Find the magnitude of 𝐀 cross 𝐁 divided by 𝐀 dot 𝐁. (A) sin of 𝜃, (B) zero, (C) cos of 𝜃, (D) tan of 𝜃, (E) one.
Alright, in this example, we have these two unknown vectors 𝐀 and 𝐁. We want to figure out what this particular combination of those vectors is equal to in terms of our answer options. To figure this out, we’ll want to recall some lesser-known identities for the cross product and the dot product of two vectors.
In general, if we have two vectors 𝐀 and 𝐁 and the angle between these two vectors is 𝜃, then we can write that the cross product of 𝐀 and 𝐁 equals the magnitude of these vectors individually multiplied together times of the sine of the angle between them in the direction that’s perpendicular to both 𝐀 and 𝐁. This implies that if we take the magnitude of 𝐀 cross 𝐁, that equals the magnitude of 𝐀 times the magnitude of 𝐁 times the sin of 𝜃. And note that this combination of vectors 𝐀 and 𝐁 matches the numerator of this fraction we’re given.
This being the numerator of our fraction, now let’s consider the denominator 𝐀 dot 𝐁. If we once again assume that 𝐀 and 𝐁 are two vectors separated by an angle 𝜃, then if we take their dot product, that’s equal to the magnitude of each vector individually multiplied together times the cosine of the angle between them. We now have an equivalent expression for the denominator of our fraction. If we divide the first equation by the second equation, we get this single resulting equation.
Notice that the left-hand side exactly matches up the expression we were given in our problem statement. Considering the right side of this equation, notice that the magnitudes of 𝐀 and 𝐁 cancel from numerator and denominator. Going further, we can recall the trigonometric identity that the sine of an angle divided by the cosine of that same angle equals the tangent of that angle. Therefore, sin 𝜃 over cos 𝜃 equals tan 𝜃. We see this listed as one of our answer options. So this is the choice we pick. The magnitude of 𝐀 cross 𝐁 divided by 𝐀 dot 𝐁 equals the tangent of the angle between these two vectors.