Lesson Video: The Scalar Product of Two Vectors | Nagwa Lesson Video: The Scalar Product of Two Vectors | Nagwa

# Lesson Video: The Scalar Product of Two Vectors Physics • First Year of Secondary School

In this video, we will learn how to calculate the scalar product of two vectors using both the components of the vectors and the magnitudes of the two vectors and the angle between them.

14:33

### Video Transcript

In this video, weβre talking about the scalar product of two vectors. Weβre going to learn a technique for combining two vectors in such a way that the outcome is a quantity that has a magnitude but no direction, that is, is a scalar.

Now, at first, we might wonder why we would ever want to do this. Why would we want to take two vectors, we could call them π and π, that both have magnitude and direction associated with them and combine them in such a way that we lose information? That is, we lose direction information. As a matter of fact, though, there are physical situations where doing this is helpful. Consider this example of a box resting on a flat surface. Letβs say that by pushing on the box with the force vector shown here, our goal is to push the box along the surface so that itβs displaced. Now, if weβre indeed able to do this, if by applying this force π, weβre able to displace the box this displacement π. Then that means that weβve done some work. We can represent this using a capital π on the box.

But then just how much work have we done? Well, it turns out that this method weβre learning of taking the scalar product of two vectors can tell us. Notice that in this situation, we have two vectors: the force weβre applying to the box and the boxβs displacement. And while the work we want to calculate depends on these vectors, the work itself is not a vector quantity but rather is a scalar quantity. So then we want to combine the force and displacement vectors in such a way that a scalar results. Weβll do this using the operation known as the scalar product. Symbolically, that operation is represented as weβve done over here, using a dot between the two vectors weβre combining.

For that reason, the scalar product of two vectors is also known as the dot product. So these two terms, scalar product and dot product, mean the same thing. Now, there are a couple of different ways to calculate a scalar or dot product. One way to do it is to use what we could call a geometric approach. Thinking from this perspective, letβs imagine these two vectors π and π that weβve named here. And letβs say that this is vector π and that this over here is vector π. If we were to calculate the scalar product of π and π, we could show this happening by moving these vectors so that the tails of each of them are overlapping. We can do that, say, by shifting vector π.

Once weβve done that, we consider the shorter of our two vectors, in this case, thatβs vector π. And then, we ask ourselves how much of vector π lies along or overlaps vector π. We can start figuring this out by sketching a line that goes from the tip of our shorter vector and then goes down to the longer vector intercepting that longer vector at a 90-degree angle. The answer to our question of how much our shorter vector, vector π, lies along our longer vector is given by calculating this distance right here. We could say that this is the overlap between our two vectors.

So, just what is this distance that weβve marked out? If we give the angle between vector π and vector π a name, say, we call it π. Then this distance weβve marked out in orange is equal to the magnitude of π times the cos of that angle. We could say that this is how much of vector π lies along vector π. And once weβve calculated this, weβre pretty close to calculating the scalar or dot product of our two vectors π and π. π dot π is equal to this overlap between vectors π and π, the magnitude of the shorter vector times the cos of the angle between them, multiplied by the magnitude of the longer vector, in this case, vector π. And we can rearrange this expression and write it this way. The magnitude of one vector times the magnitude of the other times the cos of the angle between them.

So getting back to our example of pushing this box and calculating the work we do on the box, we can say that if the angle between the force vector we exert on the box and the boxβs displacement is π. Then the work we do is equal to the magnitude of π times the magnitude of π times the cos of that angle π. This is equal to the scalar product of π and π. And notice that that product is indeed a scalar quantity. It has no direction, but it does have magnitude.

In this geometric explanation of the scalar product, weβve been assuming that one of the two vectors weβre combining is shorter than the other. But really, that doesnβt need to be the case in order to calculate the scalar product of the two vectors. Even if, for example, vectors π and vectors π were the same magnitude, we could still calculate their overlap this way. And then in multiplying that overlap by the magnitude of the other vector, in this case, vector π, we would calculate their scalar product. So, no need to focus too much on which vector is shorter or longer. This is just a way of helping us visualize what the scalar product is.

