Plane and solid analytic geometry, by William F. Osgood and William C. Graustein.

DIRECTION COMPONENTS

439

be remembered. For, by advancing the numbers of this set according to the cyclic order 1 2 3 1 2 * *, this set 2 3 becomes
3 1, i.e. the second set; and advancing the second set 3 1
cyclicly, we get the third set, 1 2.
Critique. Not all the determinants (6) are zero, for if
11m2 - 12m min2 - m2nl = n112 - n2 = 0,
then. 1:12 = m: m2= n1:n2
or ll: mi: nI = 12 2: n2
and hence the lines L1 and L2 would be the same or parallel,
which is contrary to hypothesis.
In obtaining the solution (5) of equations (4) we assumed,
tacitly, that 11m2 - 12ml =- 0; there was, then, a set of components 1, m, n, for which n = 1. If 1Im2 - 12m = 0, at least
one of the two remaining determinants cannot be zero. If,
for example, nll - n211 z- 0, there will be a set of components
1, m, n, for which m = 1, and we can find this set by putting
m =1 in (3) and solving the resulting equations for 1 and n.
EXERCISES
In each of the following exercises determine the direction
components of a line which is perpendicular to each of two lines
having the given direction components. Actually solve the
equations and then check the result by the rule of thumb.
f3, 4, 2 2 5, 6, -3; 3. 2, -1;
l, 2, 3. 2, -4, -1. - 3, -1, 2.
f2, 1, -1;, 1, 0; 6 3, 0, 2;
4. 4- 4, 2 3. 0 j, 0, 1. 1, 0, -1.
7. The two directed lines L1 and L2 passing through the
origin and having respectively the direction cosines ~V/2,
-/2, 0 and — A/2, a/V2, 0 are perpendicular to each other.
Find the direction cosines of a third line L3 through the origin,
perpendicular to both L1 and L2, and so directed that L1, L2, L3