Do you think you could find two different values of ##x## that satisfy that requirement?Indras said: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
You have ##\vec v \cdot \vec i = |\vec v| \cos \theta = \sqrt{x^2 + y^2 + z^2} \cos \theta##.Indras said: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?
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.Indras said:.
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But this method is definitely not efficient so there must be an algebraic method to find x more quickly?
The x component of a vector is the horizontal part of the vector, representing its magnitude and direction along the x-axis.
To find the x component of a vector given an angle, you can use trigonometric functions such as cosine or tangent, depending on the angle and known information about the vector.
Yes, the x component of a vector can be negative if the vector is pointing in the negative direction along the x-axis.
The x component of a vector is typically measured in the same units as the vector's magnitude, such as meters or feet.
The angle of a vector determines the direction and magnitude of its x component. A larger angle will result in a larger x component, while a smaller angle will result in a smaller x component.