A constant force is given by F = (5i + 2j − 3k) N. It causes a displacement of s = (2i + 3j + 4k) m. Calculate the work done, the magnitude of the force, the magnitude of the displacement, and the ang

The work done $W$ by a constant force $\vec{F}$ causing a displacement $\vec{s}$ is given by the dot product of the force and displacement vectors.

$$W = \vec{F} \cdot \vec{s}$$

Given $\vec{F} = (5\hat{i} + 2\hat{j} - 3\hat{k})\, \text{N}$ and $\vec{s} = (2\hat{i} + 3\hat{j} + 4\hat{k})\, \text{m}$.

$$W = (5)(2) + (2)(3) + (-3)(4)$$ $$W = 10 + 6 - 12$$ $$W = 4\, \text{J}$$

The work done by the force is $\boxed{\text{4 J}}$.

$$|\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2}$$ $$|\vec{F}| = \sqrt{5^2 + 2^2 + (-3)^2}$$ $$|\vec{F}| = \sqrt{25 + 4 + 9}$$ $$|\vec{F}| = \sqrt{38}\, \text{N}$$

$$|\vec{s}| = \sqrt{s_x^2 + s_y^2 + s_z^2}$$ $$|\vec{s}| = \sqrt{2^2 + 3^2 + 4^2}$$ $$|\vec{s}| = \sqrt{4 + 9 + 16}$$ $$|\vec{s}| = \sqrt{29}\, \text{m}$$

The angle $\theta$ between two vectors can be found using the formula:

$$\cos\theta = \frac{\vec{F} \cdot \vec{s}}{|\vec{F}| |\vec{s}|}$$

We already found $\vec{F} \cdot \vec{s} = 4\, \text{J}$.

$$\cos\theta = \frac{4}{\sqrt{38} \cdot \sqrt{29}}$$ $$\cos\theta = \frac{4}{\sqrt{38 \times 29}}$$ $$\cos\theta = \frac{4}{\sqrt{1102}}$$ $$\cos\theta \approx \frac{4}{33.196}$$ $$\cos\theta \approx 0.12049$$ $$\theta = \arccos(0.12049)$$ $$\theta \approx 83.08^\circ$$

The angle between the force and displacement is $\boxed{83.08^\circ}$.