Question Video: Using the Operators of Vectors and Dot Product Between them to Find the Value of an Expression Mathematics • Third Year of Secondary School

Video Transcript

Find the value of the magnitude of 𝚨 cross 𝚩 squared plus the magnitude of 𝚨 dot 𝚩 squared divided by two times the magnitude of 𝚨 squared times the magnitude of 𝚩 squared.

Here, we know that 𝚨 and 𝚩 are vectors. But beyond that, we don’t know anything about them. In that sense, they’re completely general. We don’t know anything about their components.

As a start point to evaluate this expression though, we see that we’re crossing as well as dotting these two vectors. In general, if we cross two vectors, called 𝚨 and 𝚩, then that cross product equals the magnitude of the first vector times the magnitude of the second vector times the sine of the angle between the two vectors, called 𝜃 here. And this whole result points in a direction that is normal or perpendicular both to vector 𝚨 and to vector 𝚩.

If we then take the absolute value of this cross product, we’re no longer calculating a vector. So we don’t have a direction associated with our result. And whatever our angle 𝜃 is, the sine of that angle must be a nonnegative number. This is so that overall the magnitude of 𝚨 cross 𝚩 is itself nonnegative.

In our given expression, we see that we’re working not with the magnitude of 𝚨 cross 𝚩 but the magnitude of 𝚨 cross 𝚩 squared. That would give us this expression. And if we consider the square of the magnitude of the sin of 𝜃, we find that either one of these operations, taking the absolute value or squaring sin 𝜃, would make the result nonnegative.

To make this value positive or zero then, we only need to apply one of these two operations. We’ll choose to remove the absolute value bars but continue to square this value. This way, we’ll end up with a nonnegative result that reflects the fact that we’re squaring the sin of 𝜃. So instead of the magnitude of 𝚨 cross 𝚩 squared, we can write this.

When it comes to the magnitude of 𝚨 dot 𝚩 squared, let’s recall that, in general, the dot product of two vectors is equal to the product of their magnitudes multiplied by the cosine of the angle between the two vectors. So then, the magnitude of 𝚨 dot 𝚩 equals the magnitude of 𝚨 times the magnitude of 𝚩 times the magnitude of the cos of 𝜃. And therefore, when we square this quantity, we get the square of the magnitude of 𝚨 times the square of the magnitude of 𝚩 times the square of the magnitude of cos 𝜃. Once again, it’s unnecessary both to take the absolute value and to square our trigonometric term. And so we can remove the absolute value bars and not change this result.

Looking back at our given expression, we see that we are adding together the magnitude of 𝚨 cross 𝚩 squared and the magnitude of 𝚨 dot 𝚩 squared. Based on what we found so far, that gives us this expression on the right-hand side here. And notice that the value of the magnitude of 𝚨 squared times the magnitude of 𝚩 squared is common to both of these terms. If we factor it out, then we see something very interesting. We have the sine squared of an angle plus the cosine squared of that same angle. Whenever this happens, whenever we have the sine squared of one angle and the cosine squared of that same angle being added together, the result equals one. And so the right-hand side of our expression simplifies to the magnitude of 𝚨 squared times the magnitude of 𝚩 squared.

And let’s recall that we’re adding together these two terms, which form the entire numerator of our fraction. Writing out our entire fraction then, we get this result. We see that here the magnitude of vector 𝚨 squared cancels in numerator and denominator, as does the magnitude of vector 𝚩 squared. This whole expression then simplifies to one over two.

For two general vectors 𝚨 and 𝚩 then, the magnitude of 𝚨 cross 𝚩 squared plus the magnitude of 𝚨 dot 𝚩 squared divided by two times the magnitude of 𝚨 squared times the magnitude of 𝚩 squared equals one-half.