Question Video: Calculating the Dot Product of Vectors Mathematics

If 𝐀 and 𝐁 are two perpendicular unit vectors, find (3𝐀 βˆ’ 𝐁) β‹… (βˆ’2𝐀 + 𝐁).


Video Transcript

If 𝐀 and 𝐁 are two perpendicular unit vectors, find three 𝐀 minus 𝐁 dotted with negative two 𝐀 plus 𝐁.

In this example, we’re calculating a dot product based on two perpendicular unit vectors, 𝐀 and 𝐁. If we think about the components of these two vectors, there’s very little we know about them specifically. Our problem statement says that they’re unit vectors and that they’re perpendicular to one another. But there are lots of ways in three-dimensional space for these conditions to be satisfied. We’re free then to choose the components of vectors 𝐀 and 𝐁 so long as they do satisfy these conditions of being unit vectors and perpendicular to one another.

If we think about three-dimensional space, we can recall that for each dimension there’s a standard unit vector that’s perpendicular to the unit vectors for the other two dimensions. This means that 𝐒 hat, for example, the unit vector in the π‘₯-direction, is perpendicular to the 𝐣 hat unit vector and the 𝐀 hat one. As we think about defining the vectors 𝐀 and 𝐁 then, the perpendicular unit vectors 𝐒 hat, 𝐣 hat, and 𝐀 hat are candidates for their components.

Just to make a selection, let’s say that 𝐀 is equal to the 𝐒 hat unit vector and 𝐁 is the 𝐣 hat unit vector. Now there’s another way to write these out that recognizes that 𝐀 and 𝐁 are three dimensional. Instead of writing that vector 𝐀 equals 𝐒 hat, we can say that it equals one 𝐒 hat plus zero 𝐣 hat plus zero 𝐀 hat. And then likewise for vector 𝐁, we can write it as zero 𝐒 hat plus one 𝐣 hat plus zero 𝐀 hat. We’ve now defined two perpendicular unit vectors that exist in three-dimensional space.

We can begin then computing the two vectors that are combined in this dot product. The first vector, three 𝐀 minus 𝐁, is equal to three multiplied by all the components of vector 𝐀 minus all the components of vector 𝐁. And since these are vectors we’re working with, we combine their components separately. For the 𝐒-component, we have three times one minus zero, or three; for the 𝐣-component, we have three times zero minus one, or negative one; and for the 𝐀-component, three times zero minus zero, or zero.

So we now have three 𝐀 minus 𝐁. We follow a similar process to compute negative two 𝐀 plus 𝐁. Once again, we combine the vectors by their components. Negative two times one plus zero is negative two. Then negative two times zero plus one is positive one. And negative two times zero plus zero is zero. Negative two 𝐀 plus 𝐁 then equals negative two 𝐒 plus 𝐣 plus zero 𝐀.

We’re now ready to calculate our dot product. When we substitute in our vector for three 𝐀 minus 𝐁 and our vector for negative two 𝐀 plus 𝐁, our next step is to multiply these vectors together by their components. For the 𝐒-component, we have three times negative two. That gives us negative six. Then for the 𝐣-component, we have negative one times one. That’s negative one. And for the 𝐀-component, we have zero times zero, which is zero.

All together then, this dot product equals negative six plus negative one plus zero, or negative seven. And note that we would get the same result regardless of how we defined 𝐀 and 𝐁 so long as they satisfied our two conditions of being perpendicular and unit vectors.

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