# Thread: Intersection of orthogonal planes

1. ## Intersection of orthogonal planes

Hello, I have been trying to tackle vectors over the weekend, but cant seem to get this one right. I have two lines, s(1,2,1) and (9,6,0) + t(0,1,-1). The orthogonal vector is v(-3,1,1) found using cross product. How can i find the equation of a line say P parallel to v(-3,1,1) which intersects both lines? Thanks.

2. Originally Posted by Oiler
Hello, I have been trying to tackle vectors over the weekend, but cant seem to get this one right. I have two lines, s(1,2,1) and (9,6,0) + t(0,1,-1). The orthogonal vector is v(-3,1,1) found using cross product. How can i find the equation of a line say P parallel to v(-3,1,1) which intersects both lines? Thanks.
1. You've 3 lines:

$l_1x,y,z)=s \cdot (1,2,1)" alt="l_1x,y,z)=s \cdot (1,2,1)" />

$l_2x,y,z)=(9,6,0) + t \cdot (0,1,-1)" alt="l_2x,y,z)=(9,6,0) + t \cdot (0,1,-1)" />

and a line $l_3$ which passes through a point of $l_1$ in the direction of (-3, 1, 1).

$l_3x,y,z)= s \cdot (1,2,1) + r \cdot (-3,1,1)" alt="l_3x,y,z)= s \cdot (1,2,1) + r \cdot (-3,1,1)" />

2. Line $l_3$ passes through a point of $l_2$ :

$s \cdot (1,2,1) + r \cdot (-3,1,1) = (9,6,0) + t \cdot (0,1,-1)$

Solve for (r, s, t).

3. For confirmation only: I've got $(r, s, t) = \left( -\frac{21}{11}, \frac{36}{11}, -\frac{15}{11} \right)$

3. An alternative:

Generic point of the first line $l_1$:

$P=(s,2s,s)\;(s\in\mathbb{R})$

Generic point of the second line $l_2$:

$Q=(9,6+t,-t)\;(t\in\mathbb{R})$

Solving the system:

$\begin{Bmatrix} \vec{PQ}\cdot (1,2,1)=0\\\vec{PQ}\cdot (0,1,-1)=0\end{matrix}$

we obtain $s_0,t_0$ , that provide two points $P_0,Q_0$ of the line $l$ which intersects and it is orthogonal to $l_1$ and $l_2$ .

Fernando Revilla

4. Hey earboth, shouldn't line 3, just be (1,2,1) + r(-3,1,1) since adding a s in front of (1,2,1) would make it a plane ? which would then be parallel to r(-3,1,1) instead of being orthogonal ? Thanks.