Question Video: Calculating the Scalar Product of Two Vectors Shown on a Grid Physics

Video Transcript

The diagram shows two vectors, 𝐀 and 𝐁. Each of the grid squares in the diagram has a side length of one. Calculate 𝐀 dot 𝐁.

We’re asked to calculate the dot product, also called the scalar product, of two vectors 𝐀 and 𝐁. These two vectors are represented on the diagram. And because each of the squares has a side length of one, we can see immediately that vector 𝐀 has a length of five units and a direction that is directly to the right. We can also see that vector 𝐁 has a length of four units and a direction that is directly upward.

Now that we know a little bit about our vectors, let’s recall how to calculate a scalar product. One way to calculate the dot product of two vectors is to multiply their corresponding vector components and then add those products. So 𝐀 dot 𝐁 will be the 𝑥-component of 𝐀 times the 𝑥-component of 𝐁 plus the 𝑦-component of 𝐀 times the 𝑦-component of 𝐁. We can also calculate the scalar product as the magnitude of 𝐀 times the magnitude of 𝐁 times the cosine of the angle between the two vectors represented by the Greek letter 𝜃.

In our diagram, the angle between 𝐀 and 𝐁 is this angle here. And as we can see, because 𝐀 and 𝐁 are parallel to the two sides of one of the grid squares, this angle must be exactly 90 degrees. Note that although it’s not immediately obvious, these two ways of calculating the scalar product actually give exactly the same answers. And in some more complicated geometric situations, we actually use the second definition as a way to define the angle between two vectors. Anyway, the fact that the angle between these two vectors is 90 degrees means that they are perpendicular, which means we can actually bypass calculation entirely by recalling that the scalar product of two perpendicular vectors is always zero.

Note that this is the number zero, not the zero vector, because a scalar product always gives a scalar. That’s why it’s called the scalar product. Anyway, we know the answer should be zero. But let’s verify this by calculating the scalar product of 𝐀 and 𝐁 in these two ways. To use the sum of the product of corresponding components, we will need to define 𝑥- and 𝑦-axes. Given our diagram, we may as well use the standard set of Cartesian axes, with 𝑥 being the horizontal axis and increasing towards the right and 𝑦 being the vertical axis and increasing upward.

Where we place the origin for these axes doesn’t matter. All that matters is that we have defined the 𝑥- and 𝑦-direction so that we can define an 𝑥- and 𝑦-component of each of our vectors. To find the 𝑥-component of each vector, we look at the number of grid squares between the tail and head of each vector along the 𝑥-direction. For vector 𝐀, the tail and the head are five units apart along the 𝑥-axis, so 𝐀 𝑥 is five. For 𝐁, the tail and head are at the exact same 𝑥-value, so they are zero 𝑥 units apart. And so 𝐁 𝑥 is zero. Doing the same for the 𝑦-components, from the tail to the head of 𝐁, we move four units in the positive 𝑦-direction. So 𝐁 𝑦 is four, while from the tail to the head of 𝐀, we don’t move any units in the 𝑦-direction at all. So 𝐀 𝑦 is zero.

Forming the scalar product from these two sets of components, we have that 𝐀 dot 𝐁 is five times zero plus zero times four. But zero times anything is zero, so five times zero is zero and zero times four is zero. And we have zero plus zero, which is zero. Now, to calculate the scalar product using magnitudes and angles, we actually already have all the information we need. The magnitude of 𝐀 is five, the magnitude of 𝐁 is four, and the angle between them is 90 degrees. So we have that 𝐀 dot 𝐁 is five times four times the cos of 90 degrees. But the cos of 90 degrees is zero. So we have five times four times zero, which, just like before, is exactly zero.

This is actually the mathematical basis for the fact that we stated earlier. Two vectors that are perpendicular meet at a right angle, and the cosine of a right angle is zero. So, no matter what the magnitudes of these vectors are, their dot product will always be zero. So, whether we calculate the scalar product of 𝐀 and 𝐁 using their vector components or by using their magnitudes and the angle between them or by simply knowing a factor of perpendicular vectors, we find that the scalar product of 𝐀 and 𝐁 is exactly zero. And this is the answer that we are looking for.