### Video Transcript

If vector π¨ equals negative two,
two, negative one; vector π© equals negative four, negative four, negative five; and
vector π equals negative four, two, four, find π¨ cross π© dotted with π¨ cross
π.

Alright, to calculate this answer,
weβll need to take two cross products, π¨ cross π© and π¨ cross π, and then a dot
product between the two resulting vectors. As we get started then, letβs
recall how to calculate a three-dimensional cross product. If we have two three-dimensional
vectors, weβve called them here π¨ and π©, then their cross product equals the
determinant of this three-by-three matrix. The top row of the matrix has the
three orthogonal unit vectors π’, π£, and π€. After that comes the corresponding
components of vector π¨ and then the corresponding components of vector π©. And note that this order in which
the last two rows are filled matters. π¨ cross π© is different from π©
cross π¨.

But anyway, this relationship gives
us a template for calculating π¨ cross π© and then π¨ cross π. π¨ cross π© is given by this
matrix, where we still need to fill in the components of π¨ and π©. Our problem statement tells us that
the components of π¨ are negative two, two, and negative one and those of vector π©
are negative four, negative four, negative five. Weβre now ready to calculate this
cross product, starting with the π’-component. Itβs equal to the determinant of
this two-by-two matrix. Two times negative five is negative
10. And from that we subtract negative
one times negative four or positive four.

Now, weβll move on to the
π£-component, which is equal to negative the determinant of this matrix. Negative two times negative five is
positive 10 and from that we subtract negative one times negative four or four, and
then lastly the π€-component equal to the determinant of this matrix. Negative two times negative four is
positive eight minus two times negative four or negative eight. Combining these components this way
gives us π¨ cross π©, which we see simplifies to negative 14π’ minus six π£ plus
16π€.

So then, weβve calculated our first
cross product π¨ cross π©. Now letβs move on to π¨ cross
π. This matrix looks identical to the
matrix for π¨ cross π© except for our last row. Note that here we have the
components of vector π, negative four, positive two, and positive four. Once again, weβll calculate this
cross product starting with the π’-component. The determinant of this two-by-two
matrix is two times four or eight minus negative one times positive two or negative
two. Our π£-component is given as
negative the determinant of this matrix, where that determinant is negative two
times four or negative eight minus negative one times negative four or positive
four.

Finally, the π€-component with this
determinant. Negative two times two is negative
four minus two times negative four or negative eight. Combined this way, these components
add up to the cross product π¨ cross π. They simplify to 10π’ plus 12π£
plus four π€. Weβre now ready to take our last
mathematical step, computing the dot product of these two cross products. Recalling that in general the dot
product of two three-dimensional vectors equals the product of their respective
components added together, we can say that the dot product of π¨ cross π© and π¨
cross π equals negative 14 times 10 plus negative six times 12 plus 16 times
four.

Notice that weβve multiplied
together the corresponding components of our two vectors. And when we add these products
together, we find theyβre equal to negative 148. And thatβs our answer. The dot product of π¨ cross π© and
π¨ cross π is negative 148.