Let \(\alpha\), \(\beta\), and \(\gamma\) be the angles the vector \(\vec{a}\) makes with the positive \(x\), \(y\), and \(z\) axes, respectively. The direction cosines are \(l = \cos\alpha\), \(m = \cos\beta\), and \(n = \cos\gamma\).Since the vector makes equal angles with all three axes, we have \(\alpha = \beta = \gamma\).Therefore, the direction cosines are equal: \(l = m = n\).The fundamental identity for direction cosines is:\[l^2 + m^2 + n^2 = 1\]Substituting \(l = m = n\):\[l^2 + l^2 + l^2 = 1\]\[3l^2 = 1\]\[l^2 = \frac{1}{3}\]\[l = \pm\frac{1}{\sqrt{3}}\]So, the direction cosines are \(l = m = n = \pm\frac{1}{\sqrt{3}}\).
A vector \(\vec{a}\) can be expressed in terms of its magnitude \(|\vec{a}|\) and its direction cosines \((l, m, n)\) as:\[\vec{a} = |\vec{a}| (l\hat{i} + m\hat{j} + n\hat{k})\]Given the magnitude of the vector is \(|\vec{a}| = 5\sqrt{3}\) units.
Case 1: Using the positive sign for the direction cosines, \(l = m = n = \frac{1}{\sqrt{3}}\).\[\vec{a}_1 = 5\sqrt{3} \left(\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}\right)\]\[\vec{a}_1 = 5\sqrt{3} \cdot \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})\]\[\vec{a}_1 = 5 (\hat{i} + \hat{j} + \hat{k}) = 5\hat{i} + 5\hat{j} + 5\hat{k}\]
Case 2: Using the negative sign for the direction cosines, \(l = m = n = -\frac{1}{\sqrt{3}}\).\[\vec{a}_2 = 5\sqrt{3} \left(-\frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}\right)\]\[\vec{a}_2 = 5\sqrt{3} \cdot \left(-\frac{1}{\sqrt{3}}\right) (\hat{i} + \hat{j} + \hat{k})\]\[\vec{a}_2 = -5 (\hat{i} + \hat{j} + \hat{k}) = -5\hat{i} - 5\hat{j} - 5\hat{k}\]
Hence the vector \(\vec{a}\) that makes equal angles with all three axes and has a magnitude of \(5\sqrt{3}\) units is:\[\vec{a} = \mathbf{5\hat{i} + 5\hat{j} + 5\hat{k}}\]or\[\vec{a} = \mathbf{-5\hat{i} - 5\hat{j} - 5\hat{k}}\]