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. R
efer to the figure. Enter an expression for the x-component vector, Dx, in terms of D, θ, and the unit vectors i and j.
1 answer:
Answer:
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
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