Question Video: Using the Operators of Vectors and Dot Product between them to Find the Value of an Expression | Nagwa Question Video: Using the Operators of Vectors and Dot Product between them to Find the Value of an Expression | Nagwa

# Question Video: Using the Operators of Vectors and Dot Product between them to Find the Value of an Expression Mathematics

Find the value of (|๐ ร ๐|ยฒ + |๐ โ ๐|ยฒ)/(2|๐|ยฒ|๐|ยฒ).

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### Video Transcript

Find the value of the magnitude of the cross product between vectors ๐ and ๐ squared plus the magnitude of the dot product between vectors ๐ and ๐ squared all divided by two multiplied by the magnitude of ๐ squared times the magnitude of ๐ squared.

In this question, weโre asked to find the value of an expression involving magnitudes, cross products, and dot products of two vectors ๐ and ๐. And there doesnโt seem to be any simple way of simplifying this expression. So weโre going to need to use the definitions of the cross product and the dot product. We recall if ๐ is the angle between two vectors ๐ and ๐, then the dot product between vectors ๐ and ๐ is equal to the magnitude of vector ๐ multiplied by the magnitude of vector ๐ times the cos of ๐.

Similarly, if ๐ is the angle between two vectors ๐ and ๐, then the cross product between vectors ๐ and ๐ is defined to be equal to the magnitude of vector ๐ times the magnitude of vector ๐ multiplied by the magnitude of the sin of ๐ times a unit vector ๐ฎ, where itโs worth noting ๐ฎ is perpendicular to vectors ๐ and ๐ and the direction of vector ๐ฎ is determined by the right-hand rule. However, in this example, we donโt need to worry about this unit vector ๐ฎ since weโre taking the magnitude of the cross product between vectors ๐ and ๐. And ๐ฎ is a unit vector. So when we take the magnitude of this cross product, weโll end up with the product of the magnitudes of the right-hand side of the equation.

So we have the magnitude of vector ๐ cross ๐ is the magnitude of ๐ times the magnitude of ๐ multiplied by the magnitude of the sin of ๐ times the magnitude of the unit vector ๐ฎ, which is just equal to one. So this simplifies to give us the magnitude of ๐ times the magnitude of ๐ multiplied by the magnitude of the sin of ๐.

Weโre now ready to apply these two results to simplify this expression. Letโs start with the first term in the numerator. Weโll replace the magnitude with the cross products of vectors ๐ and ๐ with the magnitude of vector ๐ times the magnitude of vector ๐ multiplied by the magnitude of the sin of ๐. And donโt forget, we need to square this expression. Letโs now rewrite the second term in the numerator. This time, weโre going to rewrite the magnitude of the dot product between vectors ๐ and ๐ as the magnitude of vector ๐ times the magnitude of vector ๐ multiplied by the cos of ๐. 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, weโre going to want to distribute the exponents over both sets of parentheses in the numerator. This gives us the magnitude of ๐ squared times the magnitude of ๐ squared multiplied by the magnitude of sin ๐ squared plus the magnitude of ๐ squared times the magnitude of ๐ squared times cos squared ๐ all divided by two times magnitude of ๐ squared multiplied by the magnitude of ๐ squared.

And before we continue, it is worth reiterating vectors ๐ and ๐ canโt be the zero vectors, since, in our original expression, weโre dividing by the magnitudes of ๐ and ๐. So now we can simplify this expression. First, weโll cancel the shared factor of the magnitude of ๐ squared in the numerator and denominator. Similarly, weโll cancel the shared factor of magnitude of ๐ squared in the numerator and denominator. And we can do this because we know ๐ and ๐ are not the zero vector. So the magnitude of ๐ and the magnitude of ๐ are not equal to zero.

This then leaves us with the magnitude of the sin of ๐ squared plus the cos squared of ๐ 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 ๐ all squared is just sin squared of ๐. So we can rewrite this as sin squared ๐ plus cos squared ๐ 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 ๐ฅ, the sin squared of ๐ฅ plus the cos squared of ๐ฅ 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 ๐ and ๐ squared added to the magnitude of the dot product between vectors ๐ and ๐ squared all divided by two times the magnitude of ๐ squared multiplied by the magnitude of ๐ squared is actually equal to one-half.