How can I find the vector from a given magnitude and direction?

AI Thread Summary
To find a vector from a given magnitude and direction in 3D, one can utilize the concept of direction cosines or unit vectors. If the direction is represented by a unit vector, multiplying it by the magnitude yields the desired vector. Alternatively, if angles with the coordinate axes are provided, the unit vector can be constructed using the cosine of those angles. The discussion also touches on the nature of the zero vector, clarifying that while it has no magnitude, it can still be associated with any direction. Understanding these principles allows for effective vector calculations in three-dimensional space.
jjiimmyy101
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If I'm given a vector in 3D and it asks me to find the magnitude, I just take the square root of the numbers squared. But if I'm just given a magnitude and direction how could I find the vector?
 
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jjiimmyy101 said:
If I'm given a vector in 3D and it asks me to find the magnitude, I just take the square root of the numbers squared. But if I'm just given a magnitude and direction how could I find the vector?

If u have the magnitude and the direction,then U HAVE THE VECTOR.U're probably asking how to determine the components along the (presumably reciprocal orthogonal) axis of coordinates if u'regiven the mangitude and direction.It all stays in the the word "direction" that would require someangles between the vector's direction and the coordinate axis/planes,perhaps.From here on,there's nothing that plain space geometry.
 
Suppose a vector has magnitude 8 and its direction is given to be

\frac{1}{\sqrt{3}}i + \frac{1}{\sqrt{3}}j + \frac{1}{\sqrt{3}}k

then the vector u need is

\frac{8}{\sqrt{3}}i + \frac{8}{\sqrt{3}}j + \frac{8}{\sqrt{3}}k

This should help you! :biggrin:
 
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The magnitude is a number of course, but how are you given the directions? If you are given a unit vector in that direction, just multiply by the magnitude as prasanna said.

If you are given the angles the vector makes with the coordinate axes, you can use the "direction cosines": if the angle the direction makes with the x-axis is u, with the y-axis v, and with the z axis w then cos(u)i+ cos(v)j+ cos(w)k is a unit vector in that direction. Multiply that by the magnitude.
 
The other format you might see is a magnitude and two angles. Usually one of the angles projects the vector onto the fundamental plane (latitude of some location, for example). The other gives the angle of projection relative to the x-axis (longitude of some location, for example).

(Probably not a great example unless you realize it's just assumed that everyone knows the radius of the Earth).

In this case, it would be:

r cos (lat) cos (long)_i + r cos (lat) sin (long)_j + r sin (lat)_k
 
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The magnitude is a number of course, but how are you given the directions?
I was given a magnitude (a number) and the direction is the same as this vector (and the vector was given as ?i + ?j +?k).
So I think I can go from here. I appreciate the input from all. :smile:
 
Someone argued that a zero vector has direction. Isn't this so ridiculous? As far as I know a zero vector is just a point. How can a point have directions?
 
A zero vector is a point which can have any direction!.
I remember reading something like this
 
Since...

\hat{a} = \frac{a}{|a|}

then

a = \hat{a}*|a|

simple algebra ;)
 
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I presume you're replying to jjiimmy101's original question on top? :rolleyes:

futb0l said:
Since...

\hat{a} = \frac{a}{|a|}

then

a = \hat{a}*|a|

simple algebra ;)
 
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