Vector F = 17 and is in quadrant 3 and thetaF= 30
What is the component of i along F?
What is the component of F along i?
recall that the projection of a onto b is a•b/|b|
Not quite sure what "thetaF" is supposed to mean.
a•b = |a| |b| cosθ
but what is i? because i isn't given in the problem the vectors that are given is A,B,C,F and for this problem it focuses on vector F but what is i
thetaF is 30 degrees
i is the unit vector in the +x direction -- you know: i,j,k are the three basis vectors in 3D
To find the component of vector i along vector F, we can use trigonometry. Since vector F is in the third quadrant, the x-component of F will be negative.
First, let's find the x-component of F (F_x). To do this, we use the given magnitude of F (F = 17) and the angle (θ_F = 30 degrees).
F_x = F * cos(θ_F)
Substituting the values:
F_x = 17 * cos(30)
Now, we need to find the y-component of F (F_y). Since vector F is in the third quadrant, the y-component of F will also be negative.
F_y = F * sin(θ_F)
Substituting the values:
F_y = 17 * sin(30)
Now that we have the x and y components, we can determine the component of vector i along vector F. The component of vector i along vector F is its x-component.
So, the component of i along F is -F_x.
To find the component of vector F along vector i, we simply take the x-component of F.
Therefore, the component of F along i is F_x.