Now, we mentioned earlier that the geometric approach isnβt the only one for calculating a scalar product. Thereβs also what we could call it an algebraic way of doing this. To see how this works, letβs imagine that we have these two vectors π and π not drawn out graphically. But instead theyβre written out by their component parts like this. So, vector π is defined by its π₯- and π¦-components. And likewise vector π has some magnitude in the π’-direction and some magnitude in the π£. When π and π are defined this way, we can calculate their scalar product like this. We multiply the π₯-components of each vector, and we add that product to the product of their π¦-components.

Now, regardless of which of these two approaches we use, weβll still get the same answer. Weβll still be calculating π dot π. And notice, by the way, that when we calculate a scalar product, the order in which the two vectors appear, π dot π or π dot π, doesnβt make a difference to our answer. For example, in our top equation, if we calculated π dot π, then all that would change is the order in which the magnitude of these two vectors appear in this product. And we know that regardless of that order, the product would be the same. Likewise, in our lower equation, π dot π would just involve switching the order in which these components are multiplied. But that switch wouldnβt change the final answer that we calculate. Itβs the same either way. So, we can say in both cases that π dot π is equal to π dot π.

When we go to calculate a scalar or a dot product, sometimes we encounter special cases of the ways that the two vectors weβre combining are directed. One of these cases is when the two vectors weβre considering, we can call them π and π, point in the same direction. When thatβs the case, the angle between them, weβve called that angle π, is zero degrees. This indicates that the two vectors are parallel with one another. And we can recall that the cos of zero degrees is equal to one. Itβs the maximum value that the cosine function achieves. So, we can say that when our two vectors are parallel and the angle between them is zero degrees, then their scalar product is a maximum positive value. And that comes back to the fact that the cos of zero degrees is the maximum positive value of the cosine function.

Another special case is when our two vectors π and π are at 90 degrees to one another. This means that π is 90 degrees. And we can recall that the cos of 90 degrees is zero. This tells us that for two vectors perpendicular to one another, their scalar product is zero. And this is consistent with our geometric understanding of the scalar product. When two vectors are perpendicular, they donβt overlap one another at all. This is consistent with a scalar or dot product of zero.

And then thereβs one last special case to consider. Here, vectors π and π point in opposite directions. Itβs not that theyβre parallel, but rather theyβre antiparallel. In this case, the angle between them is 180 degrees, and the cos of that angle is negative one. This is the largest negative value the cosine function can achieve. So, we can say that when two vectors are antiparallel, when the angle between them is 180 degrees. Then their scalar product is a maximum negative value.

Now, it may come as a surprise that a scalar product can be negative. After all, weβre not calculating a vector, so shouldnβt our scalar product always be nonnegative? Shouldnβt it always be positive or zero? But in fact, it is possible to have a negative scalar quantity. And if we change our box pushing arrangement a bit, we can see an example of this. Letβs say that instead of pushing forward on the box as we are now, instead weβre pulling on it. So the force vector points toward us and away from the displacement vector. Now, if thereβre some larger force acting on the box and pushing it to the right that we havenβt drawn in here. Then itβs entirely possible that even though weβre pulling to the left on the box, the box is being displaced to the right.

And we can see that this is an example of one of our special cases of the angles between our two vectors. Here, π is equal to 180 degrees, which means that the cos of that angle is negative one. And so, when we multiply that negative number by the positive magnitude of π and the positive magnitude of π, weβll get an overall negative result for the work weβre doing on the box. So, scalar quantities can be negative, and we see that happening here. Now that we know these relationships for calculating the scalar or dot product of two vectors, letβs get some practice using them through an example.

Consider the two vectors π© equals two π’ plus three π£ and πͺ equals six π’ plus four π£. Calculate π© dot πͺ.

This representation here of these two vectors tells us that weβre to calculate their scalar or dot product. And we see weβre given the two vectors π© and πͺ in their component form. So, we can start off by recalling that the scalar product of two vectors by their components is equal to the π₯-component of the first vector times the π₯-component of the second vector. Added to the π¦-component of the first vector times the π¦-component of the second. In this equation, weβve called our vectors π and π, but those are just general names for any vectors that lie in the π₯π¦-plane.

