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In this video, weβre going to learn about products of vectors, what they are, why they are useful, and how to calculate them.
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To give an idea of how useful products of vectors are in describing physical phenomena, imagine a situation where youβre moving a large heavy box across the floor.
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You push up on the box a bit such that the force vector points above the horizontal.
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Over time, youβre able to push the box along the floor such that it ends up with the horizontal displacement vector π.
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If we want to calculate how much work is done in moving the box that distance, weβll need to make use of a product of vectors.
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Or imagine a different scenario.
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In this one, weβre sending a screw into a block of wood.
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We use a screwdriver to apply a force πΉ, a displacement of π away from the screwβs axis of rotation.
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By using a particular vector product, weβre able to tell which direction, whether into or out of the board, the screw goes.
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As we start out, it can be helpful to contrast vector products with products we may be more familiar with.
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And thatβs the products of scalars.
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Recall that a scalar is a quantity that has magnitude but no direction to it.
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Examples of scalars would be numbers such as five kilograms or 10 seconds or even 65 degrees.
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Each of these values has a magnitude without an associated direction.
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If we wanna find the product of two scalars, say five kilograms and 10 seconds.
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Then we simply multiply them as weβre used to multiplying scalar values.
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The result of this multiplication would be 50 kilogram-seconds.
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Vectors, of course, are quantities that have both magnitude and direction.
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Examples of vectors might be 10 meters to the east or two kilometers down.
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Now if someone said what is the product of these two vectors, we might not be quite so confident as how to combine them together as we were with scalars.
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After all, how would we go about combining the different directions of east and down?
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It turns out that vector products come in two particular types.
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That let us navigate this tricky question of how to combine unlike things.
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The first approach we have to finding the product of two vectors, we can call them π΄ and π΅, is to take whatβs called the dot product.
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It looks like a dot placed between those two vectors.
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If we had vector π΄ and vector π΅ with their tails at the same location.
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When we take their dot product, what weβre doing is finding out how much of the vectors lie along one another.
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Thatβs the geometrical meaning of the dot product.
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And if we give the name π to the angle between these two vectors π΄ and π΅, we can write that the dot product of π΄ and π΅ is equal to the product of their magnitudes multiplied by the cos of that angle π.
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So the dot product is useful to us in finding the angle between vectors.
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With this mathematical relationship in front of us, we can try out a few simple cases to explore this equation.
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Imagine we had two vectors, πΆ and π·, that were perpendicular to one another.
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When we consider that the dot product means the overlap of two vectors β that is, how much of one lies along the other.
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By looking at this diagram, we might naturally guess that the dot product of πΆ and π· is zero since theyβre perpendicular.
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This intuition is confirmed when we consider the equation for dot product.
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We know that the cos of 90 degrees, which is the angle between πΆ and π·, is zero.
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So weβre correct in saying that this dot product itself is zero.
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Going just a bit further, imagine that we are given π΄ and π΅ in their component forms, not graphically but numerically.
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And in this case, both vectors have three dimensions to them: π, π, and π.
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If we wanted to take the dot product of these two vectors, how would we do that mathematically?
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This gets a bit of the question we talked of earlier, of how do we combine or sort out unlike types as we have with different unit vectors in this example.
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The answer to this is that we multiply our vectors according to their type.
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π΄ π₯ times π΅ π₯ plus π΄ π¦ times π΅ π¦ plus π΄ π§ times π΅ π§, keeping the different vector components separate from one another.
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Then finally add those different component products together to find the dot product, which will ultimately be a scalar.
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Both these relationships are helpful to keep in mind when weβre working with dot products.
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The second type of vector product is called a cross product.
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And it is symbolized with an x or a cross symbol in between our test vectors, π΄ and π΅.
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While the dot product creates a scalar, taking a cross product of two vectors creates a vector.
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Another difference with the dot product is that the dot product is commutative.
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Meaning that π΄ dot π΅ is equal to π΅ dot π΄.
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But the cross product is not like that.
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That is, in general, π΄ cross π΅ is not equal to π΅ cross π΄.
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So the order is very important when we calculate cross products.
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If we again had two vectors, vector π΄ and vector π΅, then we can understand the physical meaning of the cross product this way.
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When we cross π΄ into π΅, we are solving for the components that are perpendicular to the plane in which vector π΄ and vector π΅ lie.
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For example, if our two vectors π΄ and π΅ were in a plane with π- and π-components, then their cross product would have no components in that direction.
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It would be entirely either into the page or out of the page.
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So when we think of cross products, think of results that are perpendicular to the input vectors we use.
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There is a very helpful mathematical tool for calculating cross products.
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The tool involves a three-by-three matrix of values, where we take the determinant of that matrix in order to solve for the cross product.
