Let vec{a} = hat{i} + hat{j} + sqrt{2}hat{k}, | vedclass

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(D) The projection vector of $\vec{b}$ on $\vec{a}$ is given by $\left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right)\vec{a}$.Given that this projec

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$6$

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(D) The projection vector of $\vec{b}$ on $\vec{a}$ is given by $\left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right)\vec{a}$.Given that this projection vector is $\vec{a}$,we have $\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} = 1$,which implies $\vec{b} \cdot \vec{a} = |\vec{a}|^2$.Calculating $|\vec{a}|^2 = 1^2 + 1^2 + (\sqrt{2})^2 = 1 + 1 + 2 = 4$.Now,$\vec{b} \cdot \vec{a} = (b_{1}\hat{i} + b_{2}\hat{j} + \sqrt{2}\hat{k}) \cdot (\hat{i} + \hat{j} + \sqrt{2}\hat{k}) = b_{1} + b_{2} + 2$.Equating the two,$b_{1} + b_{2} + 2 = 4$,so $b_{1} + b_{2} = 2$ .....$(1)$.Given $(\vec{a} + \vec{b}) \perp \vec{c}$,we have $(\vec{a} + \vec{b}) \cdot \vec{c} = 0$.$\vec{a} + \vec{b} = (1 + b_{1})\hat{i} + (1 + b_{2})\hat{j} + 2\sqrt{2}\hat{k}$.$(\vec{a} + \vec{b}) \cdot \vec{c} = (1 + b_{1})(5) + (1 + b_{2})(1) + (2\sqrt{2})(\sqrt{2}) = 0$.$5 + 5b_{1} + 1 + b_{2} + 4 = 0 \Rightarrow 5b_{1} + b_{2} = -10$ .....$(2)$.Subtracting $(1)$ from $(2)$: $4b_{1} = -12 \Rightarrow b_{1} = -3$.Substituting $b_{1} = -3$ into $(1)$: $-3 + b_{2} = 2 \Rightarrow b_{2} = 5$.Thus,$\vec{b} = -3\hat{i} + 5\hat{j} + \sqrt{2}\hat{k}$.$|\vec{b}| = \sqrt{(-3)^2 + 5^2 + (\sqrt{2})^2} = \sqrt{9 + 25 + 2} = \sqrt{36} = 6$.

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