### Video Transcript

Given that ๐ฎ and ๐ฏ are in โ ๐ and that the inner product of ๐ฎ and ๐ฏ is equal to ๐ times the modulus of ๐ฎ plus ๐ฏ all squared minus the modulus of ๐ฎ minus ๐ฏ all squared, find the value of ๐.

An easy way for tackling any equations such as these is to directly evaluate both sides of the equation in terms of series. Since ๐ฎ and ๐ฏ are both in โ ๐, their tuples look something like this: ๐ข one, ๐ข two, all the way to ๐ข ๐, and likewise ๐ฃ one, ๐ฃ two, all the way to ๐ฃ ๐. We can use these tuples to directly evaluate both sides of the equation in terms of a series.

Letโs start with the left-hand side. In โ ๐, the inner product of ๐ฎ and ๐ฏ is just ๐ข one times ๐ฃ one plus ๐ข two times ๐ฃ two, et cetera, all the way to ๐ข ๐ times ๐ฃ ๐. We can reexpress this as a series as the sum from ๐ equals one to ๐ of ๐ข ๐ times ๐ฃ ๐.

Now letโs look at the terms on the right-hand side. The modulus of ๐ฎ plus ๐ฏ all squared is just the components of ๐ฎ and ๐ฏ added together and squared separately. So we have ๐ข one plus ๐ฃ one all squared plus ๐ข two plus ๐ฃ two all squared plus et cetera, all the way to ๐ข ๐ plus ๐ฃ ๐ all squared. We can once again express this as a series as the sum from ๐ equals one to ๐ of ๐ข ๐ plus ๐ฃ ๐ all squared.

Similarly, for the second term on the right-hand side, we have ๐ฎ minus ๐ฏ all squared equals ๐ข one minus ๐ฃ one all squared plus ๐ข two minus ๐ฃ two all squared plus et cetera, all the way to ๐ข ๐ minus ๐ฃ ๐ all squared. And again, we can reexpress this as a series as the sum from ๐ equals one to ๐ of ๐ข ๐ minus ๐ฃ ๐ all squared. So again, ignoring the ๐ for now, we can reexpress the right-hand side as the sum from ๐ equals one to ๐ of ๐ข ๐ plus ๐ฃ ๐ all squared minus ๐ข ๐ minus ๐ฃ ๐ all squared.

Expanding out the parentheses in the general term, we get ๐ข ๐ squared plus two ๐ข ๐ ๐ฃ ๐ plus ๐ฃ ๐ squared minus ๐ข ๐ squared plus two ๐ข ๐ ๐ฃ ๐ minus ๐ฃ ๐ squared. These positive and negative squares will cancel each other out. And this leaves us with the sum from ๐ equals one to ๐ of four ๐ข ๐ ๐ฃ ๐.

Now, substituting these series into the original equation, we get the sum from ๐ equals one to ๐ of ๐ข ๐ ๐ฃ ๐ is equal to ๐ times the sum from ๐ equals one to ๐ of four ๐ข ๐ ๐ฃ ๐. We can take this constant of four outside of the series summation. And by direct comparison, we can see that ๐ is equal to one-quarter.