Video: EG19M2-ALGANDGEO-Q14V2

If 𝐴 and 𝐡 are two unit vectors, then 𝐴 β‹… 𝐡 πœ– οΌΏ. [A] ]0, 1[ [B] ]βˆ’1, 1[ [C] [βˆ’1, 1] [D] ℝ⁺.

07:06

Video Transcript

If 𝐴 and 𝐡 are two unit vectors, then 𝐴 dot 𝐡 is a member of blank. The options are a) the open interval from zero to one, b) the open interval from negative one to one, c) the closed interval from negative one to one, and d) the set of positive real numbers.

Let’s take a look at this 𝐴 dot 𝐡 is a member of blank. This symbol here, which I read as β€œis a member of,” expresses membership of a set. And so in this blank, we must have a set. And this is good news because all four options are sets. The first three options are written in interval notation. And we’ll come to that later. And the fourth, ℝ superscript plus, is the set of positive real numbers. You might be used to the ℝ being written in a different script to make it clear that it refers to the set of real numbers and isn’t just some variable capital ℝ which could take any value.

Okay, so what do we know about the dot product of two vectors? Well, for any two vectors, 𝐴 and 𝐡, their dot product is a number, in fact a real number. And so certainly, the dot product of 𝐴 and 𝐡 will always be a member of the set of real numbers ℝ. But the set of real numbers ℝ is not one of our options. I think what this question is really asking is if 𝐴 and 𝐡 are two unit vectors, what is the set of possible values of 𝐴 dot 𝐡. You can pause the video and think about this question if you’d like. As the dot product of any two vectors is a real number, this set must be a subset of the real numbers. But given that 𝐴 and 𝐡 are unit vectors, we’ll find that the dot product can’t be just any real number. And so the set of possible values of 𝐴 dot 𝐡 will be smaller than the set of real numbers.

Okay, so what is a unit vector? A unit vector is a vector whose magnitude is one. And so we can see that we’re given in the question that the magnitude of 𝐴 is one. And the magnitude of 𝐡 is one. How does that affect the possible values of their dot product? Here, we want to use the geometric definition of the dot product. The dot product of 𝐴 and 𝐡 is the magnitude of 𝐴 times the magnitude of 𝐡 times the cos of πœƒ, which is the measure of the angle between the vectors. And substituting the known magnitudes of 𝐴 and 𝐡 into this formula, we find that the dot product of unit vectors 𝐴 and 𝐡 is one times one cos πœƒ, which is just cos πœƒ.

Our question becomes what is the set of possible values of cos πœƒ? In other words, what is the range of cos πœƒ. But to talk about the range of a function, we’re going to need to know the domain on which it is defined. That is the set of possible values of the inputs to our function, which is πœƒ. πœƒ can be zero degrees if 𝐴 and 𝐡 represent the same vector, all the way up to 180 degrees if they point in opposite directions. And of course, πœƒ could take any value in between. We can see then that πœƒ can take any value between zero degrees and 180 degrees inclusive. And you might know that, on this domain, cos πœƒ takes its full range of values from negative one to one, again inclusive.

You can see this by thinking about the unit circle, where cos πœƒ is the π‘₯-coordinate of a point on the circle. And as πœƒ changes from zero to 180 degrees, the π‘₯-coordinate changes from one to negative one. Or if you’d prefer, you could think about the graph of 𝑦 equals cos πœƒ when πœƒ is between zero degrees and 180 degrees. Either way, you should be able to see that cos πœƒ takes all values between cos 180 degrees, which is negative one, and cos zero degrees, which is one. And so negative one is less than or equal to cos πœƒ is less than or equal to one.

But how do we write this as a set? Well, we use interval notation. We can write that cos πœƒ is a member of the closed interval from negative one to one. This interval here is the set of real numbers between negative one and one, including negative one and one. This is the set of possible values of cos πœƒ. And so it’s the set of possible values of the dot product of two unit vectors, which was after what we’re looking for. Looking through our options, we see that this is option c. And option c is therefore the answer.

You might notice that option b looks very similar. The only difference is that the square brackets appear to be the wrong way round. Option b is the open interval from negative one to one. This is the set of real numbers between negative and one, not including negative one and one. It was therefore very important for us to get our square brackets in the right orientation, to show that both endpoints, negative one and one, were included in our interval. This is the same as she was making sure that we had less than or equal to signs here and not just less than signs. Had we had less than signs here instead, the corresponding interval would have been the one in option b, which is the wrong answer.

We’ve seen then that if 𝐴 and 𝐡 are two unit vectors, then the dot product is between negative one and one inclusive. So it could be negative one or one or any real number in between, but not any other real number. And we express this fact using set and interval notation by saying that the dot product of 𝐴 and 𝐡 is a member of the closed interval from negative one to one. The trick to this question was to use not the formula for the dot product of two vectors in terms of their components, but the geometric definition of the dot product in terms of the magnitudes of the vectors and the angle πœƒ between them.

I’d just quickly like to show another method for this question, which exploits the multiple-choice format. If you take any two unit vectors which point in opposite directions, their dot product is negative one. And so negative one is certainly a member of the set that we’re looking for. But going through the options, we see that option a, the open interval from zero to one, doesn’t contain negative one and so can’t be the answer, nor can option b, which is the open interval from negative one to one as this interval doesn’t contain either of its endpoints, negative one or one. And certainly, option d can’t be our answer because negative one isn’t a positive real number. The only option that we haven’t eliminated with this example is option c. And so that must be our answer.

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