# Find the coordinates of the points $A$ and $B$ on the lines $l_1$ and $l_2$ respectively such that $\vec{AB}$ is perpendicular to $l_1$ and $l_2$

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.

• Your method is right. What issue do you find? – Yuta Sep 9 '18 at 13:12
• 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. – John Miller Sep 9 '18 at 13:18
• There should not be any quadratic terms. Can you show some detail of your steps? – Yuta Sep 9 '18 at 13:21
• @Yuta I have edited my answer. – John Miller Sep 9 '18 at 13:50
• Since $AB\perp l_1$ and $AB\perp l_2$, dot products should be considered instead. – Yuta Sep 9 '18 at 13:53

$l_2$ 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 $l_2$.
• @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 $t$. – Bernard Sep 9 '18 at 16:22