### Video Transcript

In this video, we will learn how to
find the cross product of two vectors in the coordinate plane. We will begin by defining what we
mean by the cross product. We know that any vector has
magnitude or length and direction. One way of multiplying two vectors
is using the cross product. The cross product written 𝐚 cross
𝐛 of two vectors is another vector that is at right angles to both of them. This can be seen in the diagram
drawn.

We will now look at the geometric
meaning of the cross product. The magnitude or length of the
cross product equals the area of a parallelogram with vectors 𝐚 and 𝐛 for
sides. This is shown in the diagram. And the area of the parallelogram
and cross product alter for different values of the angle 𝜃. The cross product equals zero when
the vectors point in the same or opposite direction. The maximum value of the cross
product occurs when the vectors are perpendicular. This means that vectors 𝐚 and 𝐛
are at right angles.

Let’s now look at how we can use
this information to calculate the cross product. We can calculate the cross product
as follows. 𝐚 cross 𝐛 is equal to the
magnitude of vector 𝐚 multiplied by the magnitude of vector 𝐛 multiplied by sin of
angle 𝜃 multiplied by the unit vector 𝐧. We know that the magnitudes of
vector 𝐚 and vector 𝐛 are the lengths of these vectors. 𝜃 is the positive angle less than
or equal to 180 degrees between vector 𝐚 and vector 𝐛. 𝐧 hat is the unit vector at right
angles to both vector 𝐚 and vector 𝐛. The direction of 𝐚 cross 𝐛 is
perpendicular to the plane of vector 𝐚 and vector 𝐛 and in the right-handed
direction.

Imagine a screwdriver driving a
right-handed screw in the direction of 𝐚 cross 𝐛. In doing so, our fingers curl in
the direction of rotation of 𝐚 into 𝐛, as shown by the arrow in the diagram. Our thumb points in the direction
of 𝐚 cross 𝐛. We can therefore conclude that the
cross product of vector 𝐚 and vector 𝐛 is equal to the length of vector 𝐚
multiplied by the length of vector 𝐛 multiplied by the sin of angle 𝜃, which is
the angle between the vectors. We then multiply this by the unit
vector 𝐧 so that it heads in the correct direction.

We will now look at some
examples. And in our first question, we will
calculate the cross product in a problem involving a square.

Given that 𝐴𝐵𝐶𝐷 is a square
with side length 27 centimeters and 𝐞 hat is the unit vector perpendicular to its
plane, determine the cross product of 𝐀𝐁 and 𝐂𝐀.

In this question, we’re being asked
to calculate the cross product. And we know the cross product of
vector 𝐚 and vector 𝐛 is equal to the magnitude of vector 𝐚 multiplied by the
magnitude of vector 𝐛 multiplied by sin of angle 𝜃 multiplied by the unit vector
𝐧. 𝜃 is the angle between the two
vectors. And the unit vector 𝐧 is
perpendicular to both vector 𝐚 and vector 𝐛.

We are told that the square
𝐴𝐵𝐶𝐷 has side length 27 centimeters. As the magnitude of a vector is
equal to its length, the magnitude of vector 𝐀𝐁 is equal to 27. We also need to calculate the
magnitude of vector 𝐂𝐀. As triangle 𝐴𝐵𝐶 is a right
triangle, we can do this using the Pythagorean theorem, which states that 𝑎 squared
plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the length of the longest side
or hypotenuse.

In this question, we have the
magnitude of 𝐀𝐁 squared plus the magnitude of 𝐁𝐂 squared is equal to the
magnitude of 𝐂𝐀 squared. We know that the magnitude or
length of 𝐀𝐁 and 𝐁𝐂 is 27. 27 squared plus 27 squared is equal
to 1458. We can then square root both sides
of this equation so that the magnitude of 𝐂𝐀 is equal to 27 root two.

Our next step is to redraw our
diagram so that the tails or start points of both vectors are at the same point. We need to calculate the angle 𝜃
between vector 𝐀𝐁 and vector 𝐂𝐀. As the diagonal in a square cuts a
right angle in half and a half of 90 degrees is 45 degrees, then 𝜃 will be equal to
180 minus 45 degrees. The angle between vector 𝐀𝐁 and
vector 𝐂𝐀 is 135 degrees.

