उत्तर (2)

समाधान

इकाई वेक्टर $\hat{\mathbf{u}}=x \hat{i}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$

$\overrightarrow{p} _1=\dfrac{1}{\sqrt{2}} \hat{i}+\dfrac{1}{\sqrt{2}} \hat{k}$

$\overrightarrow{p} _2=\dfrac{1}{\sqrt{2}} \hat{j}+\dfrac{1}{\sqrt{2}} \hat{k}$

$\overrightarrow{p} _3=\dfrac{1}{\sqrt{2}} \hat{i}+\dfrac{1}{\sqrt{2}} \hat{j}$

अब $\hat{u}$ और $\overrightarrow{p} _1$ के बीच कोण $\dfrac{\pi}{2}$ है

$\hat{u} \cdot \overrightarrow{p} _1=0 \Rightarrow \dfrac{x}{\sqrt{2}}+\dfrac{z}{\sqrt{2}}=0$

$\Rightarrow x+z=0 \ldots(i)$

$\hat{u}$ और $\overrightarrow{p} _2$ के बीच कोण $\dfrac{\pi}{3}$ है

$\hat{u} \cdot \overrightarrow{p} _2=|\hat{u}| \cdot\left|\overrightarrow{p} _2\right| \cos \dfrac{\pi}{3}$

$\Rightarrow \dfrac{y}{\sqrt{2}}+\dfrac{z}{\sqrt{2}}=\dfrac{1}{2} \Rightarrow y+z=\dfrac{1}{\sqrt{2}}\ldots(ii)$

$\hat{u}$ और $\overrightarrow{p} _3$ के बीच कोण $\dfrac{2 \pi}{3}$ है

$\hat{u} \cdot \overrightarrow{p} _3=|\hat{u}| \cdot\left|\overrightarrow{p} _3\right| \cos \dfrac{2 \pi}{3}$

$\Rightarrow \dfrac{x}{\sqrt{2}}+\dfrac{4}{\sqrt{2}}=\dfrac{-1}{2} \Rightarrow x+y=\dfrac{-1}{\sqrt{2}} \ldots (iii)$

समीकरण (i), (ii) और (iii) से हम प्राप्त करते हैं

$x=\dfrac{-1}{\sqrt{2}}, \quad y=0, \quad z=\dfrac{1}{\sqrt{2}}$

इसलिए $\hat{u}-\overrightarrow{v}=\dfrac{-1}{\sqrt{2}} \hat{i}+\dfrac{1}{\sqrt{2}} \hat{k}-\dfrac{1}{\sqrt{2}} \hat{i}-\dfrac{1}{\sqrt{2}} \hat{j}-\dfrac{1}{\sqrt{2}} \hat{k}$

$\hat{u}-\overrightarrow{v}=\dfrac{-2}{\sqrt{2}} \hat{i}-\dfrac{1}{\sqrt{2}} \hat{j}$

$\therefore|\hat{u}-\overrightarrow{v}|^{2}=\left(\sqrt{\dfrac{4}{2}+\dfrac{1}{2}}\right)^{2}=\dfrac{5}{2}$