Finding the Unit Quaternion

Question

How can i take a Quaternion and find the Unit Quaternion. How can I find the Unit Quaternion (Norm of a Quaternion). The norm of a Quaternion should be equal to $1$

E.g. $a=(2-i+2j-3k)$

Here is what i have done so far

$=\frac{(2-1+2-3)}{\sqrt{(2)^2+(-1)^2+(2)^2+(-3)^2}}$

$=\frac{0}{4.242640687}$

$=0$

I know the answer isnt $0$. It should be $1$. But i dont know where i went wrong.

Answer 1

If you have a quaternion $a + bi + cj + dk$, the unit quaternion corresponding to it is

$$\frac{a + bi + cj + dk}{\sqrt{a^2 + b^2 + c^2 + d^2}}.$$

Particularly, the norm is $\sqrt{a^2+b^2+c^2+d^2}$.

Answer 2

To normalize the quaternion you do indeed divide by the norm which is $\sqrt{2^2+(-1)^2+2^2+(-3^2)}$. However, you need to divide each component by the norm rather than just the coefficients. So your quaternion becomes

$$ \tiny{ \frac{2}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}} + \frac{-i}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}} + \frac{2j}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}} + \frac{-3k}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}}} $$

Then you have norm $1$!

To show this, calculate the norm again:

$$ \sqrt{ \tiny{ \left( \frac{2}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}} \right) ^2 + \left( \frac{-1}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}} \right) ^2+ \left( \frac{2}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}} \right) ^2+ \left( \frac{-3}{\sqrt{2^2+(-1)^2+2^2+(-3^2)}} \right) ^2 } } $$

which without much analysis turns out to be exactly

$$ \sqrt{\frac{2^2+(-1)^2+2^2+(-3^2)}{2^2+(-1)^2+2^2+(-3^2)}}=\sqrt{1}=1. $$

To do that note that the denominators are all the name, and have a square root. The last expression should then have the denominator without the square root. As for the numerator, we add up the square of the latter.