Lines & Planes - Determine the distance between lines

Question

Determine the distance between the line $$(x,y,z) = (4,5,-2) + t(1,1,-1), \qquad t\in\mathbb{R}$$ and the plane $$(x,y,z) = (2,4,3)+s(3,2,0)+t(1,0,2),\qquad s,t\in\mathbb{R}.$$

Answer 1

Since this is most certainly homework, I'll just give a HINT :

If the line and the plane are not parallel their distance is $0$ because they meet in one point.

If they are parallel to compute their distance you may just compute the distance from any point on the line to the plane. For this there's a formula (that you should know or be able to find in your notes/textbook) once the plane is given in the cartesian form $ax+by+cz+d=0$.

To test whether the line and the plane are perpendicular, just check the orthogonality between the direction perpendicular to the plane and that of the line.

Answer 2

The line is parallel to the plane since $$ (1,1,-1) = \frac{1}{2}(3,2,0)-\frac{1}{2}(1,0,2), $$ hence it is enough to compute the distance of $(4,5,-2)-(2,4,3)=(2,1,-5)$ from the plane $\pi=\text{Span}\left((3,2,0),(1,0,2)\right)$. A possible approach is to find an orthonormal base of $\pi$, complete it to an orthonormal base of $\mathbb{R}^3$ and represent $(2,1,-5)$ with respect to such a base. An equivalent approach is to find the minimum of

$$ \left\|(2,1,-5)-s(3,2,0)-t(1,0,2)\right\|^2 = 30-16 s+13 s^2+16 t+6 s t+5 t^2 $$ by annihilating the partial derivatives and finding $s=\frac{8}{7},t=-\frac{16}{7}$.

It follows that the wanted distance is $\color{red}{\large\sqrt{\frac{18}{7}}}$.

Answer 3

The cross product of the two vectors $(3,2,0)$ and $(1,0,2)$ gives a normal vector $N(4,-6,-2)$ to plane $(\Pi)$. The dot product of $N$ with the directing vector of the straight line : $(1,1,-1)$ is 0 ; thus we have checked that the straight line and the plane are parallel.

It suffices now to compute the distance from any point, e.g., $P_0(x_0,y_0,z_0)=(4,5,-2)$ of the line to the plane $(\Pi)$ using the formula given in this reference:

$$d((P_0),(\Pi))=\dfrac{|a x_0+b y_0+c z_0+d|}{\sqrt{a^2+b^2+c^2}} \ \ (1)$$

if the plane has cartesian equation $ax+by+cz+d=0$.

But this cartesian equation is very easy to obtain because we have the normal vector $N$ : it is $4x-6y-2z+d=0$ with $d$ easily computed by forcing (2,4,3) to belong to this plane, yielding the following cartesian equation of the plane:

$$4x-6y-2z+22=0$$

It suffices now to plug values $a=4, b=-6, c=-2, d=22$ in (1) to obtain the result... up to you...