Vector has a magnitude of 4.40 m and is directed east. Vector has a magnitude of 3.40 m and is directed 39.0° west of north. What are (a) the magnitude and (b) the direction (counterclockwise from east) of? What are (c) the magnitude and (d) the direction (counterclockwise from east) of?

Magnitude = 4.056 m

Direction = 42.3⁰

Explanation:

The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.

The vector 4.40 is directed East. This automatically becomes a horizontal component.

But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.

Resolving the vectors should yield the horizontal and vertical components:

Horizontal components

The first component is 4.40 m

The second one is derived by resolving 3.40 to the horizontal like this 3.40 * - cos 61⁰ = - 1.648 m

Adding the horizontal component gives 4.40 m + (-1.648 m) = 2.752 m

Vertical components

Resolve 3.40 with the angle 61⁰ like this: vertical comp = 3.41 * sin 61

= 2.98 m

The magnitude is given by √[ (2.98) ² + (2.752) ²] = 4.056 m Ans

The direction us given by tan⁻¹ (2.98/2.752) = 42.3⁰ Ans