WEBVTT
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In the rectangle π΄π΅πΆπ· shown in the figure, calculate ππ cross ππ if π’, π£, and π€ form a right-hand system of unit vectors.
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Looking at our rectangle, we see itβs eight centimeters tall and 16 centimeters wide and it exists in a space defined by a right-hand system of unit vectors π’, π£, and π€.
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Relative to the rectangle, we see the π’ hat unit vector points to the right and the π£ hat unit vector points up.
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This tells us that the π€ hat unit vector points out of the screen at us.
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Anyway, we want to calculate this cross product, ππ cross ππ.
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Letβs first define these vectors.
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Vector ππ is a vector from point π· to point π΄ in our rectangle.
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And likewise, vector ππ is a vector from point π΅ to point π, the middle of our rectangle.
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Now because this rectangle lies in what we can call the π’π£-plane and because we can say that one centimeter on this plane is equal to one unit of distance, we can then write these vectors ππ and ππ according to their π’ hat and π£ hat components.
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Considering first vector ππ, we see that itβs entirely in the vertical direction.
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That means it will have no π’ hat component.
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It will, however, have a negative π£ hat component.
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And the reason is that this vector points opposite the positive π£ hat direction.
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That distance we know, the length of this vector, equals the length of the height of our rectangle, eight centimeters or just eight.
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Keeping both π’ hat and π£ hat components then, ππ equals zero π’ hat minus eight π£ hat.
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Looking next at vector ππ, we see that this vector will have both an π’ hat and a π£ hat component because it points diagonally.
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Itβs the components of this vector that weβre after, in other words, this vertical distance here and this horizontal distance here.
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Considering its horizontal component, we know that vector π points to the left.
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Thatβs in the negative π’ hat direction.
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And since it goes to the midpoint of our rectangle β which is 16 centimeters wide β it will have an π’ hat component of negative eight.
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Considering then the π£ hat component β this part of our vector β we see that it points in the positive π£ hat direction and that itβs equal in length to one-half the height of our rectangle.
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The π£-component of ππ then is positive four.
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So weβve now got our two vectors defined in terms of the unit vectors of our space.
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We can now move towards calculating their cross product.
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In general, if we have two vectors β weβll call them π and π β that lie in the π’ hat and π£ hat plane, then the cross product π cross π equals the π’-component of π times the π£-component of π minus the π£-component of π times the π’-component of π all in the π€ hat unit vector direction.
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Notice then that this cross product is perpendicular to both vectors π and π.
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And this is always the case for a cross product.
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When we go to cross ππ and ππ then, our formula tells us that this equals the π’-component of our first vector β thatβs vector ππ and that π’-component is zero β multiplied by the π£-component of our second vector β that second vector is ππ and the π£-component is four.
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Then from this we subtract the π£-component of our first vector β that vector is ππ and its π£-component is negative eight β multiplied by the π’-component of our second vector β that second vector is ππ and its π’-component is negative eight.
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And this resulting vector, we know, is in the π€ hat unit vector direction.
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Evaluating this result, we know that zero times four is zero and negative eight times negative eight is positive 64.
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Our overall outcome then is negative 64π€ hat.
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ππ cross ππ then results in a vector that points 64 units or 64 centimeters into the screen.