We can now calculate the cross
product of vector 𝐀𝐁 and vector 𝐂𝐀. 𝐀𝐁 cross 𝐂𝐀 is equal to the
magnitude of 𝐀𝐁 multiplied by the magnitude of 𝐂𝐀 multiplied by sin of angle 𝜃
multiplied by the unit vector 𝐞, as this is the unit vector perpendicular to the
plane. Substituting in our values, we have
27 multiplied by 27 root two multiplied by sin of 135 degrees multiplied by the unit
vector 𝐞. sin of 135 degrees is equal to root
two over two. We need to multiply this by 27, 27
root two, and the unit vector 𝐞. Root two multiplied by root two
over two is equal to one. So we are left with 27 multiplied
by 27 multiplied by the unit vector 𝐞. This is equal to 729𝐞. The cross product of 𝐀𝐁 and 𝐂𝐀
is equal to 729𝐞.

In our next question, we’ll find
the cross product of vectors in a rectangle.

𝐴𝐵𝐶𝐷 is a rectangle, where 𝐜
hat is a unit vector perpendicular to its plane. Find the cross product of vector
𝐂𝐌 and vector 𝐂𝐁.

We recall that the cross product of
two vectors 𝐚 and 𝐛 is equal to the magnitude of vector 𝐚 multiplied by the
magnitude of vector 𝐛 multiplied by sin of angle 𝜃 multiplied by the unit vector
𝐧. 𝜃 is the angle between the two
vectors, and the unit vector 𝐧 is perpendicular to vectors 𝐚 and 𝐛. When dealing with a two-dimensional
geometric shape, this will be perpendicular to the plane.

In this question, we need to
calculate the cross product of vector 𝐂𝐌 and vector 𝐂𝐁. As the magnitude of any vector is
its length, then the magnitude of 𝐂𝐁 is equal to 44, as the sides 𝐃𝐀 and 𝐂𝐁 on
the rectangle are equal to 44 centimeters. We can see from the diagram that
the magnitude of vector 𝐂𝐌 will be equal to one-half of the magnitude of vector
𝐂𝐀.

As triangle 𝐴𝐵𝐶 is a right
triangle, we can use the Pythagorean theorem to calculate the length of 𝐂𝐀. The Pythagorean theorem states that
𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the length of the
longest side or hypotenuse. The magnitude of vector 𝐂𝐁
squared plus the magnitude of vector 𝐁𝐀 squared will be equal to the magnitude of
vector 𝐂𝐀 squared. Substituting in our values from the
diagram, we need to calculate 44 squared plus 33 squared. This is equal to 3025. Square rooting both sides of this
equation gives us that the magnitude of vector 𝐂𝐀 is equal to 55. The length of the diagonal of the
rectangle from point 𝐶 to point 𝐴 is 55 centimeters. One-half of 55 is 27.5. Therefore, the magnitude of vector
𝐂𝐌 is 27.5.

We now need to calculate the angle
between our two vectors, which we will call 𝛼. We can do this using our trig
ratios. And we know that sin of angle 𝛼 is
equal to the opposite over the hypotenuse. The length of 𝐁𝐀 is equal to 33
centimeters, and the length of 𝐂𝐀 is equal to 55 centimeters. Therefore, sin 𝛼 is equal to 33
over 55. By dividing the numerator and
denominator by 11, this simplifies to three-fifths.

When dealing with a cross product,
we measure the angle in a counter- or anticlockwise direction. If we wanted to calculate 𝐂𝐁
cross 𝐂𝐌, then the angle 𝜃 would be as shown in the diagram such that sin 𝜃 is
equal to three-fifths. We don’t want to calculate this,
however. We want to calculate 𝐂𝐌 cross
𝐂𝐁. This means that the angle will be
negative 𝜃. We know that the sine function is
odd, which means that sin of negative 𝜃 is equal to negative sin 𝜃. This means that the value of sin 𝜃
in our example will be negative three-fifths.

This leads us to an interesting
rule about the vector or cross product. Vector multiplication is not
commutative. 𝐚 cross 𝐛 is not equal to 𝐛
cross 𝐚. However, as the sine function is
odd, 𝐚 cross 𝐛 is equal to negative 𝐛 cross 𝐚. This means that in our question,
𝐂𝐌 cross 𝐂𝐁 is equal to negative 𝐂𝐁 cross 𝐂𝐌.

Using the formula for the cross
product, 𝐂𝐌 cross 𝐂𝐁 is equal to 44 multiplied by 27.5 multiplied by negative
three-fifths multiplied by the unit vector 𝐜. This is equal to negative
726𝐜.

