### Video Transcript

In this video, we’ll 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 𝐀, 𝐁, and 𝐂 is defined as the scalar product or the dot product of vector
𝐀 with the cross product of vectors 𝐁 and 𝐂.

If we break this down, we know that
the result of the cross product of two vectors 𝐁 and 𝐂 is another vector. Let’s call that 𝐃. If we then take the scalar product
of vector 𝐀 with vector 𝐃, we know that the scalar or dot product results in a
scalar. Let’s call that 𝜆. So the scalar triple product is in
fact a scalar, hence the name scalar triple product. It’s worth noting that we don’t
actually need the parentheses since if we tried to take the dot product, that is the
scalar of product of 𝐀 with 𝐁 first, this would give us a scalar. We would then have a scalar crossed
with a vector, which we cannot do, since it’s not possible to cross a scalar with a
vector.

And so the scalar triple product is
the scalar product of vector 𝐀 with the cross product of vectors 𝐁 and 𝐂. Now let’s see what the scalar
triple product looks like in component form. If 𝐢, 𝐣, and 𝐤 are the unit
vectors in the 𝑥-, 𝑦-, and 𝑧-directions, then our three vectors are written in
component form as shown. And if we write the cross product
of vectors 𝐁 and 𝐂 in component form, we have 𝐁 cross 𝐂 is 𝐁 𝑦 𝐂 𝑧 minus 𝐁
𝑧 𝐂 𝑦 times 𝐢 minus 𝐁 𝑥 𝐂 𝑧 minus 𝐁 𝑧 𝐂 𝑥 times 𝐣 plus 𝐁 𝑥 𝐂 𝑦
minus 𝐁 𝑦 𝐂 𝑥 times 𝐤. And now if we take the scalar
product of the vector 𝐀 with this, that is, we multiply the coefficients of each of
the unit vectors together so that for example our first term is 𝐀 𝑥 times 𝐁 𝑦 𝐂
𝑧 minus 𝐁 𝑧 𝐂 𝑦.

But what does this remind us
of? Well, remember that the determinant
of a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑 is 𝑎𝑑 minus 𝑏𝑐. So for example, in our first term
𝐁 𝑦 𝐂 𝑧 minus 𝐁 𝑧 𝐂 𝑦 is actually the determinant of the two-by-two matrix
with elements 𝐁 𝑦, 𝐁 𝑧, 𝐂 𝑦, 𝐂 𝑧. And so our first term in the scalar
triple product is 𝐀 𝑥 times the determinant of the matrix with elements 𝐁 𝑦, 𝐁
𝑧, 𝐂 𝑦, 𝐂 𝑧, 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 𝐀, 𝐁, and 𝐂. So in fact, the scalar triple
product of the three vectors 𝐀, 𝐁, and 𝐂 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. Let’s now see how we can apply this
in an example.

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

We’re asked to calculate the scalar
triple product of vectors 𝐀, 𝐁, and 𝐂. 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 𝐀, 𝐁, and 𝐂,
we want to find the determinant of the matrix whose first line consists of the
components of vector 𝐀, whose second line has components of vector 𝐁, and whose
third line has the components of vector 𝐂.

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 𝐀 𝑥 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 𝑎, 𝑏, 𝑐, 𝑑 is 𝑎𝑑 minus 𝑏𝑐,
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 𝐀, 𝐁, and 𝐂 is therefore negative 41.

We’ve 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 𝐀 with 𝐂 or 𝐁 with 𝐂, the sign
changes again. And this tells us that the scalar
triple product 𝐀 dot 𝐁 cross 𝐂 is the same as 𝐁 dot 𝐂 cross 𝐀. And this is the same as 𝐂 dot 𝐀
cross 𝐁. And what this tells us is that when
the cyclic order of the three vectors remains unchanged, their scalar triple
products are equal.

Let’s consider this property of the
scalar triple product in an example.

Find the sum of the scalar triple
product 𝐢 dot 𝐣 cross 𝐤 plus 𝐣 dot 𝐤 cross 𝐢 plus 𝐤 dot 𝐢 cross 𝐣.

We’re given the sum of three scalar
triple products of the same three unit vectors 𝐢, 𝐣, and 𝐤, where these are the
unit vectors in the 𝑥-, 𝑦-, and 𝑧-directions. These three scalar triple products
are in different orders. The first is a scalar product of 𝐢
with 𝐣 cross 𝐤. The second is a scalar product of
𝐣 with 𝐤 cross 𝐢. And the third is the scalar product
of 𝐤 with 𝐢 cross 𝐣. 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 𝐢 to 𝐣 to 𝐤. And now if we look at the second
triple product, we have 𝐣 to 𝐤 to 𝐢, which is in the same cycling direction. And the third triple product goes
from 𝐤 to 𝐢 to 𝐣, 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 𝐣 cross 𝐤 is equal to 𝐢 and
similarly 𝐤 cross 𝐢 is equal to 𝐣 and that 𝐢 cross 𝐣 is equal to 𝐤. This then tells us that the scalar
product of 𝐢 with the cross product of 𝐣 and 𝐤 is simply the scalar product of 𝐢
with itself. And we know that this is simply the
modulus or the magnitude of 𝐢 squared. And since 𝐢 is a unit vector,
that’s 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, they’re 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 𝐢 with 𝐣 cross 𝐤 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.

