Square 𝐴𝐵𝐶𝐷 has side 10 centimeters. What is the dot product between the vector 𝐀𝐁 and 𝐁𝐂?
In this question, we’re given some information about the square 𝐴𝐵𝐶𝐷. We’re told the side lengths of this square are 10 centimeters. We need to use this to determine the dot product between the vectors representing two of its sides, the vector 𝐀𝐁 and the vector 𝐁𝐂.
Let’s start by sketching a picture of our square 𝐴𝐵𝐶𝐷 with side length 10 centimeters. There’s actually a few different ways we could evaluate this dot product. One way of doing this will be to write our vectors 𝐀𝐁 and 𝐁𝐂 component-wise from our diagram. For example, we can see the vector that goes from 𝐴 to 𝐵 has no horizontal component and its vertical component will be 10 centimeters. Therefore, the vector from 𝐴 to 𝐵 could be represented by having horizontal component zero and vertical component 10, because to go from point 𝐴 to point 𝐵, we increase the vertical component by 10 centimeters.
We can do exactly the same for our vector 𝐁𝐂. To go from the point 𝐵 to the point 𝐶, our vertical component will be zero. However, we increase our horizontal component by 10. So the vector from 𝐵 to 𝐶 can be represented by the vector 10, zero.
Now we can find the dot product between these two vectors directly. To do this, we need to recall how exactly we calculate the dot product between two vectors. Remember, to do this, we multiply the corresponding components together and then add the results. So we start by multiplying the first components of our vectors together. That’s zero multiplied by 10. And then we add to this the product of the second components of our vectors. That’s 10 multiplied by zero. And of course we can calculate this. It’s equal to zero.
However, this isn’t the only way we could’ve evaluated this expression. We know a formula involving the dot product between two vectors and the angle between them. We recall if 𝜃 is the angle between two vectors 𝐮 and 𝐯, then the cos of 𝜃 must be equal to the dot product between 𝐮 and 𝐯 divided by the magnitude of 𝐮 times the magnitude of 𝐯. So another way of evaluating the dot product given to us in the question is to find the magnitude of our two vectors and the angle between them. Then we can rearrange this equation and solve for the dot product.
Let’s start by finding the magnitude of our two vectors. That’s the vector 𝐀𝐁 and the vector 𝐁𝐂. This notation tells us the vector 𝐀𝐁 is the vector from 𝐴 to 𝐵 and the vector 𝐁𝐂 is the vector from 𝐵 to 𝐶. And we can see from our diagram both of these are going to be side lengths in our square. And we’re told in the question that the square has side length 10 centimeters. So we can start by saying the magnitude of 𝐀𝐁 and the magnitude of 𝐁𝐂 are going to be 10 centimeters.
Now, we need to find the angle between our two vectors. Let’s start by drawing these vectors on our diagram. We’ll start by drawing the vector 𝐀𝐁. That’s the vector from 𝐴 to 𝐵. And then we’ll also draw on the vector from 𝐵 to 𝐶.
Now, we need to be careful here. It’s very tempting to call the angle 𝐴𝐵𝐶 the angle between these two vectors. However, this would be incorrect. To find the angle between our two vectors, our vectors must start at the same point. And we can see this is not true in our diagram. The vector 𝐀𝐁 starts at 𝐴 and ends at 𝐵 and the vector 𝐁𝐂 starts at 𝐵 and ends at 𝐶. So we’re going to need to move one of our vectors. Let’s move the vector 𝐁𝐂.
Remember, a vector is an object with magnitude and direction. So if the magnitude and direction of two vectors are the same, then the vectors are the same. Because 𝐴𝐵𝐶𝐷 is a square, we know the length of 𝐴𝐷 is equal to 10 and we also know that the vectors 𝐀𝐃 and 𝐁𝐂 are parallel. Therefore, what we’ve just shown is the vector from 𝐴 to 𝐷 and the vector from 𝐵 to 𝐶 have the same magnitude and direction. They represent the same vector. So the angle between our vectors 𝐀𝐁 and 𝐁𝐂 is represented by angle 𝐷𝐴𝐵. And of course this is a right angle. So we know it’s equal to 90 degrees.
Now that we’ve found the magnitude of our vectors and the angle between them, we can substitute these values into our formula. We get the cos of 90 degrees is equal to the dot product between our vectors 𝐀𝐁, 𝐁𝐂 divided by 10 times 10. And if we start evaluating this expression, we get something interesting. The cos of 90 degrees is equal to zero. So the right-hand side of this equation must be equal to zero. And therefore, our numerator is equal to zero.
This also proves the dot product between these two vectors is equal to zero. And it illustrates a nice use of one of our properties. We know if 𝐮 and 𝐯 are perpendicular vectors, then the dot product between 𝐮 and 𝐯 will be equal to zero. And this is because 𝐮 and 𝐯 being perpendicular means the angle between them is 90 degrees. And if 𝜃 is 90 degrees, the cos of 90 degrees is equal to zero. So the right-hand side of this equation must be equal to zero. And the only way this can be true is if the dot product between 𝐮 and 𝐯 is equal to zero.
Therefore, we were able to show two different ways of finding the dot product between vectors 𝐀𝐁 and 𝐁𝐂, which were side lengths of the square 𝐴𝐵𝐶𝐷 with side length 10 centimeters. In both cases, we were able to show it was equal to zero.