a) the magnitude of the vector v1+(2xv2)
b) the unit vector in the direction of (2xv1)-v2
To a):
$\displaystyle \overrightarrow{v_1} = \langle 2,-3 \rangle$
$\displaystyle \overrightarrow{v_2} = \langle 1,5 \rangle$
then $\displaystyle \overrightarrow{v_1} + 2 \cdot \overrightarrow{v_2} = \langle 2,-3 \rangle + 2 \cdot \langle 1,5 \rangle$
Continue.
To b)
Determine
$\displaystyle 2 \cdot \overrightarrow{v_1} - \overrightarrow{v_2}$
Determine the length (the absolute value) of this vector sum. (I've got $\displaystyle |2\cdot \overrightarrow{v_1} - \overrightarrow{v_2} | = \sqrt{130}$ )
Let $\displaystyle \vec u$ denote a unit vector in the direction of $\displaystyle \vec v$. Then this unit vector is determined by: $\displaystyle \vec u = \frac{\vec v}{|\vec v|}$