I know this is a long-dead and well-answered question, but I woke this morning thinking "All of the answers involve an "if" statement, i.e., all are discontinuous functions of the inputs. Is there a continuous answer, i.e., a continuous function
$$
R: S^2 \times S^2 \to SO(3)
$$
that takes a pair of vectors $u, v$ to a rotation matrix $R(u, v)$ with the property that $R(u, v) u = v$?"
The paper that Tomas Moller and I wrote back in 1999, for instance,
uses "the coordinate vector corresponding to the smallest entry of $w$" for some vector $w$, which doesn't vary continuously as a function of $w$. And I wondered, "Did we really do as well as possible, or might there have been a continuous solution?"
The answer is "no." But the proof uses a bit of topology.
Fix the vector $u$ (set it to be $e_1$, for instance), and look at the map
$$
K : S^2 \to SO(3) : v \mapsto R(u, v).
$$
Then compose this with the map
$$
H : SO(3) \to S^2 : M \mapsto Mu.
$$
The composite map
$$
H\circ K: S^2 \to S^2
$$
is $v \mapsto R(u, v)u = v$, i.e., the identity map on $S^2$. But that means that
$$
(H\circ K)_{*} : H_2(S^2) \to H_2(S^2),
$$
the induced map on second homology, must be the identify from $\mathbb Z $ to $\mathbb Z$. But since
$$
(H\circ K)_{*} = H_{*} \circ K_{*}$$
this map must factor through $H_2(SO(3)) = 0$, which is impossible.
Thus: There's no continuous solution to the "rotate one vector to another" problem, a fact that I should have mentioned back in our original paper. Sigh. Hindshight is 20-20.
So I figured answer:
(iii) Using the distance formula, the shortest distance D from B to π1 is given by:
D = |ax + by + cz + d| / |n| where n is the normal to plane; a, b, c, d are the coefficients of the equation of the plane; x, y, z are the coordinates of the point from the plane.
π1: 8i + j - 14k = 75
B: 2i + 3j + 5k
D = |(8)(2) + (1)(3) + (-14)(5) + (-75)| / sqrt(64+1+196)
D = |-126| / sqrt(261)
D = 126/16.155 = 7.799 ~= 7.80
Best Answer
Yes, it can – if you form the determinant of
$$ \pmatrix{u_1&v_1&x_1\\u_2&v_2&x_2\\u_3&v_3&x_3}\;, $$
where $u$ and $v$ are the two given unit vectors, the coefficients of $x_1$, $x_2$ and $x_3$ form $u\times v$.