### Video Transcript

Suppose that vector π equals one,
one, three and vector π equals four, eight, negative eight fix two sides of a
triangle. What is the area of this triangle,
to the nearest hundredth?

Okay, so in this example, these two
vectors π and π, weβre told, define two sides of a triangle. If we connect the tips of these two
vectors like this, weβve got our triangle. And our question asks us to solve
for its area. We want to combine the vectors π
and π in some way that we can solve for this. And to start doing that, letβs
recall that if two such vectors define two adjacent sides of a parallelogram like we
see here, then the magnitude of the cross product of these two vectors equals that
area.

Returning to our vectors π and π
given in this problem statement, we can say that if we were to take the magnitude of
their cross product, we would solve for this entire interior area. And notice that that is twice as
big as the area of the triangle we actually want to solve for. Written mathematically, the
magnitude of π cross π is twice the area of our triangle. Therefore, if we calculate the
magnitude of π cross π divided by two, weβll solve for the value of interest.

As our first step, letβs calculate
the cross product π cross π. In general, the cross product of
two three-dimensional vectors, weβve called them π and π here, is equal to the
determinant of this three-by-three matrix. The first row of the matrix is
populated by the unit vectors π’, π£, and π€. The second and third rows has the
corresponding components of our two three-dimensional vectors. So in our scenario with the given
vectors π and π, their cross product is the determinant of this matrix with the
components of π and π yet to be filled in.

Letβs start with the components of
our first vector, vector π. We see from the problem statement
that it has components one, one, and three. And then moving on to our second
vector, we see the components of π are four, eight, and negative eight. Our task now is to compute the
determinant of this matrix. Weβll do it component by component,
starting with the π’-component. Crossing out the row and column of
our matrix that contain this element, we can say that this component is given by the
determinant of this two-by-two matrix. That determinant equals one times
negative eight or negative eight minus three times eight or 24.

We then move on to the π£-component
equal to negative the determinant of this matrix. One times negative eight once again
is negative eight. And from that, we subtract three
times four or 12. Finally, the π€-component of this
cross product equals the determinant of this matrix. One times eight is eight. And from that, we subtract one
times four or four. We find then that the cross product
of π and π is equal to negative 32π’ plus 20π£ plus four π€. Now that weβve figured this out, we
want to solve for the magnitude of this resulting vector.

Clearing a bit of space on screen,
we can remind ourselves that if we have a three-dimensional vector, the magnitude of
that vector equals the square root of the sum of the squares of each of its three
components. This tells us that the magnitude of
π cross π equals the square root of the sum of negative 32 squared, 20 squared,
and four squared. This is equal to the square root of
1440. From here, all we need to do to
calculate the area of our triangle is divide this result by two. When we enter this expression on
our calculator to the nearest hundredth, we get a result of 18.97.

And we know that this number refers
to an area. So even though we donβt know its
units, weβll report our answer as 18.97 units of area. Thatβs the size of this triangle
where vectors π and π fix two of its sides.