The vector $l_1$ is given as $$ l_1 : \mathbf{r} = \left( \begin{matrix} -3\\-4\\6 \end{matrix} \right)\; + \lambda \left( \begin{matrix} -3\\-4\\6 \end{matrix} \right) $$ and $l_2$ as $$ l_2: \frac{x-4}{-3}=\frac{y+7}{4}=-(z+3) $$

So far I've gotten an equation for each $(x,y,z)$ of $l_1$ and $l_2$ and then multiplied them to get a scalar product however I'm having issues finding the values of $\lambda$ and $\nu$ (Cartesian multiplier of $l_2$) through the simultaneous equations of finding the scalar product (when it's equal to $0$).

My approach, at least so far has given, $$ \vec{OB} = \left(\begin{matrix} 4\\ -7\\ -3 \end{matrix}\right)+\nu \left(\begin{matrix} -3\\ 4\\ -1 \end{matrix}\right) $$ $$ \vec{AB} = \vec{OB}-\vec{AB} $$ Therefore $$ \vec{AB} = \left(\begin{matrix} 1\\ 3\\ 9 \end{matrix}\right)+ \left(\begin{matrix} -3\nu - 3\lambda\\ 4\nu - 2\lambda\\ -\nu + 2\lambda \end{matrix}\right) $$ $$ x = 1 -3\nu - 3 \lambda,\; y = -3 + 4\nu - 2\lambda, \; z=-9+\nu + 2\lambda $$ Then solving for the two dot products, $$ \vec{AB} \; \cdot \; l_1, \; \vec{AB} \; \cdot \; l_2 $$ I get $$ 26\nu^2-17\lambda^2 = 0 $$ Which I have no idea how to solve. The response was too long for the comment section.

  • $\begingroup$ Your method is right. What issue do you find? $\endgroup$ – Yuta Sep 9 '18 at 13:12
  • $\begingroup$ Couldn't simplify further than having two composite quadratics, perhaps I made an arithmetic mistake but I don't see it after going through many times. $\endgroup$ – John Miller Sep 9 '18 at 13:18
  • $\begingroup$ There should not be any quadratic terms. Can you show some detail of your steps? $\endgroup$ – Yuta Sep 9 '18 at 13:21
  • $\begingroup$ @Yuta I have edited my answer. $\endgroup$ – John Miller Sep 9 '18 at 13:50
  • $\begingroup$ Since $AB\perp l_1$ and $AB\perp l_2$, dot products should be considered instead. $\endgroup$ – Yuta Sep 9 '18 at 13:53

A (slightly) different approach:

$l_2$ is defined as the intersection of the planes $x-3z=7$ and $y+4z=-19$, with normal vectors $\vec n=(1,0,-3)$ and $\vec n'=(0,1,4)$ respectively. Therefore, a directing vector of $l_2$ is the cross-product $\vec u_2=\vec n\times\vec n'$.

Let $\vec u_1=(-3,-4,6)$ the given directing vector of $l_1$. A directing vector of the common perpendicular of $l_1$ and $l_2$ is the cross-product $$\vec v=\vec u_1\times\vec u_2. $$ Thus, to find points $A$ on $l_1$ and $B$ on $l_2$ so that the line $(AB)$ is perpendicular to $l_1$ and $l_2$, we have to find a point $A$ (given by the parametric equation) and $t$ so the point $$B=A+t\vec v= A+t(\vec u_1\times\vec u_2)$$ satisfies the equations of line $l_2$.

  • $\begingroup$ @John Miller: Where did you find 3 equations and 7 variables? You should obtain the two equations which define $l_2$ and two unknowns, $\lambda$ and $t$. $\endgroup$ – Bernard Sep 9 '18 at 16:22
  • $\begingroup$ sorry I never wrote it on paper as I was in the car but I have solved the question thank you. $\endgroup$ – John Miller Sep 9 '18 at 18:08

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