### Video Transcript

In this video, weโll learn how to
calculate the scalar triple product and apply this in geometrical applications.

The scalar triple product,
sometimes also called the mixed product or the box product, is the scalar or dot
product of one vector with the cross product of another two vectors. You should already be familiar with
the scalar and cross product of vectors. The scalar triple product of three
vectors ๐, ๐, and ๐ is defined as the scalar product or the dot product of vector
๐ with the cross product of vectors ๐ and ๐.

If we break this down, we know that
the result of the cross product of two vectors ๐ and ๐ is another vector. Letโs call that ๐. If we then take the scalar product
of vector ๐ with vector ๐, we know that the scalar or dot product results in a
scalar. Letโs call that ๐. So the scalar triple product is in
fact a scalar, hence the name scalar triple product. Itโs worth noting that we donโt
actually need the parentheses since if we tried to take the dot product, that is the
scalar of product of ๐ with ๐ first, this would give us a scalar. We would then have a scalar crossed
with a vector, which we cannot do, since itโs not possible to cross a scalar with a
vector.

And so the scalar triple product is
the scalar product of vector ๐ with the cross product of vectors ๐ and ๐. Now letโs see what the scalar
triple product looks like in component form. If ๐ข, ๐ฃ, and ๐ค are the unit
vectors in the ๐ฅ-, ๐ฆ-, and ๐ง-directions, then our three vectors are written in
component form as shown. And if we write the cross product
of vectors ๐ and ๐ in component form, we have ๐ cross ๐ is ๐ ๐ฆ ๐ ๐ง minus ๐
๐ง ๐ ๐ฆ times ๐ข minus ๐ ๐ฅ ๐ ๐ง minus ๐ ๐ง ๐ ๐ฅ times ๐ฃ plus ๐ ๐ฅ ๐ ๐ฆ
minus ๐ ๐ฆ ๐ ๐ฅ times ๐ค. And now if we take the scalar
product of the vector ๐ with this, that is, we multiply the coefficients of each of
the unit vectors together so that for example our first term is ๐ ๐ฅ times ๐ ๐ฆ ๐
๐ง minus ๐ ๐ง ๐ ๐ฆ.

But what does this remind us
of? Well, remember that the determinant
of a two-by-two matrix with elements ๐, ๐, ๐, ๐ is ๐๐ minus ๐๐. So for example, in our first term
๐ ๐ฆ ๐ ๐ง minus ๐ ๐ง ๐ ๐ฆ is actually the determinant of the two-by-two matrix
with elements ๐ ๐ฆ, ๐ ๐ง, ๐ ๐ฆ, ๐ ๐ง. And so our first term in the scalar
triple product is ๐ ๐ฅ times the determinant of the matrix with elements ๐ ๐ฆ, ๐
๐ง, ๐ ๐ฆ, ๐ ๐ง, and similarly for our second two terms.

And now this should look familiar
because this is the determinant of the three-by-three matrix consisting of the
components of the vectors ๐, ๐, and ๐. So in fact, the scalar triple
product of the three vectors ๐, ๐, and ๐ is simply the determinant of the
three-by-three matrix consisting of the components of the three vectors. And of course, the determinant of a
matrix is a scalar. Letโs now see how we can apply this
in an example.

Given vectors ๐, ๐, and ๐, where
๐ has components one, five, negative five; ๐ has components two, four, three; and
๐ has components zero, five, negative four, find the scalar triple product of ๐,
๐, and ๐.

Weโre asked to calculate the scalar
triple product of vectors ๐, ๐, and ๐. And we know that this is equivalent
to calculating the determinant of the matrix consisting of the components of the
three vectors as shown. So with our vectors ๐, ๐, and ๐,
we want to find the determinant of the matrix whose first line consists of the
components of vector ๐, whose second line has components of vector ๐, and whose
third line has the components of vector ๐.

