Answer:
Magnitude 13 and S23°E
Explanation:
Given the vector:
[tex]5i-12j[/tex]
We want to find its (a)magnitude (b)direction.
Given a vector ai+bj, its magnitude and direction are calculated using the formulas:
[tex]\begin{gathered} Magnitude:r=\sqrt{a^2+b^2} \\ Direction:\theta=\tan^{-1}(\frac{b}{a}) \end{gathered}[/tex]
Therefore, for the given vector:
[tex]\begin{gathered} r=\sqrt{5^2+(-12)^2}=\sqrt{25+144}=\sqrt{169}=13 \\ \theta=\tan^{-1}(-\frac{12}{5})=-67.4\degree\approx-67\degree \end{gathered}[/tex]
However, from the diagram of the vector below:
Since the angle is in Quadrant IV:
[tex]-67.4\degree=360-67=293\degree=S23\degree E[/tex]
The magnitude of the vector is 13, and its direction is S23°E.
The last option is correct.
