How to find the equation of a line which intersects these lines at 90 degrees?

Question

How to find the equation of a line which intersects these lines at 90 degrees?

$p\equiv \dfrac{x}{2}=\dfrac{y+1}{0}=\dfrac{z-2}{1}$

$q\equiv \dfrac{x-1}{1}=\dfrac{y-2}{1}=\dfrac{z+5}{0}$

Since the third line has to be vertical on both of these lines, it has to be that its vector of direction is $\mathbf{c=}\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 1 \\ 1 & 1 & 0 \\ \notag \end{vmatrix}$ and I get $\mathbf{c}=-\mathbf{i}+\mathbf{j}+2\mathbf{k}$. And now I'm stuck.

Answer 1

Let $$\mathbf r_1=\mathbf u_1+s\mathbf v_1$$$$\mathbf r_2=\mathbf u_2+t\mathbf v_2$$the position vectors of the generic points of the lines.

Then solve the system $$\begin {cases} (\mathbf r_1-\mathbf r_2) \cdot \mathbf v_1=0 \\ \\ (\mathbf r_1-\mathbf r_2) \cdot \mathbf v_2=0\end {cases}$$

Finally substitutes the values of $s$ and $t$ to find the intersection points and so the equation of the line passing through them.

Answer 2

In general, a line in 3 dimensions given by $\frac{x-x_0}{A}= \frac{y- y_0}{B}= \frac{z- z_0}{B}$ is the same as that given by $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$ where t is the common value of those fractions. Here, with $\frac{x}{2}= \frac{y- 1}{0}= \frac{z- 2}{1}$ is the same as x= 2t, y= 1, z= t+ 2. This has "direction vector" <2, 0, 1>, the denominators of those fractions.

Similarly, $\frac{x- 1}{1}= \frac{y- 2}{1}= \frac{z+ 5}{0}$ is the same as x= t+ 1, y= t+ 2, z= 5. This has "direction vector" <1, 1, 0>.

Any vector that is perpendicular is in the direction of the cross product of those vectors.

But since these lines are skew- they do not intersect and are not parallel, you will need to find the unique point on each line where there is a common perpendicular.