Question

Find the magnitude and direction angle θ, to the nearest tenth of a degree, for the given vector v. v = -4i - 3

Answer 1

I got 5; 216.9deg Nor real confident!!

Answer 2

arctan(4/3)=53.1301023542

Answer 3

-4i-3
standard form
-3 - 4i
^ ^
a b components
\(\large \bf || < -3, -4> || = \sqrt{(-3)^2+(-4)^2} \implies \sqrt{25} = 5\)

Answer 4

$$\bf
tan(\theta) = \cfrac{b}{a} \implies \frac{-4}{-3}\\
tan^{-1}(tan(\theta)) = tan^{-1}\left(\frac{-4}{-3}\right)\\
\theta = 53.13^o
$$

Answer 5

thanks to you both

Answer 6

yw