To determine the vector \( \mathbf{a} \), given that it forms equal angles with the three coordinate axes and possesses a magnitude of \( 5\sqrt{3} \) units.
1. Direction Cosines:
A vector that forms equal angles with the x-, y-, and z-axes has equal direction cosines. Let \( l \) represent this common direction cosine.
The property that the sum of the squares of direction cosines equals 1 yields:
\[l^2 + l^2 + l^2 = 1 \Rightarrow 3l^2 = 1 \Rightarrow l^2 = \frac{1}{3} \Rightarrow l = \frac{1}{\sqrt{3}}\]
2. Unit Vector:
The unit vector aligned with \( \mathbf{a} \) is:
\[\hat{\mathbf{a}} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle\]
3. Scalar Multiplication:
Multiplying the unit vector by the given magnitude \( 5\sqrt{3} \) results in:
\[\mathbf{a} = 5\sqrt{3} \cdot \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle = \left\langle 5, 5, 5 \right\rangle\]
Result:
The vector \( \mathbf{a} \) is \( \boxed{\langle 5, 5, 5 \rangle} \.