Question

Find the magnitude and direction of the vector. Round the length to the nearest tenth and the degree to the nearest unit. Diagram is not drawn to scale.

https://www.connexus.com/content/media/456845-2232011-25155-PM-1704478853.png

Answer 1

since its a right triangle, you can solve this using pythagorean theorem:

c^2= a^2+b^2

where a is 6 and b is 4.

Answer 2

then you can use the inverse trig function to find the angle:

example:

sin^(-1)(theta)= (4/c)
or
cos^(-1)(theta)= (6/c)
or
tan^(-1)(theta)= (4/6)

Answer 3

the magnitude of a vector is the distance between your 2 points. \[\left| PQ \right|=\sqrt{6^2 +4^2 }=2 \sqrt {13}=7.21\]
The direction of a vector is the measure of the angle it makes with a horizontal line.
one of the formula to find its direction:\[\tan \theta = \frac{\Delta y}{\Delta x}\]
\[\theta = \tan^{-1} (\frac{4-0}{6-0})\]\[\theta = 33.69^{o}\]therefore the vector has a direction of about 33.69