Video Transcript
Find the value of the magnitude of
the cross product between vectors 𝐀 and 𝐁 squared plus the magnitude of the dot
product between vectors 𝐀 and 𝐁 squared all divided by two multiplied by the
magnitude of 𝐀 squared times the magnitude of 𝐁 squared.
In this question, we’re asked to
find the value of an expression involving magnitudes, cross products, and dot
products of two vectors 𝐀 and 𝐁. And there doesn’t seem to be any
simple way of simplifying this expression. So we’re going to need to use the
definitions of the cross product and the dot product. We recall if 𝜃 is the angle
between two vectors 𝐀 and 𝐁, then the dot product between vectors 𝐀 and 𝐁 is
equal to the magnitude of vector 𝐀 multiplied by the magnitude of vector 𝐁 times
the cos of 𝜃.
Similarly, if 𝜃 is the angle
between two vectors 𝐀 and 𝐁, then the cross product between vectors 𝐀 and 𝐁 is
defined to be equal to the magnitude of vector 𝐀 times the magnitude of vector 𝐁
multiplied by the magnitude of the sin of 𝜃 times a unit vector 𝐮, where it’s
worth noting 𝐮 is perpendicular to vectors 𝐀 and 𝐁 and the direction of vector 𝐮
is determined by the right-hand rule. However, in this example, we don’t
need to worry about this unit vector 𝐮 since we’re taking the magnitude of the
cross product between vectors 𝐀 and 𝐁. And 𝐮 is a unit vector. So when we take the magnitude of
this cross product, we’ll end up with the product of the magnitudes of the
right-hand side of the equation.
So we have the magnitude of vector
𝐀 cross 𝐁 is the magnitude of 𝐀 times the magnitude of 𝐁 multiplied by the
magnitude of the sin of 𝜃 times the magnitude of the unit vector 𝐮, which is just
equal to one. So this simplifies to give us the
magnitude of 𝐀 times the magnitude of 𝐁 multiplied by the magnitude of the sin of
𝜃.
We’re now ready to apply these two
results to simplify this expression. Let’s start with the first term in
the numerator. We’ll replace the magnitude with
the cross products of vectors 𝐀 and 𝐁 with the magnitude of vector 𝐀 times the
magnitude of vector 𝐁 multiplied by the magnitude of the sin of 𝜃. And don’t forget, we need to square
this expression. Let’s now rewrite the second term
in the numerator. This time, we’re going to rewrite
the magnitude of the dot product between vectors 𝐀 and 𝐁 as the magnitude of
vector 𝐀 times the magnitude of vector 𝐁 multiplied by the cos of 𝜃. And once again, we need to square
this expression.
And now, we leave the denominator
unchanged. This gives us the following
expression. We now want to simplify this even
further. And to do this, we’re going to want
to distribute the exponents over both sets of parentheses in the numerator. This gives us the magnitude of 𝐀
squared times the magnitude of 𝐁 squared multiplied by the magnitude of sin 𝜃
squared plus the magnitude of 𝐀 squared times the magnitude of 𝐁 squared times cos
squared 𝜃 all divided by two times magnitude of 𝐀 squared multiplied by the
magnitude of 𝐁 squared.
And before we continue, it is worth
reiterating vectors 𝐀 and 𝐁 can’t be the zero vectors, since, in our original
expression, we’re dividing by the magnitudes of 𝐀 and 𝐁. So now we can simplify this
expression. First, we’ll cancel the shared
factor of the magnitude of 𝐀 squared in the numerator and denominator. Similarly, we’ll cancel the shared
factor of magnitude of 𝐁 squared in the numerator and denominator. And we can do this because we know
𝐀 and 𝐁 are not the zero vector. So the magnitude of 𝐀 and the
magnitude of 𝐁 are not equal to zero.
This then leaves us with the
magnitude of the sin of 𝜃 squared plus the cos squared of 𝜃 all divided by
two. And we can simplify this
further. Taking the magnitude of a number
and then squaring it is the same as just squaring the number. So the magnitude of sin of 𝜃 all
squared is just sin squared of 𝜃. So we can rewrite this as sin
squared 𝜃 plus cos squared 𝜃 all divided by two. But now we can see the numerator of
this expression could be simplified by using the Pythagorean identity. For any angle 𝑥, the sin squared
of 𝑥 plus the cos squared of 𝑥 is equal to one. So we write in the numerator as one
gives us one-half, which is our final answer.
Therefore, we were able to show the
magnitude of the cross product between vectors 𝐀 and 𝐁 squared added to the
magnitude of the dot product between vectors 𝐀 and 𝐁 squared all divided by two
times the magnitude of 𝐀 squared multiplied by the magnitude of 𝐁 squared is
actually equal to one-half.