And remember to calculate the
determinant of a three-by-three matrix using the first row as our pivot, we take the
first top-left element ๐ ๐ฅ and multiply this by the determinant of the two-by-two
matrix in the bottom-right corner. We then take the negative of the
second element in the top row and multiply this by the determinant shown. And finally, we add the third
element in the top row, multiply it by the determinant formed by the four elements
in the bottom-left-hand corner.

In our case, this translates into
one times the two-by-two matrix with elements four, three, five, negative four minus
five times the determinant of the matrix with elements two, three, zero, and
negative four plus a negative five times the determinant of the two-by-two matrix
with elements two, four, zero, five. And now remembering that the
determinant of a two-by-two matrix with elements ๐, ๐, ๐, ๐ is ๐๐ minus ๐๐,
we have one times four times negative four minus three times five minus five times
two times negative four minus three times zero minus five again times two times five
minus four times zero.

That is one times negative 16 minus
15 minus five times negative eight minus zero minus five times 10 minus zero. Evaluating this gives us negative
31 plus 40 minus 50, which is negative 41. The scalar triple product of
vectors ๐, ๐, and ๐ is therefore negative 41.

Weโve seen that the scalar triple
product of three vectors is equivalent to the determinant of the three-by-three
matrix consisting of their components. And it turns out that if we switch
two of the vectors around, this changes the sign of our result.

And this is, of course what we
would expect since if we switch around two rows in our determinant, this changes the
sign. If we were then to perform a
further permutation, for example, to switch ๐ with ๐ or ๐ with ๐, the sign
changes again. And this tells us that the scalar
triple product ๐ dot ๐ cross ๐ is the same as ๐ dot ๐ cross ๐. And this is the same as ๐ dot ๐
cross ๐. And what this tells us is that when
the cyclic order of the three vectors remains unchanged, their scalar triple
products are equal.

Letโs consider this property of the
scalar triple product in an example.

Find the sum of the scalar triple
product ๐ข dot ๐ฃ cross ๐ค plus ๐ฃ dot ๐ค cross ๐ข plus ๐ค dot ๐ข cross ๐ฃ.

Weโre given the sum of three scalar
triple products of the same three unit vectors ๐ข, ๐ฃ, and ๐ค, where these are the
unit vectors in the ๐ฅ-, ๐ฆ-, and ๐ง-directions. These three scalar triple products
are in different orders. The first is a scalar product of ๐ข
with ๐ฃ cross ๐ค. The second is a scalar product of
๐ฃ with ๐ค cross ๐ข. And the third is the scalar product
of ๐ค with ๐ข cross ๐ฃ. We do know, however, that because
these three scalar triple products have the same three vectors, the absolute values
of the scalar triple products will be the same.

We also know that only if the three
vectors are in the same cyclic order will they have the same sign. In our case, the cyclic order of
the first triple product is ๐ข to ๐ฃ to ๐ค. And now if we look at the second
triple product, we have ๐ฃ to ๐ค to ๐ข, which is in the same cycling direction. And the third triple product goes
from ๐ค to ๐ข to ๐ฃ, which again is in the same cyclic direction.

And because these are all in the
positive direction, if we find one of the scalar triple products, we simply multiply
our result by three, since they all have the same magnitude. Now using our cyclic diagram on the
right, from the cyclic properties, this tells us that ๐ฃ cross ๐ค is equal to ๐ข and
similarly ๐ค cross ๐ข is equal to ๐ฃ and that ๐ข cross ๐ฃ is equal to ๐ค. This then tells us that the scalar
product of ๐ข with the cross product of ๐ฃ and ๐ค is simply the scalar product of ๐ข
with itself. And we know that this is simply the
modulus or the magnitude of ๐ข squared. And since ๐ข is a unit vector,
thatโs equal to one.

