Two lines in space

Find the normal vector which is the Cross Product of (1, 0, -1) x (1, 1, 2)

n = i(0+1) -j(2-(-1)) + k(1-0) = i-3j+k

Whose magnitude is sqrt(11)

Distance between the lines is

((1-2),(2-1),(3-1))

Minimum distance is this distance projected on the normal.

|((1-2),(2-1),(3-1).n)/sqrt(11)|

= | ((-4, 1, 2).(1,-3, 1))/sqrt(11)|

= | (-4-3+2)/sqrt(11)| = 5/sqrt(11)

$|\text{Normalize}[\{1,0,-1\}\times \{1,1,2\}].(\{1,2,3\}-\{5,1,1\})|\to \frac{5}{\sqrt{11}}$ which gives the answer $16$ .

My linear algebra hated this method because he never understood how it worked. To him it was magic . The basis of the method is that two vectors are orthogonal if their dot (inner product) is 0. The part that the instructor was having problems with was understanding that if the direction vector that is prependicular to a plane is normalized (i.e., has a length of 1), then the expression used to define a plane $\bm{n}.(\bm{x}-\bm{p})$ , where $\bm{n}$ is the vector that is normal to the plane and $\bm{p}$ is a point on the plane, gives the signed distance of the point $\bm{x}$ to the plane, which evaluates to $0$ when $\bm{x}$ is in the plane.

Let P be an arbitrary point on line p and Q be an arbitrary point on line q. Then, the coordinates of P can be written P(1 + t, 2, 3-t) for some t. Similarly, we can write Q(5+s, 1+s, 1 + 2s). Therefore, the vector PQ can be written as the difference PQ = (4 + s - t, s - 1, 2*s + t 0 2). For the distance to be minimized, we need this vector PQ to be perpendicular to both line p and line q. That is, we need the dot product between PQ and the vectors parallel to p and q to be zero. This requires:

PQ dot (1, 0, -1) = 0 and

PQ dot (1, 1, 2) = 0.

Expanding the dot product and simplifying gives the equations:

s + 2*t = 6

6*s + t = 1.

The solution to this system is t = 35/11 and s = -4/11.

Substituting these values of t and s into the equations of p and q shows that the points that are closest to each other on each respective line are:

(46/11, 22/11, -2/11) and (51/11, 7/11,3/11).

Now we can use the distance formula / Pythagorean Theorem to find that the distance between these two points is 5/sqrt(11) so that a + b = 5 + 11 = 16