Finding magnitudes...

[Velvet]

Hello!

Could someone explain in detail how to find the magnitude and direction angle of the vector v when v= 6i - 6j?

Thank you very much in advance!

~V.

 

[stapel]

Isn't the magnitude the square root of the sum of the squares of the vector components? And don't you use trig to find the angle/direction?

(It's been a while, which is why I'm not sure. See what your book says.)

Eliz.

 

[pka]

Think of the point (6,−6).
How far is it from the origin? That is the magnitude of the vector!

Draw the line through that point and the origin.
What angle does it make with the positive x-axis? That is thedirection angle of the vector!

 

[Velvet]

Thanks

 

[soroban]

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 \(\displaystyle P(6,\,-6)\).

        |
        |
    - - + - - - + - -
       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!]