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The three common unit vectors π, π, and π are the head of each column.
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And the two vectors weβre crossing together, π΄ and π΅, have their components listed as the last two rows of this matrix.
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If we were to calculate the determinant of this matrix, weβd find a result with an π-, π-, and π-component.
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In other words, our result would be a vector.
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The particular value for each of those three components depends on the values of π΄ and π΅.
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But notice as we look at the three different component results, the components of π΄ and π΅ that go into each one are perpendicular to that component direction.
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For example, with π, the only components of π΄ and π΅ that affect π are in the π- and π-directions.
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In a similar way, itβs the components in the π- and π- directions that determine the magnitude of the π-component.
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And itβs the components of π΄ and π΅ in the π- and π-direction that determine the magnitude of the π-component.
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This confirms what we spoke of earlier where the cross product has to do with results perpendicular to the direction of the input vectors.
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With all that as background, letβs try a couple of example problems having to do with vector products.
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Find the angle between the two vectors π² equals 2.0π’ plus 4.0π£ plus 8.0π€ and π³ equals 6.0π’ plus 4.0π£ plus 6.0π€.
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If we call the angle between these two vectors π, itβs that we want to solve for.
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To start out, we can recall that the dot product between two vectors, π and π, is equal to the product of their magnitudes times the cosine of the angle between them.
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And we can recall further that the dot product is also equal to the product of the π₯-components plus the product of the π¦-components plus the product of the π§-components of our two vectors.
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Bringing those two relationships together, we can write that, in our case, the product of the π₯-components of our vectors plus the product of the π¦-components plus the product of the π§-components.
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Is equal to the product of their magnitudes multiplied by the cos of π, the angle between them.
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If we divide both sides of this equation by the magnitude of π² times the magnitude of π³ and then take the inverse cosine of both sides.
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So π, the angle weβre interested in, is equal to the inverse cos of π¦ π₯, π§ π₯ plus π¦ π¦, π§ π¦ plus π¦ π§, π§ π§ divided by the product of the magnitudes of our two vectors.
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We can recall that the magnitude of a vector, when it has three dimensions, is equal to the square root of the π₯-dimension squared plus the π¦-dimension squared plus the π§-dimension squared.
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If we insert the magnitude expansion for both π¦ and π§, then we now have an expression for π entirely in terms of the components of our two vectors.
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In the case of our vector π², 2.0 is π¦ π₯, 4.0 is π¦ π¦, and 8.0 is π¦ π§.
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And for π³, 6.0 is π§ π₯, 4.0 is π§ π¦, and 6.0 is π§ π§.
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When we plug each of these values in where it fits in our equation.
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With all these values plugged in, when we enter this expression on our calculator, we find π is 28 degrees.
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Thatβs the angle between our two vectors π² and π³.
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Thatβs an example using the dot product combination of vectors.
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Now letβs try an example using the cross product.
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Calculate the cross product of π² equals two π’ plus four π£ plus eight π€ and π³ equals six π’ plus four π£ plus two π€.
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In this example, we want to solve for π² cross π³, where π² and π³ are both three-dimensional vectors.
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We can begin our solution by recalling the mathematical definition for a cross product.
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The cross product of two vectors, we can call them π and π, is equal to the determinant of a three-by-three matrix , where the columns are headed by the three unit vectors π’, π£, and π€.
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And the last two rows are populated by the respective components of vectors π and π.
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We can apply this formula to our vectors π² and π³.
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As we set up our matrix, we know the top row will be the π’, π£, and π€ unit vectors.
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The next row will be the respective components of the vector π².
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Looking at π², we see that itβs π₯-component is two, its π¦-component is four, and its π§-component is eight.
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So we write those values in to our matrix.
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In the next row, weβll write the components of π³.
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The π₯-component of π³ is six.
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Its π¦-component is four.
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And its π§-component is two.
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Now weβre ready to calculate the determinant of this matrix and solve for the cross product of π² cross π³.
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When we compute this determinant, we find that the π’-component is eight minus 32, or negative 24.
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The π£-component is four minus 48, or negative 44.
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And the π€-component is eight minus 24, which equals negative 16.
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Our overall cross product then is negative 24π’ minus 44π£ minus 16π€.
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Thatβs the cross product of the vectors π² and π³.
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To summarize, products of vectors are useful because they help us to understand and themselves reveal physical phenomena.
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These could include the direction to point a windmill or the best way to get power out of an engine.
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There are two types of products of vectors that weβve learned: the dot product, or the scalar product, and the cross product, which produces a vector.
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And finally, when calculating vector products, matrices can be helpful.
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For example, the dot product of π and π can be expressed as a combination of two matrices.
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And the cross product of π and π can be expressed as the determinant of a three-by-three matrix.