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"Give the equations of two lines that meet at the point $(2, -3, 5)$ and which meet at right angles, but do not use that point in either of the equations."

I am having a bad time with this one. I figured out the first line is the vector $(4,-6,10)$, that vector goes through the point, and then I was trying to find a perpendicular vector to $(2,-3,5)$. But if I do the rule of $u \cdot v = 0$, the lines are going to meet at $(0,0,0)$ and I want them to meet at $(2,-3,5)$, an help please?

QuantumSuperfield
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Julio
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  • A vector doesn't go "through a point" and a line is not the same as a vector. –  Mar 18 '15 at 00:19
  • Oh, so what can I do? – Julio Mar 18 '15 at 00:21
  • I just started the Intersections unit, but according to my notes, there are 3 types, parametric, vector, and symmetric equation. (x-x1)/a = (y-y1)/b = (z-z1)/c – Julio Mar 18 '15 at 00:25
  • Like that [x, y, z] = [ax, ay, az] + t[bx, by, bz] – Julio Mar 18 '15 at 00:28
  • 2 points in my plane? or 2 lines? – Julio Mar 18 '15 at 00:29
  • Ohhhhh, so in that case one of my lines could be, [x, y, z] = [2x, -3y, 5z] + t[bx, by, bz] I don't know the other coordinate, but the "t" should be the one pointing a 90 degree direction from the other line. – Julio Mar 18 '15 at 00:33
  • I am sorry, I am really trying. But thanks for taking the time to explain me this, I really appreciate your help, since I am homeschooled I am pretty much on my own. – Julio Mar 18 '15 at 00:36

1 Answers1

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The general equation of a line is given by $\mathbf r(t) = \mathbf r_0 + t\mathbf v$. In particular, when our vector space is $\Bbb R^3$, this equation is just $(x,y,z) = (x_0, y_0, z_0) + t(v_1, v_2, v_3)$.

Here $(x_0, y_0, z_0)$ is any point on your line, $(v_1, v_2, v_3)$ is any vector parallel to your line, and $t$ is your variable.

We know a point that is on both lines, don't we? It's $(2,-3,5)$.

Basically now all you have to do is find $2$ nonzero vectors which are perpendicular to each other to be the $\mathbf v$'s in your $2$ equations. I suggest just choosing one randomly and then using the definition of orthogonality to find the other one.

Definition: Two vectors, $\mathbf a$ and $\mathbf b$ are orthogonal if and only if $\mathbf a \cdot \mathbf b = \langle \mathbf a, \mathbf b\rangle =0$ (I'm not sure what notation you use for the dot product so I put both of the common ones).

So once you've chosen any vector, say $(1,1,1)$, you just need to plug that into this definition to find any other vector orthogonal to it: $(1,1,1)\cdot(a,b,c)=0$. Just find any nonzero solution to this.

Once you have those two vectors you just plug in. So your two equations will be $(x,y,z) = (2,-3,5) + t(u_1, u_2, u_3)$ and $(x,y,z)=(2,-3,5)+s(v_1, v_2, v_3)$ where $(u_1, u_2, u_3)$ and $(v_1, v_2, v_3)$ are your two orthogonal (perpendicular) vectors.

However, the question specifically asks that you not include the vector $(2,-3,5)$ in your equations. So basically, you just need one more point on each of these lines to replace the $(2,-3,5)$ with. So how would you find those two vectors?

  • (1,1,1)⋅(1,1,-2)=0 Those can be my vectors. – Julio Mar 18 '15 at 00:46
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    Sounds good to me. –  Mar 18 '15 at 00:48
  • Can I use a multiple of 2,-3,5? How can I not include that point if that is the point in which my 2 vectors intersect. – Julio Mar 18 '15 at 00:48
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    You can't use a multiple of a point unless your two lines happen to go through the origin (and neither one of these do). How can you find any other point on your lines if you already have equations for your lines? –  Mar 18 '15 at 00:49
  • [x,y,z]=(2,-3,5)+t(1,1,1) and [x,y,z]=(2,-3,5)+s(1,1,-2). Is that valid? – Julio Mar 18 '15 at 00:51
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    Or here's another question that might lead you to the solution -- how could you either prove or disprove that the point $(6,2,4)$ is on one of your lines? –  Mar 18 '15 at 00:51
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    Those are valid equations for your lines, but your problem asks you not to use the point $(2,-3,5)$ in your solutions, so you need to replace that point in both of your equations with $2$ other points that happen to be on your lines. Remember, there are an infinite number of equations which describe these same two lines. –  Mar 18 '15 at 00:52
  • The multiple of my lines? Like
    [x,y,z]=(1,1,1)+t(2,2,2) and [x,y,z]=(1,1,-2)+t(2,2,-4)     – Julio Mar 18 '15 at 00:54
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    Try to answer my other question. How could you prove or disprove that $(6,2,4)$ is a point on the line $(x,y,z)=(2,-3,5)+t(1,1,1)$? –  Mar 18 '15 at 00:55
  • The collinear equation? See if they are coplanar or collinear? – Julio Mar 18 '15 at 00:58
  • Let us continue this discussion in chat. – Julio Mar 18 '15 at 01:02