Problem: If $\vec u$ is a unit vector and $s$ is a positive scalar satisfying $s\vec u + \left\langle 2,-3,-5\right\rangle =\vec 0$, find $s$ and $\vec u$.
Not sure exactly how to approach this problem.
I would assume $s\vec u$ has to equal $\left\langle -2, 3, 5\right\rangle$. Not sure how to find unit vector $\vec u$ if a unit vector is just equal to $1$ usually (like $i$, $j$ and $k$).
Thanks for any help
$s\vec u+\langle2,-3,-5\rangle=0$ is equivalent to $s\vec u=-\langle2,-3,-5\rangle=\langle-2,3,5\rangle=\vec{v}$ and $\vec u=\frac {\vec v}s$. This implies $\left\|\vec u\right\|=\left\|\frac{\vec v}s\right\|=\frac{\left\|\vec v\right\|}{s^2}$.
The magnitude of $\vec v$ is $\left\|\vec{v}\right\|=(-2)^2+3^2+5^2=38$
In order for $\left\|\vec u\right\|=1$, $\frac{\left\|\vec v\right\|}{s^2}=1$, thus $s=\pm\sqrt{\left\|\vec v\right\|}=\pm\sqrt{38}$. As $s$ is positive, $s=\sqrt {38}$
Subtitute the value of $s$ and the vector $\vec v$ into $\vec u=\frac {\vec v}s$.
$\vec u={\langle-2,3,5\rangle\over \sqrt{38}}=\left\langle\frac{-2}{\sqrt{38}},\frac{3}{\sqrt{38}},\frac{5}{\sqrt{38}}\right\rangle$