mathrm{c}} = \overline{0}.$$

This implies that $$\overline{\mathrm{a}} + \overline{\mathrm{b}} = -\overline{\mathrm{c}}$$. Taking the magnitudes on both sides, we get:

$$|\overline{\mathrm{a}} + \overline{\mathrm{b}}| = |\overline{\mathrm{c}}|.$$

Since magnitudes are always non-negative, and given the magnitudes of the vectors $$|\overline{\mathrm{a}}| = 3$$, $$|\overline{\mathrm{b}}| = 5$$, and $$|\overline{\mathrm{c}}| = 7$$, we can apply the law of cosines:

$$|\overline{\mathrm{a}} + \overline{\mathrm{b}}|^2 = |\overline{\mathrm{a}}|^2 + |\overline{\mathrm{b}}|^2 + 2 |\overline{\mathrm{a}}| |\overline{\mathrm{b}}| \cos{\theta},$$

where $$\theta$$ is the angle between $$\overline{\mathrm{a}}$$ and $$\overline{\mathrm{b}}$$. Plugging in the given magnitudes:

$$7^2 = 3^2 + 5^2 + 2 \cdot 3 \cdot 5 \cdot \cos{\theta}.$$

Simplify this equation:

$$49 = 9 + 25 + 30 \cos{\theta}.$$

Combine like terms:

$$49 = 34 + 30 \cos{\theta}.$$

Isolate the cosine term:

$$49 - 34 = 30 \cos{\theta},$$

$$15 = 30 \cos{\theta},$$

Therefore:

$$\cos{\theta} = \frac{1}{2}.$$

The angle whose cosine is $$\frac{1}{2}$$ is:

$$\theta = \frac{\pi}{3}.$$

Thus, the angle between the vectors $$\overline{\mathrm{a}}$$ and $$\overline{\mathrm{b}}$$ is:

Option A: $$\left(\frac{\pi}{3}\right)^{\mathrm{c}}$$

Vector Algebra

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