The problem is as following : Find the line that contains the point A(-4,-5,3) and intersects the lines:
p: x=3t-1 , y=-2t-3 , z=-t+2 and the line q: x=2t+2 , y=2t-1 , z=-5t+1
I have been dealing with a good amount of problems with lines and planes in 3D, but I have no idea how to approach this one. I have no idea how to get the direction vector that's missing or just another point on the line I have to find. Any help would be appreciated. Thanks in prior.
If a pair of lines intersect, then they are coplanar. The line you’re looking for therefore lies on the intersection of the two planes defined by $A$ and the two given lines. Knowing this, there are several ways to go about finding it. A straightforward method takes advantage of the fact that you’re working in $\mathbb R^3$: the solution must be perpendicular to both plane normals, so its direction vector is the cross product of those normals. To compute the two normals, you need two points on each of the two lines, which you can generate by choosing convenient values of $t$. You’re given a point on the line, namely $A$, so once you have its direction vector, you’re done. Equivalently, compute one of the these two planes and the intersection of the other line with it. The line that you’re looking for must pass through this point and $A$.
$L_P:\overrightarrow r = - \hat i - 3\hat j + 2\hat k + s\left( {3\hat i - 2\hat j - \hat k} \right) = \vec c + s\vec b$
$L_Q:\overrightarrow r = 2\hat i - \hat j + \hat k + t\left( {2\hat i + 2\hat j - 5\hat k} \right) = \vec a + t\vec d$
$L_R:\overrightarrow r = - 4\hat i - 5\hat j + 3\hat k + u\left( {a\hat i + b\hat j + c\hat k} \right) = \vec e + u\vec f$
$\vec f \times \vec d = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ a&b&c\\ 2&2&{ - 5} \end{array}} \right| = - \left( {5b + 2c} \right)\hat i + \left( {5a + 2c} \right)\hat j + \left( {2a - 2b} \right)\hat k \& \vec e - \vec a = - 6\hat i - 4\hat j + 2\hat k$
$\left( {\vec e - \vec a} \right).\left( {\vec f \times \vec d} \right) = 0 \Rightarrow 6\left( {5b + 2c} \right) - 4\left( {5a + 2c} \right) + 2\left( {2a - 2b} \right) = 0$
$30b + 12c - 20a - 8c + 4a - 4b = 0 \Rightarrow - 16a + 26b + 4c = 0...(i)$
$\vec f \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ a&b&c\\ 3&{ - 2}&{ - 1} \end{array}} \right| = \left( { - b + 2c} \right)\hat i + \left( {a + 3c} \right)\hat j - \left( {2a + 3b} \right)\hat k$
$\vec e - \vec c = - 3\hat i - 2\hat j + \hat k$
$\left( {\vec e - \vec c} \right).\left( {\vec f \times \vec b} \right) = 0 \Rightarrow - 3\left( { - b + 2c} \right) - 2\left( {a + 3c} \right) - \left( {2a + 3b} \right) = 0$
$3b - 6c - 2a - 6c - 2a - 3b = 0 \Rightarrow - 4a - 12c = 0...(ii)$
$c = k;a = - 3k$
$ \Rightarrow - 16a + 26b + 4c = 0 \Rightarrow - 16a \times \left( { - 3k} \right) + 26b + 4k = 0 \Rightarrow b = - \frac{{52k}}{{26}} \Rightarrow b = - 2k$
$\left\langle {a,b,c} \right\rangle = \left\langle { - 3, - 2,1} \right\rangle $
$L_R:\overrightarrow r = - 4\hat i - 5\hat j + 3\hat k + u\left( { - 3\hat i - 2\hat j + \hat k} \right) \Rightarrow {L_R}:\frac{{x + 4}}{{ - 3}} = \frac{{y + 5}}{{ - 2}} = \frac{{z - 3}}{1}$
This is the final equation
Let's say you have $$\left \lbrace \begin{aligned} p_x(t) &= 3 t - 1 \\ p_y(t) &= -2 t - 3 \\ p_z(t) &= -t + 2 \end{aligned} \right . , \quad \left \lbrace \begin{aligned} q_x(t) &= 2 t + 1 \\ q_y(t) &= 2 t - 1 \\ q_z(t) &= -5 t + 1 \end{aligned} \right .$$
The line passes through $(-4, -5, 3)$, but we do not know its direction. Let's say its direction components are $x$, $y$, and $z$. The line is then $$\left \lbrace \begin{aligned} L_x(t) &= x t - 4 \\ L_y(t) &= y t - 5 \\ L_z(t) &= z t + 3 \end{aligned} \right .$$
When the three intersect, the $t$ is specific to each line. Therefore, we need to use e.g. $p(t_1) = L(t_2)$ for the point where line $p$ intersects line $L$, and $q(t_3) = L(t_4)$ for the point where line $q$ intersects line $L$. Essentially, we get six equations with seven unknowns $t_1$, $t_2$, $t_3$, $t_4$, $x$, $y$, and $z$: $$\left \lbrace \begin{aligned} p_x(t_1) &= L_x(t_2) \\ p_y(t_1) &= L_y(t_2) \\ p_z(t_1) &= L_z(t_2) \\ q_x(t_3) &= L_x(t_4) \\ q_y(t_3) &= L_y(t_4) \\ q_z(t_3) &= L_z(t_4) \end{aligned} \right .$$ We can fix the extra unknown by controlling the magnitude of $(x, y, z)$, i.e. $\sqrt{x^2 + y^2 + z^2}$.
The easiest way to do that is to use $t_2 = 1$. This is because $t_2$ cannot be zero, as $p$ does not pass through $(-4, -5, 3)$. Fixing $t_2$ to a nonzero value simply fixes the magnitude of $(x, y, z)$.
Alternatively, you can add a seventh equation, for example $x^2 + y^2 + z^2 = 1$, or you can solve the system of equations, leaving $t_2$ as a free variable (so one or more of the other variables will depend on $t_2$).
While the system of six equations with six (or seven) unknowns is a bit large, the equations are linear and quite simple if using $t_2$, and should be solvable using back substitution, or other standard methods.
If I ask Maple for the solution, using $t_2 = 1$, I get $$\left \lbrace \begin{aligned} t_1 &= -\frac{2}{23} \\ t_3 &= -\frac{1}{13} \\ t_4 &= \frac{23}{13} \\ x &= \frac{63}{23} \\ y &= \frac{50}{23} \\ z &= -\frac{21}{23} \end{aligned} \right .$$ so the equation of the line sought is $$\left \lbrace \begin{aligned} L_x(t) &= \frac{63}{23} t - 4 \\ L_y(t) &= \frac{50}{23} t - 5 \\ L_z(t) &= -\frac{21}{23} t + 3 \end{aligned} \right .$$ i.e. $L(t) = (t 63/23 - 4, t 50/23 - 5, - t 22/13 + 3)$.
Line $L$ intersects line $p$ at $(-29/23, -65/23, 48/23) \approx (-1.261, -2.826, 2.087)$, and line $q$ at $(11/13, -15/13, 18/13) \approx (0.846, -1.154, 1.384)$.
