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Basically I have to find a line that passes through a point and intersects two lines.

The Point $P=(6,-2,-2)$

The first Line is

$$x = 1+t$$ $$y = 5 + 2t$$ $$z = 3 +4t$$

The second line is

$$x = 3 + 3s $$ $$y = 1 + s $$ $$z = 4 + 2s $$

I have to express the answer in parametric form:

$$x = 6 + Aw $$ $$y = -2 + Bw $$ $$z = -2 + Cw $$

I basically have to find $A$, $B$ and $C$.

Thanks!

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  • $\begingroup$ I have no idea how to start it, I'm clueless at this point, the teacher gave us a suggestion, like find the perpendicular from first line to the third line and the second line to the third line and [A,B,C] would equal the perpendicular from first line to third and second line to third $\endgroup$ – Shawn Edward Nov 8 '15 at 20:43
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The plane containing all lines through $P$ and some point on the first line has normal vector $$ \vec n_{\alpha}= \begin{pmatrix} 1-6\\ 5-(-2)\\ 3-(-2) \end{pmatrix} \times \begin{pmatrix} 1\\ 2\\ 4 \end{pmatrix} $$ When this cross product has been computed, a similar computation can be carried out to determine a normal vector $\vec n_{\beta}$ for the plane containing the point and the second line. Then a direction vector for the line of intersection can be found as $$ \vec r=\vec n_{\alpha}\times\vec n_{\beta} $$ and the coordinates of $\vec r$ should contain the figures $A,B,C$ that you seek.


I did just that and ended up with $\vec r=\langle 1,2,4\rangle^T$ after having reduced the size of the vector. If this is correct, the problem has no solutions since then the line through $P$ never intersects the first line, as it is parallel to it.

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  • $\begingroup$ in the first bracket, you are subtracting from Point P from the first line and then doing the cross product of that to (1,2,4)? $\endgroup$ – Shawn Edward Nov 8 '15 at 21:03
  • $\begingroup$ @ShawnEdward: Yes. Let $Q=(1,5,3)$ denote immediately known point on the first line. Then $\overrightarrow{PQ}$ and $\langle 1,2,4\rangle$ are vectors in the first mentioned plane, so their cross product becomes a normal vector for that plane. $\endgroup$ – String Nov 8 '15 at 21:06
  • $\begingroup$ When you finding PQ isn't it Q - P, so like in the first bracket, wouldn't it be 1-6, 5- (-2) and so on..? $\endgroup$ – Shawn Edward Nov 8 '15 at 21:17
  • $\begingroup$ @ShawnEdward: Yes, you are right about that :o) Note that $\overrightarrow{QP}$ works equally well. $\endgroup$ – String Nov 8 '15 at 21:24
  • $\begingroup$ for some reason I didn't get the answer you got $\endgroup$ – Shawn Edward Nov 8 '15 at 23:17
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Hint: define a plane containing the point and first line, the same for 2nd line

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