Hello, Velvet!
Could someone explain in detail how to find the magnitude and direction angle
of the vector \(\displaystyle \vec{v}\) when \(\displaystyle \vec{v}\:=\:6i\,-\,6j\)?
Think of the two coefficients as coordinates of a point: \(\displaystyle P(6,\,-6)\)
The vector \(\displaystyle \vec{v}\) goes from \(\displaystyle O(0,0)\) to \(\displaystyle P(6,\,-6)\).
Code:
|
|
- - + - - - + - -
O| \θ :
| \ :
| \ :
| *P
| (6,-6)
The magnitude is the <u>length</u> of vector \(\displaystyle OP:\;\;|OP|\:=\:\sqrt{6^2\,+\,(-6)^2}\:=\:\sqrt{72}\:=\:6\sqrt{2}\)
The direction angle is the angle \(\displaystyle \vec{OP}\) makes with the positive x-axis, \(\displaystyle \theta\).
. . We can see that we have a 45-45-90 right triangle.
. . Therefore, the direction angle is: -\(\displaystyle 45^o\) or -\(\displaystyle \frac{\pi}{4}\) radians.
[Edit: Too fast for me, pka!]