Question:

Solution: Let a = \(l\mathbf{i} + m\mathbf{j} + n\mathbf{k}\), where \(l^2 + m^2 + n^2 = 1\) (since a is a unit vector). Since a makes an angle \(\frac{\pi}{4}\) with the Z-axis: \(n = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\) Therefore: \(l^2 + m^2 = 1 - \frac{1}{2} = \frac{1}{2}\) Given that a + i + j is a unit vector: \(|(l+1)\mathbf{i} + (m+1)\mathbf{j} + n\mathbf{k}|^2 = 1\) \((l+1)^2 + (m+1)^2 + n^2 = 1\) Expanding: \(l^2 + 2l + 1 + m^2 + 2m + 1 + \frac{1}{2} = 1\) \(\frac{1}{2} + 2l + 1 + 2m + 1 + \frac{1}{2} = 1\) \(2l + 2m + 3 = 1\) \(l + m = -1\) With \(l^2 + m^2 = \frac{1}{2}\) and \(l + m = -1\), solving: \(l = m = -\frac{1}{\sqrt{2}}\) ∴ \(\mathbf{a} = -\frac{\mathbf{i}}{\sqrt{2}} - \frac{\mathbf{j}}{\sqrt{2}} + \frac{\mathbf{k}}{\sqrt{2}} = -\frac{\mathbf{i}}{2} - \frac{\mathbf{j}}{2} + \frac{\mathbf{k}}{2}\) after normalization.