The magnitute of a vector can be calculated by calculating the distance beteen its endpoints.
The vector starts at point (2, 5) and ends at point (-3, -2), so the distance between them is:
[tex]\begin{gathered} d=\sqrt[]{(-3-2)^2+(-2-5)^2} \\ d=\sqrt[]{(-5)^2+(-7)^2} \\ d=\sqrt[]{25+49} \\ d=\sqrt[]{74} \\ d\approx8.602 \end{gathered}[/tex]To calculate the direction, we must calculate the angle from the x-axis, just like in the unit circle.
Firstly, we calculate using the arc tangent to calculate the angle in the first quadrant, but then we need to figure out the actual value considering the quadrant the vector is.
Firstly, we calculate the x and y differences from the start point to the endpoint:
[tex]\begin{gathered} \Delta x=-3-2 \\ \Delta x=-5 \\ \Delta y=-2-5 \\ \Delta y=-7 \end{gathered}[/tex]So, we calculate the arctanent of the absolute values of these differences:
[tex]\begin{gathered} \theta=\arctan (\frac{|\Delta y|}{|\Delta x|})^{} \\ \theta=\arctan (\frac{7}{5}) \\ \theta=54.46232\ldots\degree \end{gathered}[/tex]Now, since the differences of the x and y coordinates of the endpoint are negative, this means that the vector is in the third quadrant, so we need the equivalent angle that gives this tangent but is in the third quadrant.
The tangent values repeat every 180°, so the value that gives the same tangent and is in the third quadrant is:
[tex]\begin{gathered} \theta=54.46232\ldots\degree+180\degree=234.46232\ldots\degree \\ \theta\approx234.462\degree \end{gathered}[/tex]So, the vector has magnitute of approximately 8.602 and a direction of approximately 234.462°, which second alternative.corresponds to the