It’s 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 they’re all in the same cyclic
direction. So by finding one of our
determinants, multiply this by three. And again our answer is three.

Let’s now consider another property
of the scalar triple product concerned with coplanar vectors. Recall that the cross product of a
vector 𝐁 with another vector 𝐂 is another vector that’s perpendicular to the plane
defined by the two vectors 𝐁 and 𝐂. Now if the vector 𝐀 is in the same
plane as 𝐁 and 𝐂, then 𝐀 must also be perpendicular to 𝐁 cross 𝐂. 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 𝐀, 𝐁, and 𝐂 are in the same plane, then 𝐀 is
perpendicular to 𝐁 cross 𝐂 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 𝐀, 𝐁, and 𝐂 must be coplanar. Let’s use this property in our next
example.

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

We’re given four points, which
we’re told are coplanar; that is, they all lie in the same plane. And we’re asked to find the value
of the unknown constant 𝑘. To do this, we’re 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 we’re not simply taking the scalar product with the zero vector in our
scalar triple product.

Let’s label our points 𝐀, 𝐁, 𝐂,
and 𝐃. And we can take any two of the
vectors formed by the points 𝐀, 𝐁, and 𝐂. Let’s choose 𝐁𝐀 and 𝐁𝐂. We know that 𝐁𝐀 is equal to 𝐎𝐀
minus 𝐎𝐁. And subtracting like-for-like
components, we have negative two, two, negative eight. Similarly, 𝐁𝐂 is equal 𝐎𝐂 minus
𝐎𝐁. And that’s equal to negative four,
one, negative 10. We know that the cross product of
𝐁𝐀 with 𝐁𝐂 is given by the determinant shown, where 𝐢, 𝐣, and 𝐤 are the unit
vectors in the 𝑥-, 𝑦-, and 𝑧-direction so that our determinant is 𝐢 times two
times negative 10 minus negative eight times one, which is 𝐢 times the determinant
of the two-by-two matrix in the bottom corner, minus 𝐣 times the determinant of the
matrix with elements negative two, negative eight, negative four, and negative 10
plus 𝐤 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 𝑘. We’re told that all four points 𝐀,
𝐁, 𝐂, and 𝐃 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. That’s with the cross product of
𝐁𝐀 and 𝐁𝐂. Now if all four points lie in the
plane, then the vector 𝐀𝐃 must also lie in the plane. The vector 𝐀𝐃 is given by 𝐎𝐃
minus 𝐎𝐀. And that is negative five, negative
10, 𝑘 plus two.

So taking the scalar triple product
of 𝐀𝐃 with 𝐁𝐀 and 𝐁𝐂, 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 𝑘 plus two times six. That is 60 minus 120 plus six 𝑘
plus 12, which simplifies to six 𝑘 minus 48. If the vectors are coplanar, this
must be equal to zero. Adding 48 to both sides, this gives
us six 𝑘 is equal to 48. And dividing both sides by six,
this gives us 𝑘 is equal to eight. Hence, the value of 𝑘 for which
the four given points all lie in a single plane is 𝑘 is equal to eight.

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

So we have the volume of the
parallelepiped is the magnitude of the cross product of 𝐁 and 𝐂 times the
magnitude of the vector 𝐀 times the cos of 𝜃. And by definition, this is the
magnitude of the scalar triple product of 𝐀 with 𝐁 cross 𝐂. It’s worth noting that if the
orientation of 𝐁 cross 𝐂 changed to down, then the angle between 𝐀 and 𝐁 cross
𝐂 would be 𝜃 prime, which is 180 minus 𝜃. And since cos of 180 minus 𝜃 is
negative cos 𝜃, we have the absolute value of cos 𝜃 prime is the absolute value of
cos 𝜃. And so, by taking the magnitude of
the scalar triple product, we have the volume of the parallelepiped, whatever the
orientation of 𝐁 cross 𝐂. Let’s 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 𝐮 is equal to one, one, three; 𝐯 is the
vector two, one, four; and 𝐰 is the vector five, one, negative two.

The parallelepiped as defined is
spun by the vectors 𝐮, 𝐯, and 𝐰. 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 𝐮, 𝐯, and 𝐰 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
𝐮, 𝐯, and 𝐰. 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 𝐮, 𝐯, and 𝐰. 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 𝑎, 𝑏, 𝑐, 𝑑 is 𝑎𝑑 minus 𝑏𝑐,
this evaluates to the magnitude of negative six plus 24 minus nine, which is
nine. The volume of the parallelepiped
with adjacent sides 𝐮, 𝐯, and 𝐰 is therefore nine cubic units.

Let’s complete this video by noting
some of the key points we’ve covered. We know that the scalar triple
product of three vectors 𝐀, 𝐁, and 𝐂 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 𝐀, 𝐁, and 𝐂. Scalar triple products are equal if
the cyclic order is unchanged. If the vectors 𝐀, 𝐁, and 𝐂 are
coplanar, then their scalar triple product is equal to zero. And finally, that the volume of the
parallelepiped spanned by the vectors 𝐀, 𝐁, and 𝐂 is given by the magnitude of
their scalar triple product.