Let vec{alpha}=4 hat{i}+3 hat{j}+5 hat{k} and | vedclass

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(C) Let $\vec{\beta}_1 = \lambda \vec{\alpha}$.Since $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$,we have $\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_

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$7$

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(C) Let $\vec{\beta}_1 = \lambda \vec{\alpha}$.Since $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$,we have $\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = \vec{\beta} - \lambda \vec{\alpha}$.Substituting the vectors,$\vec{\beta}_2 = (\hat{i} + 2\hat{j} - 4\hat{k}) - \lambda(4\hat{i} + 3\hat{j} + 5\hat{k}) = (1 - 4\lambda)\hat{i} + (2 - 3\lambda)\hat{j} - (4 + 5\lambda)\hat{k}$.Since $\vec{\beta}_2 \perp \vec{\alpha}$,their dot product is zero: $\vec{\beta}_2 \cdot \vec{\alpha} = 0$.$4(1 - 4\lambda) + 3(2 - 3\lambda) + 5(-4 - 5\lambda) = 0$.$4 - 16\lambda + 6 - 9\lambda - 20 - 25\lambda = 0$.$-50\lambda - 10 = 0 \Rightarrow \lambda = -\frac{1}{5}$.Now,$\vec{\beta}_2 = (1 - 4(-\frac{1}{5}))\hat{i} + (2 - 3(-\frac{1}{5}))\hat{j} - (4 + 5(-\frac{1}{5}))\hat{k} = \frac{9}{5}\hat{i} + \frac{13}{5}\hat{j} - 3\hat{k}$.Thus,$5\vec{\beta}_2 = 9\hat{i} + 13\hat{j} - 15\hat{k}$.Finally,$5\vec{\beta}_2 \cdot (\hat{i} + \hat{j} + \hat{k}) = 9(1) + 13(1) - 15(1) = 9 + 13 - 15 = 7$.

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