Let a unit vector hat{u}=x hat{i}+y hat{j}+z | vedclass

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(B) Given unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$.Let $\overrightarrow{p}_1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$,$\overr

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$\frac{5}{2}$

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(B) Given unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$.Let $\overrightarrow{p}_1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$,$\overrightarrow{p}_2=\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,and $\overrightarrow{p}_3=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$.Since $\hat{u} \cdot \overrightarrow{p}_1 = |\hat{u}| |\overrightarrow{p}_1| \cos(\frac{\pi}{2}) = 0$,we have $\frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \Rightarrow x+z=0$ $(i)$.Since $\hat{u} \cdot \overrightarrow{p}_2 = |\hat{u}| |\overrightarrow{p}_2| \cos(\frac{\pi}{3}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$,we have $\frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \Rightarrow y+z = \frac{1}{\sqrt{2}}$ $(ii)$.Since $\hat{u} \cdot \overrightarrow{p}_3 = |\hat{u}| |\overrightarrow{p}_3| \cos(\frac{2\pi}{3}) = 1 \cdot 1 \cdot (-\frac{1}{2}) = -\frac{1}{2}$,we have $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \Rightarrow x+y = -\frac{1}{\sqrt{2}}$ $(iii)$.Adding $(i), (ii), (iii)$,we get $2(x+y+z) = 0 \Rightarrow x+y+z = 0$.Subtracting $(ii)$ from this,$x = -\frac{1}{\sqrt{2}}$. Subtracting $(iii)$ from this,$z = \frac{1}{\sqrt{2}}$. Subtracting $(i)$ from this,$y = 0$.Thus $\hat{u} = -\frac{1}{\sqrt{2}} \hat{i} + 0 \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$.Then $\hat{u}-\overrightarrow{v} = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \hat{i} + (0 - \frac{1}{\sqrt{2}}) \hat{j} + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \hat{k} = -\frac{2}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} = -\sqrt{2} \hat{i} - \frac{1}{\sqrt{2}} \hat{j}$.$|\hat{u}-\overrightarrow{v}|^2 = (-\sqrt{2})^2 + (-\frac{1}{\sqrt{2}})^2 = 2 + \frac{1}{2} = \frac{5}{2}$.

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