### Video Transcript

If vector π equals three, four,
negative four; vector π equals two, five, negative four; and vector π equals
negative four, negative four, two, find π minus π cross π minus π.

Okay, so here we have these three
vectors π, π, and π written out by their components. So, for example, considering vector
π, this is its π’-component, this is its π£-component, and this is its
π€-component. Our goal is to calculate this cross
product, where the two vectors involved are differences between two of our π, π,
and π vectors.

What we can do here is give these
differences their own names. Weβll call π minus π vector π,
and π minus π weβll call vector π. What weβll eventually do then is
cross π and π. But before that, weβll need to
figure out what each of these vectors are. Starting with vector π, we know
that thatβs equal to vector π, which is three, four, negative four, minus vector
π, which is two, five, negative four. Working component by component,
three minus two is one, four minus five is negative one, and negative four minus
negative four is zero. So thatβs vector π.

And vector π is equal to vector π
here minus vector π here. Negative four minus three is
negative seven, negative four minus four is negative eight, and then two minus
negative four is positive six. So we now have the two vectors π
and π. And weβll write them off to the
side that will combine in our cross product. To do that, letβs recall this rule
for the three-dimensional cross product of two vectors. Weβve called them π and π
here. This cross product equals the
determinant of a three-by-three matrix.

In the first row, we have our unit
vectors, in the second the corresponding components of vector π, and in the third
those of vector π. So when we set up the cross product
of π and π, weβll once again have π’, π£, and π€ unit vectors at the top of each
column. And in the next row, weβll write in
the components of vector π. Those are one, negative one, and
zero. Lastly, the components of vector
π: negative seven, negative eight, positive six.

Computing this cross product is now
a matter of calculating the determinant of this matrix. Beginning with the π’-component, if
we cross out the row and column that contain this element, then this component is
equal to the determinant of the two-by-two matrix remaining. This is equal to negative one times
six or negative six minus zero times negative eight, or zero. So the π’-component of our
resulting vector is negative six.

And then we move on to calculate
the π£-component. This has a value of negative one
times one times six, or six, minus zero times negative seven, or zero. This comes out to negative six
π£. And lastly, for our π€-component,
this is equal to one times negative eight, or negative eight, minus negative one
times negative seven, which comes out to negative seven. Our π€-component then simplifies to
negative 15.

And now weβve computed all three
components of a resulting cross product. The vector we end up with is
negative six π’ minus six π£ minus 15π€. Going back to our original three
vectors π, π, and π, this equals π minus π cross π minus π.