Answer:
Part (i) the magnitude of the vector is 5.315 m
Part (ii) the direction of the vector is 138.8⁰
Explanation:
Given;
y-component Ay = 3.50 m
x -component Ax = -4.00 m
Vector representation = (-4i + 3.5j)
Part (i) the magnitude of the vector
Magnitude of the vector = √(-4)² +(3.5)²
Magnitude of the vector = √28.25
= 5.315 m
Part (ii) the direction (in degrees counterclockwise from the +x-axis) of the vector.
If we make a rough sketch of this vector, the direction of this vector lies in second quadrant.
That is; 90 < θ < 180
Let's solve for θ
tan θ = y/x
tan θ = 3.5/-4
tan θ = - 0.875
θ = tan⁻¹ (-0.875)
θ = - 41.2⁰
Since θ lies in the second quadrant,
θ = - 41.2⁰ = 180 - 41.2 = 138.8⁰
The direction of the vector is 138.8⁰