Q6) Find the vector component of \( u \) along \( a \) and the vector component of \( u \) orthogonal to \( a \) and then find their norms. (a) \( \mathbf{u}=(1,-2), \mathbf{a}=(-4,-3) \) (b) \( \mathbf{u}=(3,0,4), \mathbf{a}=(2,3,3) \)

To find the vector component of \( \mathbf{u} \) along \( \mathbf{a} \) and the vector component of \( \mathbf{u} \) orthogonal to \( \mathbf{a} \), we can use the following formulas: 1. The component of \( \mathbf{u} \) along \( \mathbf{a} \) is given by: \[ \text{proj}_{\mathbf{a}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \] 2. The component of \( \mathbf{u} \) orthogonal to \( \mathbf{a} \) is given by: \[ \mathbf{u}_{\perp} = \mathbf{u} - \text{proj}_{\mathbf{a}} \mathbf{u} \] 3. The norms of the components can be found using: \[ \|\text{proj}_{\mathbf{a}} \mathbf{u}\| \quad \text{and} \quad \|\mathbf{u}_{\perp}\| \] Let's solve each part step by step. ### (a) \( \mathbf{u} = (1, -2), \mathbf{a} = (-4, -3) \) 1. **Calculate \( \mathbf{u} \cdot \mathbf{a} \)**: \[ \mathbf{u} \cdot \mathbf{a} = 1 \cdot (-4) + (-2) \cdot (-3) = -4 + 6 = 2 \] 2. **Calculate \( \mathbf{a} \cdot \mathbf{a} \)**: \[ \mathbf{a} \cdot \mathbf{a} = (-4)^2 + (-3)^2 = 16 + 9 = 25 \] 3. **Calculate the projection of \( \mathbf{u} \) onto \( \mathbf{a} \)**: \[ \text{proj}_{\mathbf{a}} \mathbf{u} = \frac{2}{25} \mathbf{a} = \frac{2}{25} (-4, -3) = \left(-\frac{8}{25}, -\frac{6}{25}\right) \] 4. **Calculate the orthogonal component**: \[ \mathbf{u}_{\perp} = \mathbf{u} - \text{proj}_{\mathbf{a}} \mathbf{u} = (1, -2) - \left(-\frac{8}{25}, -\frac{6}{25}\right) = \left(1 + \frac{8}{25}, -2 + \frac{6}{25}\right) = \left(\frac{25}{25} + \frac{8}{25}, -\frac{50}{25} + \frac{6}{25}\right) = \left(\frac{33}{25}, -\frac{44}{25}\right) \] 5. **Calculate the norms**: - Norm of the projection: \[ \|\text{proj}_{\mathbf{a}} \mathbf{u}\| = \sqrt{\left(-\frac{8}{25}\right)^2 + \left(-\frac{6}{25}\right)^2} = \sqrt{\frac{64}{625} + \frac{36}{625}} = \sqrt{\frac{100}{625}} = \sqrt{\frac{1}{6.25}} = \frac{10}{25} = \frac{2}{5} \] - Norm of the orthogonal component: \[ \|\mathbf{u}_{\perp}\| = \sqrt{\left(\frac{33}{25}\right)^2 + \left(-\frac{44}{25}\right)^2} = \sqrt{\frac{1089}{625} + \frac{1936}{625}} = \sqrt{\frac{3025}{625}} = \sqrt{4.84} = \frac{55}{25} = \frac{11}{5} \] ### (b) \( \mathbf{u} = (3, 0, 4), \mathbf{a} = (2, 3, 3) \) 1. **Calculate \( \mathbf{u} \cdot \mathbf{a} \)**: \[ \mathbf{u} \cdot \mathbf{a} = 3 \cdot