# Lesson Video: Cross Product in 2D Mathematics

In this video, we will learn how to find the cross product of two vectors in the coordinate plane.

17:59

### 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.