Question

Help with Precalc Vectors? Medal and fan v = (-7,8) Find the magnitude of vector v. Express your answer in radical form sqrt of 113 is what I got for that. second part: use the magnitude and the angle of vector v to write it in the form (lvlcostheta, lvl sin theta). I don't know how to determine the single angle for one vector. Earlier in a different part of the problem, I was asked to find the angle between two vectors, but I don't think you use that here since no second vector is provided. So how would I answer the second part? Thanks(:

Answer 1

Hello Kaviipik!
Lets find the mag of v = (-7,8). Since we have one order pair we are finding the mag from the origin.
Yes we are using Pythagorean Theorem to calculate the magnitude
$a^2 + b^2 = c^2$ 

$\sqrt{v^2} = \sqrt{(-7)^2 + (8)^2}$
$v = \sqrt{(-7)^2 + (8)^2}$
$v = \sqrt{49 + 64}$
$v = \sqrt{113}$
$v = \sqrt{113} = 10.6301458$
Now  we use 
$\theta = tan^{-1} (\frac{b}{a})$
$\theta = tan^{-1} (\frac{8}{-7})= 48.8$
$(\sqrt{113} \times cos(48.8) , \sqrt{113} \times sin(48.8) )$

Answer 2

$\bf v=(\stackrel{a}{-7}~,~\stackrel{b}{8})\qquad ||v||=\sqrt{(-7)^2+8^2}\implies ||v||=\sqrt{113} \\\\\\ \begin{cases} x=rcos(\theta )\\ y=rsin(\theta )\\ \theta =tan^{-1}\left( \frac{b}{a} \right)\\ ||v||=r=\sqrt{a^2+b^2} \end{cases}\qquad \theta =tan^{-1}\left( \frac{8}{-7} \right)\implies \theta \approx 48.8^o \\\\\\ \left[ \sqrt{113}\cdot cos(48.8^o)~~,~~ \sqrt{113}\cdot sin(48.8^o)\right]$