_{0}, y_{0}, z_{0}> + t<x,y,z>[/itex]

## The Attempt at a Solution

First I find a point common to both planes, this will be [itex]P_{0}[/itex].

Set [itex]y = 0[/itex] and add the plane equations:

[itex] -2x + 0y - z - 1 = 0 [/itex]

[itex] 1x + 0y + z - 0 = 0[/itex]

Resulting in: [itex]-x - 1 = 0 [/itex] so [itex]x = -1 , z = 1

[/itex] and [itex]P_{0} = (-1, 0, 1)[/itex].

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:

[itex]<-2, 3, -1> X <1, 1, 1> = <4, 1, -5>[/itex]

Now I just plug the obtained info into the equation for a line:

[itex]\vec{r}(t) = <-1, 0, 1> + t<4, 1, -5> -> \vec{r}(t) = <-1 + 4t, t, 1 - 5t>[/itex]

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