To find the projection of the sum of vectors $\mathbf{a}$ and $\mathbf{b}$ on a vector perpendicular to their plane, follow these steps:

Calculate $\mathbf{a} + \mathbf{b}$:

$ \mathbf{a} = 3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}, \quad \mathbf{b} = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $

$ \mathbf{a} + \mathbf{b} = (3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}) + (2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2 \hat{\mathbf{k}}) $

$ = 5 \hat{\mathbf{i}} + 5 \hat{\mathbf{j}} - 7 \hat{\mathbf{k}} $

Find $\mathbf{a} \times \mathbf{b}$ (the cross product):

$ \mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & -5 \\ 2 & 1 & -2 \end{array} \right| $

$ = \hat{\mathbf{i}}(-8+9) - \hat{\mathbf{j}}(-6+10) + \hat{\mathbf{k}}(3-8) $

$ = \hat{\mathbf{i}} - 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}} $

Normalize the perpendicular vector $\mathbf{n} = -3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}$:

First, find the magnitude $|\mathbf{n}|$:

$ |\mathbf{n}| = \sqrt{(-3)^2 + (-4)^2 + (-5)^2} = \sqrt{9+16+25} = \sqrt{50} = 5 \sqrt{2} $

The unit vector $\hat{\mathbf{n}}$ is:

$ \hat{\mathbf{n}} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{-3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} - 5 \hat{\mathbf{k}}}{5 \sqrt{2}} $

$ = \frac{-3}{5 \sqrt{2}} \hat{\mathbf{i}} - \frac{4}{5 \sqrt{2}} \hat{\mathbf{j}} - \frac{5}{5 \sqrt{2}} \hat{\mathbf{k}} $

Projection of $\mathbf{a} + \mathbf{b}$ on $\hat{\mathbf{n}}$:

$ (\mathbf{a} + \mathbf{b}) \cdot \hat{\mathbf{n}} = (5 \hat{\mathbf{i}} + 5 \hat{\mathbf{j}} - 7 \hat{\mathbf{k}}) \cdot \left(\frac{-3}{5 \sqrt{2}} \hat{\mathbf{i}} - \frac{4}{5 \sqrt{2}} \hat{\mathbf{j}} - \frac{5}{5 \sqrt{2}} \hat{\mathbf{k}}\right) $

Calculate the dot product, which results in 0. Thus, the projection of the sum of the vectors $\mathbf{a}$ and $\mathbf{b}$ on the vector perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$ is 0.

Vector Algebra

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