Let $\,\vec v = \langle a,b\rangle\,$ be a vector.
Depending upon the signs (plus or minus) of $\,a\,$ and $\,b\,,$
the vector $\,\vec v\,$ is one of the four vectors shown below.
(To match the diagram, suppose that $\,a\,$ and $\,b\,$ are both nonzero.)
In all four cases, the length (size, magnitude) of $\,\vec v\,$ is the hypotenuse of a triangle with sides of length $\,|a|\,$ and $\,|b|\,.$ Recall that $\,\|\vec v\|\,$ denotes the length of $\,\vec v\,.$ We have: $$ \begin{alignat}{2} \cssId{s7}{\|\vec v\|^2} \quad &=\quad \cssId{s8}{|a|^2 + |b|^2} &&\qquad \cssId{s9}{\text{(by the Pythagorean Theorem)}}\cr &=\quad \cssId{s10}{a^2 + b^2} &&\qquad \cssId{s11}{\text{($x^2 = |x|^2\,,$ since they have the same size and sign)}}\cr \end{alignat} $$ Take the square root of both sides, and use the fact that $\,\|\vec v\|\ge 0\,.$ The result is the formula for the length of $\,\vec v = \langle a,b\rangle\,$:
$$
\|\vec v\| = \sqrt{a^2 + b^2} \qquad \text{(vector length formula)}
$$
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![]() is one of these four vectors:
$$
\|\vec v\| = \sqrt{a^2 + b^2}
$$
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Let $\,a\,,$ $\,b\,,$ and $\,k\,$ be real numbers.
Let $\,\vec v = \langle a,b\rangle\,.$
Then,
$$
\begin{alignat}{2}
\cssId{s56}{\|k\vec v\|} \quad&=\quad \cssId{s57}{\|\,k\langle a,b\rangle\,\|} \qquad\qquad&&\cssId{s58}{\text{(definition of $\,\vec v\,$)}}\cr
&=\quad \cssId{s59}{\|\,\langle ka,kb \rangle\,\|} \qquad\qquad&&\cssId{s60}{\text{(multiply a vector by a scalar)}}\cr
&=\quad \cssId{s61}{\sqrt{(ka)^2 + (kb)^2}} \qquad\qquad&&\cssId{s62}{\text{(the vector length formula)}}\cr
&=\quad \cssId{s63}{\sqrt{k^2a^2 + k^2b^2}} \qquad\qquad&&\cssId{s64}{\text{(exponent law, squaring a product)}}\cr
&=\quad \cssId{s65}{\sqrt{k^2(a^2 + b^2)} } \qquad\qquad&&\cssId{s66}{\text{(factor)}}\cr
&=\quad \cssId{s67}{\sqrt{k^2}\sqrt{a^2 + b^2}} \qquad\qquad&&\cssId{s68}{\text{(property of radicals)}}\cr
&=\quad \cssId{s69}{|k| \cdot \sqrt{a^2 + b^2}} \qquad\qquad&&\cssId{s70}{\text{($\ \sqrt{x^2}=|x|\ $)}}\cr
&=\quad \cssId{s71}{|k|\cdot \|\vec v\| } \qquad\qquad&&\cssId{s72}{\text{(the vector length formula)}}
\end{alignat}
$$
So, $\,\|k\vec v\| = |k|\cdot \|\vec v\|\,.$
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