In our final question in this
video, we will calculate the area of a triangle using vectors.

Find the area of a triangle 𝐴𝐵𝐶,
where 𝐴 has coordinates negative eight, negative nine; 𝐵 has coordinates negative
seven, negative eight; and 𝐶 has coordinates nine, negative two.

We could begin by trying to draw
the points on a coordinate grid. However, as points 𝐴 and 𝐵 are so
close together, it will be difficult to draw this accurately. As a result, we will just sketch
the triangle roughly as shown.

We recall that the area of any
triangle is equal to half the area of a parallelogram, where the lengths of the
sides of the parallelogram are equal to two of the lengths of the triangle. We recall that the area of any
parallelogram is equal to the magnitude of the cross product of vector 𝐚 and vector
𝐛.

In this question, the area of
parallelogram 𝐴𝐵𝐶𝐷 is equal to the magnitude of the cross product of vectors
𝐁𝐀 and 𝐁𝐂. The area of triangle 𝐴𝐵𝐶 is
therefore equal to a half of this. We know that the magnitude of the
cross product is equal to the magnitude of vector 𝐚 multiplied by the magnitude of
vector 𝐛 multiplied by the magnitude of sin 𝜃, where 𝜃 is the angle between
vectors 𝐚 and 𝐛.

The area of triangle 𝐴𝐵𝐶 is
therefore equal to a half multiplied by the magnitude of vector 𝐁𝐀 multiplied by
the magnitude of vector 𝐁𝐂 multiplied by the magnitude of sin 𝜃. The magnitude of any vector is
equal to its length. Therefore, the magnitude of vector
𝐁𝐀 is equal to the square root of negative eight minus negative seven all squared
plus negative nine minus negative eight all squared. This is equal to the square root of
two.

We can repeat this process to
calculate the magnitude of vector 𝐁𝐂. This is equal to nine minus
negative seven all squared plus negative two minus negative eight all squared. This is equal to the square root of
292, which is equal to two root 73.

In order to calculate the angle 𝜃
in our triangle, we will firstly need to calculate the length of 𝐴𝐶 and then use
the cosine rule. The length of 𝐴𝐶 is equal to the
magnitude of vector 𝐀𝐂. And this is equal to the square
root of nine minus negative eight all squared plus negative two minus negative nine
all squared. This is equal to the square root of
338, which is equal to 13 root two.

The cosine rule states that the cos
of angle 𝐵 is equal to 𝑎 squared plus 𝑐 squared minus 𝑏 squared all divided by
two 𝑎𝑐. In this question, the sides 𝑎, 𝑏,
and 𝑐 are equal to two root 73, 13 root two, and root two, respectively. Angle 𝐵 is the angle 𝜃 we are
trying to calculate. Substituting in our values, we see
that the cos of 𝜃 is equal to two root 73 squared plus root two squared minus 13
root two squared all divided by two multiplied by two root 73 multiplied by root
two.

Typing the right-hand side of the
equation into our calculator gives us negative 0.9103 and so on. We can then take the inverse cosine
of both sides of this equation such that 𝜃 is equal to 155.556 and so on
degrees.

We can now substitute these values
back into our equation to calculate the area of the triangle. It is important at this stage that
we do not round the angle 155.556 and so on degrees. Typing this into the calculator
gives us an answer of five. Therefore, the area of triangle
𝐴𝐵𝐶 is equal to five square units.

We will now summarize the key
points from this video. We saw in this video that the cross
product of two vectors is another vector. However, the magnitude of this
cross product is a scalar. 𝐚 cross 𝐛 is equal to the
magnitude of vector 𝐚 multiplied by the magnitude of vector 𝐛 multiplied by sin of
angle 𝜃 multiplied by the unit vector 𝐧, where 𝜃 is the angle between vectors 𝐚
and 𝐛. The magnitude of the cross product
is equal to the magnitude of vector 𝐚 multiplied by the magnitude of vector 𝐛
multiplied by the magnitude of sin 𝜃. This is also equal to the area of a
parallelogram with sides vector 𝐚 and vector 𝐛.

We know that a parallelogram can be
split into two equal-sized triangles. Therefore, the area of a triangle
is equal to a half of the magnitude of the cross product of vector 𝐚 and vector
𝐛. Finally, we saw that the cross
product was not communicative. However, 𝐚 cross 𝐛 is equal to
negative 𝐛 cross 𝐚.