Source: Extra Practice

A force vector F=(5i^+3j^) N\vec{F} = (5\hat{i} + 3\hat{j})\text{ N} acts on a particle. What is the component of this force along the direction of the vector d=3i^+4j^\vec{d} = 3\hat{i} + 4\hat{j}?

Options

Option A is correct

5.4 N5.4\text{ N}

Option B

4.2 N4.2\text{ N}

Option C

3.0 N3.0\text{ N}

Option D

2.7 N2.7\text{ N}

Explanation

The component of F\vec{F} along the direction of d\vec{d} is given by Fd=Fd^=FddF_d = \vec{F} \cdot \hat{d} = \frac{\vec{F} \cdot \vec{d}}{|\vec{d}|}. First, we find the magnitude of d\vec{d}: d=32+42=5|\vec{d}| = \sqrt{3^2 + 4^2} = 5. Next, we compute the dot product: Fd=(5)(3)+(3)(4)=15+12=27\vec{F} \cdot \vec{d} = (5)(3) + (3)(4) = 15 + 12 = 27. Finally, Fd=275=5.4 NF_d = \frac{27}{5} = 5.4\text{ N}. Thus, option 1 is correct, and the other options do not match this calculation.