A displacement vector points in a direction of θ = 23° left of the positive y-axis. The magnitude of this vector is D = 155 m. Refer to the figure. Enter an expression for the x-component vector, Dx, in terms of D, θ, and the unit vectors i and j.

Dₓ = -155 sin 23° i + 0 j

Explanation:

The diagram showing the vector has been attached to this response.

As shown in the diagram,

The vector D has an x-component (also called horizontal component) of -D sinθ i. i.e

Dₓ = -D sin θ i   [The negative sign shows that D lies in the negative x direction]

Where;

D = magnitude of D = 155m

θ = direction of D = 23°

Therefore;

Dₓ = -155 sin 23° i

Since Dₓ represents the x component, its unit vector, j component has a value of 0.

Therefore, Dₓ can be written in terms of D, θ and the unit vectors i and j as follows;

Dₓ = -155 sin 23° i + 0 j

a straight line sloped