Answer:
[tex]||v||\approx21.21[/tex]
[tex]\theta =-45^{\circ}[/tex]
Step-by-step explanation:
Given a vector of the form:
[tex]v=ai+bj\\\\a\in R, b\in R[/tex]
Its magnitude can be found using pythagorean theorem:
[tex]||v||=\sqrt{a^2+b^2}[/tex]
And its direction angle is given by:
[tex]\theta =arctan (\frac{b}{a} )[/tex]
In this sense for the vector:
[tex]v=15i-15j[/tex]
Its magnitude is:
[tex]||v||=\sqrt{(15)^2+(15)^2} =15\sqrt{2} = 21.21320344[/tex]
and its direction angle is:
[tex]\theta =arctan(\frac{-15}{15} )=-45^{\circ}[/tex]
