Let vector A = 4i^ + 4j^, vector B = -2i^ - 5j^, and vector F = vector A - 5(vector B).
a) Write vector F in component form.
vector F = ?
b) What is the magnitude of vector F? F = ?
c) What is the direction of vector F?
theta = ?
Thank you!
F=A-5B=4i+4j-(-10i-25j)=14i+29j
checkthat.
a) To write the vector F in component form, we need to subtract 5 times the components of vector B from the components of vector A.
Given:
Vector A = 4i^ + 4j^
Vector B = -2i^ - 5j^
To find vector F:
F = A - 5B
To subtract vectors, we subtract the corresponding components.
F = (4i^ + 4j^) - 5(-2i^ - 5j^)
Distribute the -5 to the components of vector B:
F = 4i^ + 4j^ + 10i^ + 25j^
Combine like terms:
F = (4 + 10)i^ + (4 + 25)j^
Simplify:
F = 14i^ + 29j^
Therefore, vector F in component form is F = 14i^ + 29j^.
b) To find the magnitude of vector F, we use the formula:
|F| = √(Fx^2 + Fy^2)
Given:
Fx = 14
Fy = 29
Substituting the values into the formula:
|F| = √(14^2 + 29^2)
Calculating:
|F| = √(196 + 841)
|F| = √1037
So, the magnitude of vector F is |F| = √1037.
c) To find the direction of vector F, we use the formula:
θ = tan^(-1)(Fy/Fx)
Given:
Fx = 14
Fy = 29
Substituting these values into the formula:
θ = tan^(-1)(29/14)
Calculating:
θ ≈ 1.09 radians or ≈ 62.22 degrees
Therefore, the direction of vector F is θ ≈ 1.09 radians or ≈ 62.22 degrees.