Video: MATH-ALG+GEO-2018-S1-Q14v2

MATH-ALG+GEO-2018-S1-Q14v2

06:26

Video Transcript

In the Cartesian π‘₯𝑦-plane, πœƒ is the measure of the angle between vectors 𝐴 and 𝐡. Write the magnitude of the cross product of 𝐴 and 𝐡 over the dot product of 𝐴 and 𝐡 in terms of πœƒ.

This expression that we’d like to find in terms of πœƒ involves the cross products of 𝐴 and 𝐡 and also their dot product. And we can write each of these products in terms of πœƒ, the measure of the angle between 𝐴 and 𝐡, by using their geometric definitions.

The dot product of 𝐴 and 𝐡 is the magnitude of 𝐴 times the magnitude of 𝐡 times the cosine of the angle πœƒ between them. The geometric definition of the cross product is slightly more tricky. The cross product of 𝐴 and 𝐡 is the magnitude of 𝐴 times the magnitude of 𝐡 times sine πœƒ times another vector 𝐢. And so, we see that while the dot product or scalar product gives us a number or scalar quantity, the cross product β€” also known as the vector product β€” gives us a vector quantity. This vector 𝐢 is unit vector perpendicular to both 𝐴 and 𝐡. This vector 𝐢 gives us the direction of the cross product. And the rest of the definition β€” this magnitude of 𝐴 times magnitude of 𝐡 times sine πœƒ β€” gives us the magnitude of the cross product.

Now for the purposes of our question, we’re interested in the magnitude of this cross product. And as 𝐢 is unit vector, that means that its magnitude is one. Surely, the magnitude of the cross product is just this constant multiplier. The magnitude of 𝐴 times the magnitude of 𝐡 times sine πœƒ. And it turns out that this is true. The magnitude of 𝐴 cross 𝐡 is the magnitude of 𝐴 times the magnitude of 𝐡 times the sine of the angle πœƒ between them. But our justification for this fact is incomplete. We’ll come back to it at the end of the video.

Before we do that though, let’s answer our question. Writing this fraction in terms of πœƒ, we found the magnitude of 𝐴 cross 𝐡 in terms of πœƒ. It’s the magnitude of 𝐴 times the magnitude of 𝐡 times sine πœƒ. And similarly for 𝐴 dot 𝐡, we found that it is the magnitude of 𝐴 times the magnitude of 𝐡 times cos πœƒ. And now we seen in our fraction that the magnitudes of 𝐴 in the numerator and denominator cancel, as too the magnitudes of 𝐡. And we’re left with just sin πœƒ over cos πœƒ. And sin πœƒ over cos πœƒ is just tan πœƒ. We have therefore succeeded in writing the magnitude of the cross product of 𝐴 and 𝐡 over the dot product of 𝐴 and 𝐡 in terms of πœƒ.

The key idea here was to use the geometric definitions of the dot product and cross product and not the formulas for them in terms of the components of the vectors involved. But I’d still like to justify the expression we found for the magnitude of 𝐴 cross 𝐡. We found that the cross product of 𝐴 and 𝐡 was some number π‘˜ times unit vector. And so, we said that its magnitude must be π‘˜. But this isn’t always true. For example, when π‘˜ is negative two and when the unit vector is 𝑖, the unit vector in the π‘₯-direction, we see that the magnitude is two and not negative two. And in fact, this would be the same whatever unit vector we chose.

The real fact that we should be using is that if we take a vector 𝑣 and multiply it by a constant π‘˜, then its magnitude is the absolute value of π‘˜ times the magnitude of 𝑣. We can use this formula to correct ourselves then. The magnitude of π‘˜ times the unit vector is the absolute value of π‘˜ times the magnitude of the unit vector. And of course, the magnitude of the unit vector is just one. So we’re left with just the absolute value of π‘˜. So should the magnitude of the cross product of 𝐴 and 𝐡 actually be the absolute value of the magnitude of 𝐴 times the magnitude of 𝐡 times sine πœƒ? Well, no. It turns out the magnitude of 𝐴 times the magnitude of 𝐡 times sine πœƒ is always greater than or equal to zero. And so, the absolute value of this quantity is just equal to the quantity itself. We don’t need to take the absolute value here.

Let’s go through this carefully. To show that the absolute value of the magnitude of 𝐴 times the magnitude of 𝐡 times sine πœƒ is just itself, we use the fact that the absolute value of a product of vectors is equal to the product of the absolute values of those vectors. And so, we have the absolute value of the magnitude of 𝐴 times the absolute value of the magnitude of the 𝐡 times the absolute value of sine πœƒ. Now, the magnitude of 𝐴 and the magnitude of 𝐡 like the magnitude of any vector is greater than or equal to zero. And the absolute value of something greater than or equal to zero is just that something itself.

The only thing left to do here is to show that the absolute value of sin πœƒ is just sin πœƒ. This is the tricky bit. Of course, in general, sin πœƒ can be negative. But remember that we’re told in the question that πœƒ is the measure of the angle between vectors 𝐴 and 𝐡. And this angle is always taken to be the nonreflex angle between them. The greatest that πœƒ can be is 180 degrees. And of course, the least it can be is zero degrees. And on this domain, when πœƒ is between zero degrees and 180 degrees, sin πœƒ is always greater than or equal to zero. As a result, the absolute value of sin πœƒ is just sin πœƒ.

So as claimed, the absolute value of the magnitude of 𝐴 times the magnitude of 𝐡 times sin πœƒ is just the magnitude of 𝐴 times the magnitude of 𝐡 times sin πœƒ. This justifies the lack of absolute value signs we saw in the expression for the magnitude of 𝐴 cross 𝐡. And it’s well worth remembering this magnitude. It’s also have an importance to remember that we can’t always just drop the absolute value signs as we saw with the case of negative two times 𝑖. We could only do this for the cross product because the constant multiplier in the geometric definition of the cross product turns down to always be greater than or equal to zero.

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