Question

Consider a vector described by the following components:x-component= +5.5 my-component-8.2mWhat is the magnitude of the vector and its direction relative to the x-axis?

Answer 1

Let the vector be:

$$\mathbf{v} = (v_x,v_y)=(5.5,-8.2)\space (m)$$

Then, by definition, its magnitude is:

$$v = \sqrt{v_x^2+v_y^2} = \sqrt{5.5^2 +(-8.2)^2} \approx 9.9\space (m)$$

 Its direction relative to the x-axis (angle with positive x-axis) is:

$$\theta =\arctan\left(\dfrac{v_y}{v_x} \right) = \arctan\left(\dfrac{-8.2}{5.5} \right) \approx -56.1\degree$$

Answer. Magnitude: 9.9 m, direction to the x-axis: -56.1 degrees.