Equation of a line given point, perpendicular line and intersecting line

Question

My problem states:

Find the equation of the line $r$ knowing that:

$r$ passes through the point B = (-1, 2, -3)

$r$ is perpendicular to the line $s$ = (-1+6λ, -3-2λ, 2-3λ)

$r$ intersects with the line $s'$ = (-1+3λ, -1+2λ, 3-5λ)

So if $r$ is in a plane perpendicular to the line $s$, I can use the direction vector of $s$ as the normal vector of that plane. I could also find the point A where $s$ intersects that plane, and if I take B to be contained in the plane as well, I can find $r$ with A and B.

However, I'm not at all sure if this idea is correct, and furthermore, I don't know if it's an appropriate method at all because I have no idea how to relate it to the third idea of $r$ intersecting $s'$... thus I have arrived at SE to ask for help.

Answer 1

That’s not quite correct, but the question could’ve been worded better. The second criterion isn’t meant to imply that $r$ and $s$ intersect; only that their direction vectors are perpendicular. So $r$ does lie in the plane through $B$ perpendicular to $s$, but you still need a second point for $r$. That’s where $s'$ comes in: the point where it intersects this plane will give you a second point for $r$.

There’s another, equivalent, way to use $s'$: since $r$ and $s'$ intersect, they must be coplanar, so $r$ also lies in the plane defined by $s'$ and $B$, and is in fact the intersection of the two planes. This allows you to compute the direction vector of $r$ without solving any equations: $r$ is perpendicular to the normals of both planes, so their cross product is parallel to the line. The direction vector of $s$ is normal to the first plane, as you’ve already noted, and a normal to the second can be obtained via another cross product: $(3,2,-5)\times(B-(-1,-1,3))$

Answer 2

Since you know how to figure out the plane that is perpendicular to line S, which contains point B, you are most of the way there.

Observe that line s' is not perpendicular to line S which means it must intersect the plane somewhere. Find that point and you have the second point for your line.

Hope this helps, but not too much.

Ced

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Followup:

$$ \vec d_r = (-1+3\lambda, -1+2\lambda, 3-5\lambda) - (-1, 2, -3) = (0, -3, 6) + (3,2,-5)\lambda $$

$$ \vec n = ( 6, -2, -3 ) $$

$$ \vec d_r \cdot \vec n = 0 $$

$$ -12 + 29 \lambda = 0 $$

$$ \lambda = \frac{12}{29} $$