Let a unit vector u=x i+y j+z k make angles π 2, π 3 and 2π 3 with the vectors 1 √2 i+ 1 √2 k, 1 √2 j+ 1 √2 k and 1 √2 i+ 1 √2 j respectively. If v=…

Unit vector $\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}$u^=xi^+yj^​+zk^

$${\vec{p}}_{1}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k},{\vec{p}}_{2}=\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$$p​1​=2​1​i^+2​1​k^,p​2​=2​1​j^​+2​1​k^

$${\vec{p}}_{3}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$$p​3​=2​1​i^+2​1​j^​

Now angle between $\hat{u}$u^ and ${\vec{p}}_{1}=\frac{\pi }{2}$p​1​=2π​

$$\hat{u}\cdot {\vec{p}}_{1}=0\Rightarrow \frac{x}{\sqrt{2}}+\frac{z}{\sqrt{2}}=0$$u^⋅p​1​=0⇒2​x​+2​z​=0

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

Angle between $\hat{u}$u^ and ${\vec{p}}_{2}=\frac{\pi }{3}$p​2​=3π​

$$\hat{u}\cdot {\vec{p}}_{2}=|\hat{u}|\cdot |{\vec{p}}_{2}|\mathrm{cos}\frac{\pi }{3}$$u^⋅p​2​=∣u^∣⋅∣p​2​∣cos3π​

$$\hat{u}\cdot {\vec{p}}_{2}=\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2}\ldots (\mathrm{ii})$$u^⋅p​2​=2​y​+2​z​=21​…(ii)

Angle between $\hat{u}$u^ and ${\vec{p}}_{3}=\frac{2\pi }{3}$p​3​=32π​

$$\hat{u}\cdot {\vec{p}}_{3}=|\hat{u}|\cdot |{\vec{p}}_{3}|\mathrm{cos}\frac{2\pi }{3}$$u^⋅p​3​=∣u^∣⋅∣p​3​∣cos32π​

$$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{-1}{2}\Rightarrow x+y=\frac{-1}{\sqrt{2}}\ldots (\mathrm{iii})$$⇒2​x​+2​y​=2−1​⇒x+y=2​−1​…(iii)

from equation (i), (ii) and (iii) we get

$$x=\frac{-1}{\sqrt{2}},y=0,z=\frac{1}{\sqrt{2}}$$x=2​−1​,y=0,z=2​1​

Thus $\hat{u}-\vec{v}=\frac{-1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}-\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}$u^−v=2​−1​i^+2​1​k^−2​1​i^−2​1​j^​−2​1​k^

$$\Rightarrow \hat{u}-\vec{v}=\frac{-2}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}$$⇒u^−v=2​−2​i^−2​1​j^​

$$\therefore |\hat{u}-\vec{v}{|}^{2}={(\sqrt{\frac{4}{2}+\frac{1}{2}})}^{2}=\frac{5}{2}$$∴∣u^−v∣2=(24​+21​​)2=25​