# Find equation of a plane containing two lines

1. Oct 8, 2009

### hachi_roku

1. The problem statement, all variables and given/known data
find the equation of the plane that contains the line r_1(t) = (t,2t,3t) and r_2(t)=(3t,t,8t)

2. Relevant equations

3. The attempt at a solution
i don't know where to start...my book does not have an example similar. can somebody just point me to the right direction?

2. Oct 8, 2009

### Staff: Mentor

Each of your line equations defines a vector with the same direction as the line. Find the cross product of these vectors to get a third vector, say <a, b, c>. That will be a normal to the plane. Find a point on either of your given lines, say (x0, y0, z0).

Use the normal and the given point to write the equation of the plane as a(x - x0) + b(y - y0) + c(z - z0) = 0.

3. Oct 8, 2009

### Donaldos

$$r_1=t(1,2,3)$$

$$r_2=t(3,1,8)$$

Both $$\vec{v}_1=\left(\begin{array}{c} 1 \\ 2 \\ 3\end{array}\right)$$ and $$\vec{v}_2=\left(\begin{array}{c} 3 \\ 1 \\ 8\end{array}\right)$$ are in the plane.

So $$\vec{n}=\vec{v}_1 \times \vec{v}_2=\left(\begin{array}{c} 13 \\ 1 \\ -5\end{array}\right)$$ is normal to the plane.

Hence the equation:

$$13x+y-5z=0$$

4. Oct 8, 2009

### Staff: Mentor

Donaldos,
It is the policy of this forum to provide help to a poster, but not to give a complete answer to someone's problem.

5. Oct 8, 2009

### hachi_roku

ok i got the answer, but when i did the cross product my answers sitll have the t in them

my ans is 13t2x+t2y-5t2z = 0

the books answer is exactly that but without the t's

6. Oct 8, 2009

### Donaldos

I'm sorry. I'll keep that in mind.

7. Oct 8, 2009

### hachi_roku

i see donaldos wrote r with the t outside before doing the determinant...what happens to that?

8. Oct 8, 2009

### Staff: Mentor

All you need is any old vector that is parallel to the line, so any multiple of the vector will still be parallel. The t multiplier can be any real value, so it's convenient to let t = 1.