Find two vectors parallel to $v$ with $3$ times the magnitude of $v$

Question

This is $v=\left \langle 1,1\right \rangle$ , the result i got for the parallel vector and that which the textbook provides wrere different. these are the results from the textbook $\left \langle 3,3 \right \rangle$ and $\left \langle -3,-3 \right \rangle$. My approach to this problem was to first find the unit vector by diving $\left \langle 1,1 \right \rangle$ by its magnitude then multiplying it by $3$. This gives me $\left \langle \dfrac{3}{\sqrt2},\dfrac{3}{\sqrt2} \right \rangle$ not $\left \langle 3, 3 \right \rangle $

what am i doing wrong here?

Answer 1

Your result has a magnitude of exactly $3$, not three times the magnitude of $v$ as the problem required. When you created a unit vector you basically threw away the original magnitude and replaced it with $1$.

For any vector $\vec v$, $\|c\vec v\|=|c|\|\vec v\|$. Since you need to end up with $3\|\vec v\|$, the solutions are $\pm3\vec v$.

Answer 2

You can solve it as follows: Let $u = \left \langle x,y \right \rangle$ be the desire vector, and since it is parallel to $v$, we have $x = y$. The condition $|u| = 3|v|$ implies $\sqrt{x^2+y^2} = |x|\sqrt{2}=3\sqrt{2}\implies x = y = \pm 3$. Thus you have $2$ vectors: $\left \langle 3,3 \right \rangle$ and $\left \langle -3,-3 \right \rangle$ .