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 at right. (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: $$ \begin{alignat}{2} \\vec v\^2 \quad &=\quad a^2 + b^2 &&\qquad \text{(by the Pythagorean Theorem)}\cr &=\quad a^2 + b^2 &&\qquad \text{($x^2 = x^2\,$, since they have the same size and sign)}\cr \end{alignat} $$ 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 $\,\vec v = \langle a,b\rangle\,$: $$ \\vec v\ = \sqrt{a^2 + b^2} \qquad \text{(vector length formula)} $$ 
is one of these four vectors: $$ \\vec v\ = \sqrt{a^2 + b^2} $$ 
Let $\,a\,$, $\,b\,$, and $\,k\,$ be real numbers.
Let $\,\vec v = \langle a,b\rangle\,$.
Then,
$$
\begin{alignat}{2}
\k\vec v\ \quad&=\quad \\,k\langle a,b\rangle\,\ \qquad\qquad&&\text{(definition of $\,\vec v\,$)}\cr
&=\quad \\,\langle ka,kb \rangle\,\ \qquad\qquad&&\text{(multiply a vector by a scalar)}\cr
&=\quad \sqrt{(ka)^2 + (kb)^2} \qquad\qquad&&\text{(the vector length formula)}\cr
&=\quad \sqrt{k^2a^2 + k^2b^2} \qquad\qquad&&\text{(exponent law, squaring a product)}\cr
&=\quad \sqrt{k^2(a^2 + b^2)} \qquad\qquad&&\text{(factor)}\cr
&=\quad \sqrt{k^2}\sqrt{a^2 + b^2} \qquad\qquad&&\text{(property of radicals)}\cr
&=\quad k \cdot \sqrt{a^2 + b^2} \qquad\qquad&&\text{($\ \sqrt{x^2}=x\ $)}\cr
&=\quad k\cdot \\vec v\ \qquad\qquad&&\text{(the vector length formula)}
\end{alignat}
$$
So, $\,\k\vec v\ = k\cdot \\vec v\\,$.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
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