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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.

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2 Answers 2

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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: $(3,2,-5)\times(B-(-1,-1,3))$

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  • $\begingroup$ Hi amd, thanks for your help. I've reached the conclusion that the plane I'm constructing has the general equation $6x-2y-3z-1=0$. I'm now substituting the equation for $s'$ into the equation of the plane, but the values I'm getting for λ are suspiciously ugly. I'm reaching the conclusion that λ = 12/29 when I substitute the variables in the equation of my plane with -1+3λ, -1+2λ, 3-5λ. $\endgroup$
    – Jake S
    Commented May 4, 2018 at 2:52
  • $\begingroup$ @JakeS, There is a slightly different way to do it. Let $ \vec d_r = \vec {s'} (λ) - B $ be the direction vector of line r and $ \vec n $ be the direction vector of the perpendiclar line (also the normal to the plane). When $ \vec d_r \cdot \vec n = 0 $ the two are perpendicular. I also get $ λ = 12/29 $. Carry on. Who says all answers have to be "neat"? $\endgroup$ Commented May 4, 2018 at 3:08
  • $\begingroup$ @JakeS You’ve got a sign error in the plane equation in your comment. That aside, $12/29$ doesn’t look all that ugly to me. $\endgroup$
    – amd
    Commented May 4, 2018 at 5:23
  • $\begingroup$ Much thanks to both of you. I end up reaching the parametrized equation of $r$ to be (-1 + λ(-36/29), 2 + λ(63/29), -3 + λ(-114/29)). I substituted into the equation of $s'$, used the point I found along with $B$ to find the direction vector of $r$, and constructed my equation for $r$. Much thanks to you both! $\endgroup$
    – Jake S
    Commented May 4, 2018 at 13:34
  • $\begingroup$ @JakeS See my recent update for a way to compute the direction vector directly, without solving any equations. $\endgroup$
    – amd
    Commented May 4, 2018 at 19:38
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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


Followup:

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

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

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

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

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

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  • $\begingroup$ Thanks for your help Ced! See my comment on amd's answer for how I've progressed and how I'm stuck, haha. $\endgroup$
    – Jake S
    Commented May 4, 2018 at 2:53
  • $\begingroup$ @JakeS. See my followup. $\endgroup$ Commented May 4, 2018 at 3:17

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