And recall that all of our scalar
triple products have the same magnitude. And since all three are in the same
direction, theyโre all equal. So we can say that the sum of our
three scalar triple products is three times one of the scalar triple products. And since we found that the scalar
triple product of ๐ข with ๐ฃ cross ๐ค is equal to one, we have three times one,
which is equal to three. The sum of our three scalar triple
product is therefore three.

Itโs worth noting that we could
also have used the determinant method to work this out. We again use the fact that their
magnitudes are all equal to one and that theyโre all in the same cyclic
direction. So by finding one of our
determinants, multiply this by three. And again our answer is three.

Letโs now consider another property
of the scalar triple product concerned with coplanar vectors. Recall that the cross product of a
vector ๐ with another vector ๐ is another vector thatโs perpendicular to the plane
defined by the two vectors ๐ and ๐. Now if the vector ๐ is in the same
plane as ๐ and ๐, then ๐ must also be perpendicular to ๐ cross ๐. Now if we think about the scalar
product of two vectors, if the vectors are perpendicular, then their scalar product
is equal to zero because the angle between them is 90 degrees and the cos of 90
degrees is equal to zero.

So now if we go back to our scalar
triple product, if our vectors ๐, ๐, and ๐ are in the same plane, then ๐ is
perpendicular to ๐ cross ๐ and their scalar triple product is equal to zero. And the converse is also true. If the scalar triple product is
equal to zero, then the vectors ๐, ๐, and ๐ must be coplanar. Letโs use this property in our next
example.

Find the value of ๐ for which the
four points one, seven, negative two; three, five, six; negative one, six, negative
four; and negative four, negative three, ๐ all lie in a single plane.

Weโre given four points, which
weโre told are coplanar; that is, they all lie in the same plane. And weโre asked to find the value
of the unknown constant ๐. To do this, weโre going to use the
scalar triple product since we know that if the scalar triple product of three
vectors is zero, then the three vectors must be coplanar. Before we do this, however, we need
to check that our three known points are noncolinear. We do this because we need to be
sure that weโre not simply taking the scalar product with the zero vector in our
scalar triple product.

Letโs label our points ๐, ๐, ๐,
and ๐. And we can take any two of the
vectors formed by the points ๐, ๐, and ๐. Letโs choose ๐๐ and ๐๐. We know that ๐๐ is equal to ๐๐
minus ๐๐. And subtracting like-for-like
components, we have negative two, two, negative eight. Similarly, ๐๐ is equal ๐๐ minus
๐๐. And thatโs equal to negative four,
one, negative 10. We know that the cross product of
๐๐ with ๐๐ is given by the determinant shown, where ๐ข, ๐ฃ, and ๐ค are the unit
vectors in the ๐ฅ-, ๐ฆ-, and ๐ง-direction so that our determinant is ๐ข times two
times negative 10 minus negative eight times one, which is ๐ข times the determinant
of the two-by-two matrix in the bottom corner, minus ๐ฃ times the determinant of the
matrix with elements negative two, negative eight, negative four, and negative 10
plus ๐ค times the determinant of the matrix with elements negative two, two,
negative four, and one.

Making some room, this evaluates to
negative 12, 12, six, which is not equal to the zero vector. So we know that our three points
are not colinear. And we can move on to using the
scalar triple product to find the value of ๐. Weโre told that all four points ๐,
๐, ๐, and ๐ are coplanar. And we know that the scalar triple
product of three coplanar vectors is equal to zero. We already have part of our scalar
triple product. Thatโs with the cross product of
๐๐ and ๐๐. Now if all four points lie in the
plane, then the vector ๐๐ must also lie in the plane. The vector ๐๐ is given by ๐๐
minus ๐๐. And that is negative five, negative
10, ๐ plus two.

So taking the scalar triple product
of ๐๐ with ๐๐ and ๐๐, we simply have the scalar product of the two vectors
shown. This evaluates to negative five
times negative 12 plus negative 10 times 12 plus ๐ plus two times six. That is 60 minus 120 plus six ๐
plus 12, which simplifies to six ๐ minus 48. If the vectors are coplanar, this
must be equal to zero. Adding 48 to both sides, this gives
us six ๐ is equal to 48. And dividing both sides by six,
this gives us ๐ is equal to eight. Hence, the value of ๐ for which
the four given points all lie in a single plane is ๐ is equal to eight.

