# Question Video: Inner Products of Vectors - The Polarization Identity Mathematics

Given that ๐ฎ and ๐ฏ โ โ^(๐) and that <๐ฎ, ๐ฏ> = ๐[|๐ฎ + ๐ฏ|ยฒ โ |๐ฎ โ ๐ฏ|ยฒ], find the value of ๐.

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### Video Transcript

Given that ๐ฎ and ๐ฏ are in โ ๐ and that the inner product of ๐ฎ and ๐ฏ is equal to ๐ times the modulus of ๐ฎ plus ๐ฏ all squared minus the modulus of ๐ฎ minus ๐ฏ 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 ๐ฎ and ๐ฏ are both in โ ๐, their tuples look something like this: ๐ข one, ๐ข two, all the way to ๐ข ๐, and likewise ๐ฃ one, ๐ฃ two, all the way to ๐ฃ ๐. We can use these tuples to directly evaluate both sides of the equation in terms of a series.

Letโs start with the left-hand side. In โ ๐, the inner product of ๐ฎ and ๐ฏ is just ๐ข one times ๐ฃ one plus ๐ข two times ๐ฃ two, et cetera, all the way to ๐ข ๐ times ๐ฃ ๐. We can reexpress this as a series as the sum from ๐ equals one to ๐ of ๐ข ๐ times ๐ฃ ๐.

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

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

Expanding out the parentheses in the general term, we get ๐ข ๐ squared plus two ๐ข ๐ ๐ฃ ๐ plus ๐ฃ ๐ squared minus ๐ข ๐ squared plus two ๐ข ๐ ๐ฃ ๐ minus ๐ฃ 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 ๐ข ๐ ๐ฃ ๐.

Now, substituting these series into the original equation, we get the sum from ๐ equals one to ๐ of ๐ข ๐ ๐ฃ ๐ is equal to ๐ times the sum from ๐ equals one to ๐ of four ๐ข ๐ ๐ฃ 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.