**Problem : **
Find a vector which is perpendicular to both
*u* = (3, 0, 2)
and
*v* = (1, 1, 1)
.

The vector

*u*×*v*
is perpendicular to both

*u*
and

*v*
, so we need only compute
the cross product to do this problem. From the component formula,

*u*×*v* = (- 2, - 1, 3)
. Using the dot product we can check that this vector is indeed
perpendicular to

*u*
and

*v*
.

**Problem : **
A triangle has two sides of length 5 and 6. If the triangle's area is 12, what
is the angle between these two sides?

The given sides of the triangle can be thought of as two vectors

*u*
and

*v*
with magnitudes 5 and 6, respectively. From the geometric formula for the cross
product, we know that the area of the parallelogram defined by these vectors is
given by

*A* = | *u*|| *v*| sin*θ* = 30 sin*θ*
. The area

*A*
of the parallelogram
is exactly twice the area of the triangle in question. Hence we can solve for

sin*θ* = 24/30 = 4/5
.