Finding the x component of a vector given this angle

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Homework Statement:
This is probably a simple question for most of you but I can't seem to figure out this format of solving a vector when there are two unknowns (Unknown magnitude and unknown x). Thank you!
Relevant Equations:
V = xi + 5.9j + 9.8k
If V makes an angle with i of 33.7 degrees, what is x? x is positive
A = |A|cos33.7

A = |A|0.83195
 

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PeroK
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Homework Statement:: This is probably a simple question for most of you but I can't seem to figure out this format of solving a vector when there are two unknowns (Unknown magnitude and unknown x). Thank you!
Relevant Equations:: V = xi + 5.9j + 9.8k
If V makes an angle with i of 33.7 degrees, what is x? x is positive

A = |A|cos33.7

A = |A|0.83195
Do you think you could find two different values of ##x## that satisfy that requirement?
 
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Thanks!

They want the x component in decimal form. It definitely involves some algebra. For example I can solve it through process of elimination by guessing for x and solving until the magnitude divides by x and equals 0.83195.
So for this question it'd be 20.615 = √(17.15)^2 + (5.9)^2 + (9.8)^2
17.15 / 20.615 = .8319
Which gives me 33.7 degrees

But this method is definitely not efficient so there must be an algebraic method to find x more quickly?
 
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Thanks!

They want the x component in decimal form. It definitely involves some algebra. For example I can solve it through process of elimination by guessing for x and solving until the magnitude divides by x and equals 0.83195.
So for this question it'd be 20.615 = √(17.15)^2 + (5.9)^2 + (9.8)^2
17.15 / 20.615 = .8319
Which gives me 33.7 degrees

But this method is definitely not efficient so there must be an algebraic method to find x more quickly?
You have ##\vec v \cdot \vec i = |\vec v| \cos \theta = \sqrt{x^2 + y^2 + z^2} \cos \theta##.

Can you find another expression for ##\vec v \cdot \vec i ##?
 
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Steve4Physics
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But this method is definitely not efficient so there must be an algebraic method to find x more quickly?
In this question θ is called the vector's ‘direction angle’ with respect to the x-axis. And cosθ is called the ‘direction cosine’ with respect to the x-axis.

With x, y and z the components of v, the length (magnitude) of v is |v| = √(x² + y² + z²).

If you can’t draw your own, a search will give many helpful diagrams explaining (3D) direction angles/cosines.

Using your diagram, can you express cosθ in terms of x and |v|? Finding x is then just algebra (solving a quadratic).
 

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