Consider the four points A : (2, 4, 0), B : (3, 1, 1), C : (1, 1, 3), D : (0, 5, 1). Find the
distance between the lines (AB) and (CD), i.e. the distance between the closest points on
these two lines, by using two different methods:
a) Find a pair of planes which are parallel to both lines, with the first plane containing the
line (AB) and the second plane containing the line (CD). Then find the distance between
these two planes
First, let's find AB then CD
$$AB = (1,-3,1)$$
$$CD = (-1,4,-2)$$
now let's find the planes P1 AND P2:
$$P1 = -x + 4y -2z = 14 $$
$$P2 = x -3y + z = -14 $$
How can I find the distance between two planes?
Best Answer
If the following method isn't correct, please let me know in the comments because I'm a little rusty on $3$D geometry.
To build two parallel planes including each of these two lines, we'll use the cross product of $AB$ and $CD$,
$$\mathbf{\hat n}=\mathbf{AB\times CD}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -3 & 1\\ -1 & 4 & -2 \end{vmatrix}=2\mathbf i+\mathbf j+\mathbf k,$$
and then for each plane to include each line we'll dot it with the vector $\mathbf{x-x_o}$ which lies on the plane, $\mathbf{x_o}$ being a point of each line, and impose it to be $0$ so that the vector included in the plane and $\mathbf{\hat n}$ are $\perp$: $$\pi_{AB}:\mathbf{\hat n\cdot(x-A)}=0\iff 2x+y+z=8$$ $$\pi_{CD}:\mathbf{\hat n\cdot(x-D)}=0\iff 2x+y+z=6$$
Now what's left is defining another line perpendicular to the two planes and which passes through any point of the two planes, for instance, $\mathbf A$: $\boldsymbol \ell=\mathbf{A+\lambda\hat n}$. This way, what's left is finding the point where this line intersects with $\pi_{CD}$ in view that we already know it intersects $\pi_{AB}$ at $\mathbf A$.
So:
$$2(2+2\lambda)+(4+\lambda)+\lambda=6\iff\lambda=-\dfrac{1}{3}\Longrightarrow \mathbf{P_{CD}}=\mathbf{A}-\dfrac{1}{3}(2,1,1).$$
Thus, the distance you're willing to find is
$$d(\mathbf{A,P_{CD}})=||\mathbf{P_{CD}-A}||=\dfrac{1}{3}||(2,1,1)||=\dfrac{1}{3}\sqrt{2^2+1^2+1^2}=\sqrt{\dfrac{2}{3}}\approx 0.8165\ \text{u}$$