So I have this maths question that I can't seem to wrap my head around. I have two lines:

Line 1 $= r = 3i+2j+7k + X(i-j+3)$

Line 2 $= r = 6i+5j+2k + Y(2i+j-k)$

I have found the point of intersection when $X=-1$ and $Y=-2$, and thus the point of intersection to be $2i+3j+4k$

Now I have been told that a vector $(i+aj+bk)$ is perpendicular to both, and to find $a$ and $b$. I understand how this is, but can't seem to find the answer. I can't use dot product with two unknowns, so I would really appreciate some help.

P.S I'm not sure how you're properly supposed to structure these questions with the fancy format so if anyone could link me to how I am supposed to lay questions out please do.

  • $\begingroup$ Hello @Harvey Stanfield, here is the link that explains how to use MathJax. $\endgroup$ – Ernie060 Oct 22 '18 at 21:34

Found out the answer, I was just calculating the wrong thing over and over.

I used the dot product with each of the directional vectors of my two lines and equated them to 0, thus making two simultaneous equations both with terms of a and b.

Then the process is simple.

  • $\begingroup$ Your method works. My idea was the desired vector should be some scalar multiple of the cross product of the direction vectors. Then adjust to match first component $i$. [should give same answer, but either way OK.] (+1 on your answer) $\endgroup$ – coffeemath Oct 22 '18 at 22:17
  • $\begingroup$ Please write out all of your work, so that others may see the solution. Otherwise, this isn't an answer. $\endgroup$ – Chickenmancer Oct 22 '18 at 22:47

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