Finding Magnitude and Direction of Three Dimensional Vectors

[ƒ(x)]

Homework Statement

Find magnitude and direction

\stackrel{\rightarrow}{C} = -2i-3j+4k

Homework Equations

Trig. Functions
Pythagorean Theorem

The Attempt at a Solution

C = \sqrt{4+9+16} = \sqrt{29}

I don't know what to do for the angles because I'm not sure where the reference point is supposed to be.

 

[kuruman]

The magnitude is OK. As far as the "direction" is concerned, you need two angles to specify direction. If you imagine the tip of the vector being on the surface of a sphere (like the Earth), then you need to give the "latitude" angle θ and the "longitude" (measured from the x axis) angle φ.

 

[ƒ(x)]

Ah, ok.

If θ is the latitude one, and A is the angle from the Z-axis to the vector, then:

θ = 90° - A

cos(A) = \sqrt{13/29}
A = cos^-1(\sqrt{13/29}) = 47.97°

θ = 42.03°

If Ø is the longitude one, then:

Ø = 90 + tan^-1(2/4)
Ø = 116.57° <---- Where I have a problem

My teacher says that the answer is 112°, but whenever I do it I get 116.57°. What am I missing?

 

[kuruman]

If you project the tip of the vector onto the xy plane, you get the point (-2, -3, 0). Draw a line from the origin to that point. What is the angle it makes with respect to the x-axis?

 

[ƒ(x)]

Isn't that the situation though?

In this case, Ø = 180 - arctan(2/4)

 

[ƒ(x)]

Guys, I need to know if I'm doing this correctly. The homework assignment is due tomorrow.

 

[ƒ(x)]

If you project the tip of the vector onto the xy plane, you get the point (-2, -3, 0). Draw a line from the origin to that point. What is the angle it makes with respect to the x-axis?

Ok, so that would be arctan (-3/-2) or 56.3 degrees. Which means that the angle from the x-axis would be 236.3 degrees going clockwise or 123.7 degrees going counter clockwise.

 

[kuruman]

Sounds OK. Good luck with your homework.