$\displaystyle || \vec{v} || = \sqrt{ 4^2 + 2^2 + (-5)^2 } = \sqrt{16 + 4 + 25 } = \sqrt{45}$

This can be simplified:

$\displaystyle \sqrt{3 \cdot 3 \cdot 5 } = 3 \sqrt {5}$

$\displaystyle || \vec{v} || = \sqrt{ 2^2 + (-3)^2 + (-6)^2 } = \sqrt{4 + 9 + 36 } = \sqrt{49} = 7$

$\displaystyle || \vec{v} || = \sqrt{3^2 + (-2)^2 + (-2)^2 } = \sqrt{9+4+4} = \sqrt{17}$

$\displaystyle || \vec{v} || = \sqrt{ 4^2 + (-1)^2 + 3^2 } = \sqrt{16+1+9 } = \sqrt{26}$

$\displaystyle || \vec{v} || = \sqrt{ 3^2 + 4^2 + 5^2 } = \sqrt{9+16+25} = \sqrt{50 }$

This can be simplified:

$\displaystyle \sqrt{2 \cdot 5 \cdot 5 } = 5 \sqrt{2 }$

$\displaystyle \left \| \textbf{u} \right \|$, the norm, or length, of vector $\displaystyle \textbf{u}$, is equal to the square root of the sum of the squares of its elements. Therefore, 

$\displaystyle \left \| \textbf{u} \right \| = \sqrt{2^{2}+3^{2}+a^{2}+1^{2}}$

$\displaystyle = \sqrt{4+9+a^{2}+1}$

$\displaystyle = \sqrt{ a^{2}+14}$

Set this equal to 4:

$\displaystyle \sqrt{ a^{2}+14} = 4$

$\displaystyle a^{2}+14 = 16$

$\displaystyle a^{2}= 2$

$\displaystyle a = \pm \sqrt{2}$

$\displaystyle \left \| \textbf{u} \right \|$, the norm, or length, of vector $\displaystyle \textbf{u}$, is equal to the square root of the sum of the squares of its elements. Therefore, 

$\displaystyle \left \| \textbf{u} \right \|= \sqrt{(-3)^{2}+ 2 ^{2}+ x ^{2} +x ^{2}}$

$\displaystyle =\sqrt{9+4+ x ^{2} +x ^{2}}$

$\displaystyle =\sqrt{2 x ^{2} +13}$

$\displaystyle \textbf{v}$ is a unit vector if and only if its norm, or length, $\displaystyle \left \| \textbf{v} \right \|$ - the square root of the sum of the squares of its elements - is equal to 1. Find the length using this definition: 

$\displaystyle \left \| \textbf{v} \right \| = \sqrt{\left ( \frac{1}{3} \right )^{2}+ \left ( -\frac{1}{3} \right )^{2}+ \left ( \frac{1}{3} \right )^{2}}$

$\displaystyle = \sqrt{\frac{1}{9}+ \frac{1}{9} + \frac{1}{9} }$

$\displaystyle = \sqrt{\frac{1}{3} }$

$\displaystyle \left \| \textbf{v} \right \| \ne 1$, so $\displaystyle \textbf{v}$ is not a unit vector.

$\displaystyle \textbf{v}$ is a unit vector if and only if its norm, or length, $\displaystyle \left \| \textbf{v} \right \|$ - the square root of the sum of the squares of its elements - is equal to 1. Find the length using this definition: 

$\displaystyle \left \| \textbf{v} \right \| = \sqrt{\left ( \frac{1}{2} \right )^{2}+ \left ( \frac{\sqrt{2}}{2} \right ) ^{2}+ \left ( -\frac{1}{2} \right ) ^{2}}$

$\displaystyle =\sqrt{ \frac{1}{4} + \frac{2}{4} + \frac{1}{4} }$

$\displaystyle =\sqrt{1}$

$\displaystyle =1$

$\displaystyle \textbf{v}$ is a unit vector.

This function's range is $\displaystyle \mathbb{R}$, the set of all real numbers. In short, this is set of all possible "distances between two given numbers" in elementary linear algebra. $\displaystyle h$ would not be a norm. For example, $\displaystyle h(1,1,1) = \sqrt{1+1+1} =\sqrt{3}$, which is not a rational number (part of $\displaystyle \mathbb{Q}$). Similarly, $\displaystyle g$ is also not a norm. We have$\displaystyle g(1,1) = \sqrt{1+1} = \sqrt{2}$, which is not a natural number.