In this example, what we want to calculate is π© dot πͺ. And to do it, we can follow this prescription for combining the components of these vectors. First, we take the π₯-component of our first vector, thatβs π©, and the π₯-component of that vector is two. And we multiply this by the π₯-component of our second vector. That second vector is πͺ and that π₯-component is six. So, we have two times six. And to that, we add the π¦-component of our first vector. That first vector is π© and that π¦-component is three multiplied by the π¦-component of our second vector. That second vector is πͺ and that π¦-component is four.

So what we have then is π© dot πͺ is equal to two times six plus three times four. Two times six is 12 and so is three times four. So, our final answer is 24. And notice that, indeed, this answer is a scalar quantity. It has a magnitude but no direction.

Letβs look now at a second example exercise.

The diagram shows two vectors, π and π. Each of the grid squares in the diagram has a side length of one. Calculate π dot π.

We see over in our diagram our two vectors, π and π, and that theyβre laid out on a grid spacing. Weβre told that each one of these grid squares has a side length of one. Weβre not told the units of these lengths, but simply that the side lengths can be represented by one single unit, whatever our unit is. Knowing this, we want to calculate the scalar product of π and π.

Now, we can recall that a scalar product involves combining two vectors. So, weβre off to a good start there because π and π are vectors. And we can recall further that, mathematically, the scalar product of two general vectors, π and π, is equal to the product of their π₯-components plus the product of their π¦-components. Now, for our two specific vectors, also called π and π, we donβt yet know their π₯- and π¦-components, but we can use this grid to find out.

We can start by laying down coordinate axes on this grid. Letβs say that for our origin, we pick the location where the tails of vectors π and π overlap. So, weβll say that this is our π₯-axis, and this is our π¦. Relative to these axes, we can define the π₯- and π¦-components of our two vectors. Just as a side note, we could pick any orientation for our π₯- and π¦-axes so long as theyβre perpendicular to one another and quantified vectors π and π that way. And our answer would come out the same.

Using these specific π₯- and π¦-axes though, letβs write out the components of vector π. We can see that along the π₯-axis, vector π extends one, two, three units. So, that means vector π is equal to three π’, three units in the π₯-direction, plus some amount in the π¦-direction. Starting again at the origin, we count up one, two, three units and see that this is the vertical extent of vector π. Therefore, we can write vector π as three π’ plus three π£. And now, weβll do the same thing for vector π. The π₯-component of vector π is equal to one, two, three, four, five, six units and its π¦-component is equal to one unit. And we can write that as one π£ or simply π£ so that vector π overall is equal to six π’ plus π£.

Now that we know the components of our two vectors, we can use this relationship to solve for their scalar product. π dot π is equal to the π₯-component of vector π, we see that π₯-component is three, multiplied by the π₯-component of vector π. And we see that π₯-component is six. So, we have three times six. And to that, we add the π¦-component of vector π. That π¦-component is three multiplied by the π¦-component of vector π. And that π¦-component as we saw is one. So, π dot π is equal to three times six plus three times one, and that is equal to 18 plus three or 21. This is π dot π, also called the scalar or dot product of π and π.

Letβs take a moment now to summarize what weβve learned about the scalar product of two vectors. In this lesson, we saw that the scalar, also called the dot, product of two vectors results in a scalar quantity. We saw that one way to calculate a scalar product, represented symbolically like this, is to multiply together the magnitudes of the two vectors involved times the cosine of the angle between the vectors. And a second, what we called algebraic way to calculate a scalar product is to combine vectors by their component parts. The product of the π₯-components plus the product of the π¦-components.

And lastly, we looked at what we called some special cases of the orientation between the two vectors involved. When π is parallel to π, then the angle between them is zero degrees and their scalar product is a maximum positive value. When π is perpendicular to π though, the angle between them is 90 degrees, thereβs no overlap between the two vectors, and their scalar product is zero. And, lastly, when π is antiparallel to π, that is, the angle between them is 180 degrees, then the scalar or dot product of these two vectors has a maximum negative value. This is a summary of the scalar product of two vectors.

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