We will begin by finding the magnitude of a vector, denoted |v|.
The formula we can use is
[tex]|v|=\sqrt{a^2+b^2}[/tex]where a and b represent the vector components. Since we are given the vector <-11,-5>, we will let a be -11 and b is -5.
Substituting those values, we have
[tex]\begin{gathered} |<-11,-1>|=\sqrt{(-11)^2+(-5)^2} \\ \sqrt{121+25} \\ \sqrt{146} \\ \approx12.083 \end{gathered}[/tex]So far, your answer is either the first option or the second option.
Next, we want to find the direction of the vector. We can use another helpful formula:
[tex]\tan\theta=\frac{b}{a}[/tex]Substituting our original values for a and b, we have:
[tex]\tan\theta=\frac{-5}{-11}[/tex]Be careful here! Since the both the a-value and b-value are negative, we are going to be in the third quadrant. After finding our angle (which will be in quadrant 1), we will need to add 180 degrees.
Take the inverse tangent of both sides to get the angle:
[tex]\begin{gathered} \theta=\tan^{-1}(\frac{-5}{-11}) \\ \theta\approx24^{\circ} \end{gathered}[/tex]We'll add 180 degrees to get our final angle:
[tex]24+180=204[/tex]Since our final angle is 204 degrees, the correct answer is the second option.