The final property of the scalar
triple product that weโre going to look at arises from what it means
geometrically. Remember that the cross product ๐
cross ๐ is a vector perpendicular to the plane defined by the vectors ๐ and
๐. The magnitude of this cross product
is the area of the parallelogram spanned by the vectors ๐ and ๐. Now if we add another vector ๐ and
consider the parallelepiped spanned in three dimensions by the vectors ๐, ๐, and
๐, we know that its volume is the area of the parallelogram spanned by the vectors
๐ and ๐ multiplied by the perpendicular height โ. We also know that this height โ is
equal to the magnitude of the vector ๐ times the cos of ๐, which is the acute
angle between ๐ and โ.

So we have the volume of the
parallelepiped is the magnitude of the cross product of ๐ and ๐ times the
magnitude of the vector ๐ times the cos of ๐. And by definition, this is the
magnitude of the scalar triple product of ๐ with ๐ cross ๐. Itโs worth noting that if the
orientation of ๐ cross ๐ changed to down, then the angle between ๐ and ๐ cross
๐ would be ๐ prime, which is 180 minus ๐. And since cos of 180 minus ๐ is
negative cos ๐, we have the absolute value of cos ๐ prime is the absolute value of
cos ๐. And so, by taking the magnitude of
the scalar triple product, we have the volume of the parallelepiped, whatever the
orientation of ๐ cross ๐. Letโs now look at an example of
finding the volume of a parallelepiped using the scalar triple product.

Find the volume of the
parallelepiped with the adjacent sides ๐ฎ is equal to one, one, three; ๐ฏ is the
vector two, one, four; and ๐ฐ is the vector five, one, negative two.

The parallelepiped as defined is
spun by the vectors ๐ฎ, ๐ฏ, and ๐ฐ. And we know that to find the volume
of such a parallelepiped, we can use the scalar triple product. That is, the volume of the
parallelepiped with adjacent sides ๐ฎ, ๐ฏ, and ๐ฐ is the magnitude of the scalar
triple product. We also know that the scalar triple
product is the determinant of the matrix whose rows are the elements of the vectors
๐ฎ, ๐ฏ, and ๐ฐ. So in fact, the volume is the
magnitude of this.

In our case, then this is the
magnitude of the determinant of the matrix whose elements are one, one, three; two,
one, four; and five, one, negative two. That is where the rows are our
vectors ๐ฎ, ๐ฏ, and ๐ฐ. That is one times the determinant
of the two-by-two matrix with elements one, four, one, negative two minus one times
the two-by-two matrix with elements two, four, five, and negative two plus three
times the determinant of the two-by-two matrix with elements two, one, five, and
one.

And using the fact that the
determinant of a two-by-two matrix with elements ๐, ๐, ๐, ๐ is ๐๐ minus ๐๐,
this evaluates to the magnitude of negative six plus 24 minus nine, which is
nine. The volume of the parallelepiped
with adjacent sides ๐ฎ, ๐ฏ, and ๐ฐ is therefore nine cubic units.

Letโs complete this video by noting
some of the key points weโve covered. We know that the scalar triple
product of three vectors ๐, ๐, and ๐ is a scalar. The scalar triple product is
equivalent to the determinant of the three-by-three matrix whose rows are the
component of the vectors ๐, ๐, and ๐. Scalar triple products are equal if
the cyclic order is unchanged. If the vectors ๐, ๐, and ๐ are
coplanar, then their scalar triple product is equal to zero. And finally, that the volume of the
parallelepiped spanned by the vectors ๐, ๐, and ๐ is given by the magnitude of
their scalar triple product.