Intersection between line passing through 2 points and line passing through a point in the direction

Question

I am stuck on this question during my exam review:

Let L1 be the line passing through $$P(-1,3,-2), Q=(5,-3,10).$$ Let L2 be the line passing through $$(4,0,3)$$ in the direction of $$v= \begin{bmatrix} 3\\ -1\\ 1\\ \end{bmatrix}. $$ Do the lines intersect? If so, find their point of intersection.

Attempt: I first found the directional vector of L1, which equals n=(6,-6,12). Then, using the coordinates from P, I created a system of equations as follows: $$-1+6t = 4+3s$$

$$-3-6t=-s$$

$$-2+12t=3+s$$

and found that they do not intersect. Is this correct?

Answer 1

A line has infinitely many normal vectors in $R^3$, you meant $(6,-6,12)$ is the directional vector of $L1$.This is the correct value. However, I think there is a typo in your set up of equations:

It should be: $$-1+6t = 4+3s$$

$$3-6t=-s$$

$$-2+12t=3+s$$ This means the two lines intersect at $(1,1,2)$

Answer 2

Your second equation doesn’t match the description of the first lines. Since what you’ve written for $P$ and $Q$ do match your stated value for $\mathbf n$, (which, by the way is the line’s direction vector, not a normal to it), I conclude that it’s the equation that’s in error. It should be $3-6t=-s$, and the resulting system has the solution $s=-1$, $t=\frac13$, therefore the two lines intersect at $(1,1,2)$.