The lines l1 and l2 have equations r= 6i -3j + s(3i - 4j - 2k ) and r= 2i -j - 4k + t(i-3j-k )

The position vector of a point P on line L1 is 3i+j+2k. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. Find the position vector of Q.

This is the part I'm having trouble with :

Find, in the form r = a + kb + uc, an equation of the plane which passes through P and is perpendicular to L1.

Okay, so I need two linearly independent vectors lying in the plane. One is PQ, as it is perpendicular to l1, which means that it lies in the plane. Another can be the cross product of the direction vector of l1, and any other vector. The resultant vector will be perpendicular to l1, and will also lie in the plane. But this will give me a different answer depending on the second vector I use.. Why is this not correct?

  • $\begingroup$ I don't quite understand you question. Technically, for a given line, there is only one plane that is perpendicular to it and passes through a given point. Then, there are infinitely many lines (vectors?) that lie in that plane and pass through that point. Here, I believe what you want to find is a line. Since a vector can translate freely while not changing itself, you can not say a vector "passes through" a point. $\endgroup$
    – Huang
    May 21, 2017 at 23:47
  • $\begingroup$ Yeah, I meant a line. Sorry. $\endgroup$
    – user440261
    May 22, 2017 at 0:29

1 Answer 1


So you have:

$$ L1: r=6i-3j+0k+s(3i-4j-2k) L2: r=2i-j-4k+t(i-3j-k) $$

Where $s$ and $t$ are the scalar parameter of your lines.

Obviously P is at the L1 line by choosing s=-1: $$ P: 3i+j+2k $$

PQ should be perpendicular to L1 and L2, hence the cross product between the directions will serve as new perpendicular direction: $$ PQ: cross([3, -4, -2],[1, -3, -1])=[-2, -1, -5] $$

Hence, you obtain the perpendicular line, starting from P. This line passes through P: $$ L_{PQ}=P+tPQ=P: 3i+j+2k+t(-2i-1j-5k) $$

Now we need to locate the point in $L_{PQ}$ passing through L2:

$$ L2: r=2i-j-4k+s(i-3j-k) L_{PQ}=3i+j+2k+t(-2i-1j-5k) $$


$$ 2+s=3-2t\\ -1-3s=1-t\\ -4-s=2-5t\\ $$


$$ 1+3: -2=5-7t, t=1\\ 1,(1+3): s=-1\\ 2,1,(1+3): 2=0 $$

Thus, a line perpendicular both to L1 and L2, passing through P do not cross with a Q in L2.

The plane passing through P and perpendicular to L1. Indeed several solutions exist, but lets take the easiest ones from the cross product of the L1 direction and the x, y and z axis.

$$ B1=cross([3, -4, -2],[1, 0, 0])=2[0, -1, 2]\\ B2=cross([3, -4, -2],[0, 1, 0])=[2, 0, 3]\\ B3=cross([3, -4, -2],[0, 0, 1])=[-4, -3, 0] $$

All B1, B2 & B3 serve us. Hence we take B1 and B2, plus P, thus the plane is defined. Note we took B1/2:

$$ F: P+sB1+tB2= 3i+j+2k + s(0i-j+2k) + t(2i+0j+3kj) $$

Edit: note that vectors B1 and B2 and given answer C1=[-22, -19, 5] and C2=PQ=[2 ,-1 ,5] are related through:

$$ C1=19B1-11B2\\ C2=B1+B2 $$

Thus, both C1,C2 are linearly dependent with B1,B2, hence defining the same solution: $$ F: 3i+j+2k + s(0i-j+2k) + t(2i+0j+3kj)\\ = 3i+j+2k + (19s'+t')(0i-j+2k) + (-11s'+t')(2i+0j+3kj)\\ = 3i+j+2k + s'(-22i-19j+5k) + t'(2i-1j+5kj) $$

  • $\begingroup$ The answer to the last part ( the part I want ) is wrong. The given answer is: r = 3i + j +2k + k(-22i -19k + 5k ) + u( 2i-j+5k) The two vectors used are the direction vector of PQ, and the cross product PQ x L1. Also, how do you get that the direction of the L1 vector is 6i -30j? It should be 3i - 4j -2k $\endgroup$
    – user440261
    May 22, 2017 at 0:35
  • $\begingroup$ I will check asap $\endgroup$
    – Brethlosze
    May 22, 2017 at 0:40
  • $\begingroup$ Checked, now this is corrected. Note the given solution for the plane is the same as the given. $\endgroup$
    – Brethlosze
    May 22, 2017 at 1:15
  • $\begingroup$ Thank you! I have a question though. What if you used one vector as the direction vector of the line PQ i.e (2,1,5)? $\endgroup$
    – user440261
    May 22, 2017 at 4:06
  • $\begingroup$ check the edit :) $\endgroup$
    – Brethlosze
    May 22, 2017 at 4:37

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