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JEE Main 29-Jan-2024 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2024 Paper - Shift 24 Marks
MCQ
Let a unit vector $\hat{ u }=x \hat{ i }+\hat{ y }+ zk$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{ i }+\frac{1}{\sqrt{2}} \hat{ k }, \frac{1}{\sqrt{2}} \hat{ j }+\frac{1}{\sqrt{2}} \hat{ k }$ and $ \frac{1}{\sqrt{2}} \hat{ i }+\frac{1}{\sqrt{2}} \hat{ j } $ respectively. If $\overrightarrow{ v }=\frac{1}{\sqrt{2}} \hat{ i }+\frac{1}{\sqrt{2}} \hat{ j }+\frac{1}{\sqrt{2}} \hat{ k },$ then $|\hat{ u }-\overrightarrow{ v }|^2$ is equal to
  • A
    $\frac{11}{2}$
  • B
    $\frac{5}{2}$
  • C
    9
  • D
    7
✓

Answer

$(B)$ Unit vector $\hat{ u }= x \hat{ i }+ y \hat{ j }+ z \hat{ k }$
$\overrightarrow{p}_1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \overrightarrow{p}_2$
$=\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$
$\overrightarrow{p}_3=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$
Now angle between $\hat{ u }$ and $\overrightarrow{ p }_1=\frac{\pi}{2}$
$\hat{u} \cdot \overrightarrow{p}_1=0$
$ \Rightarrow \frac{x}{\sqrt{2}}+\frac{z}{\sqrt{2}}=0$
$\Rightarrow x+z=0 \ldots \text { (i) }$
Angle between $\hat{ u }$ and $\overrightarrow{ p }_2=\frac{\pi}{3}$
$\hat{u} \cdot \overrightarrow{p}_2=|\hat{u}| \cdot\left|\overrightarrow{p}_2\right| \cos \frac{\pi}{3}$
$\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} $
$\Rightarrow y+z=\frac{1}{\sqrt{2}}$
Angle between $\hat{u}$ and $\overrightarrow{ p }_3=\frac{2 \pi}{3}$
$\hat{u} \cdot \overrightarrow{p}_3=|\hat{u}| \cdot\left|\overrightarrow{p}_3\right| \cos \frac{2 \pi}{3}$
$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} $
$\Rightarrow x+y=\frac{-1}{\sqrt{2}}$
from equation $(i), (ii)$ and $(iii)$ we get
$x=\frac{-1}{\sqrt{2}} y=0 z=\frac{1}{\sqrt{2}}$
Thus $\hat{ u }-\overrightarrow{ v }=\frac{-1}{\sqrt{2}} \hat{ i }+\frac{1}{\sqrt{2}} \hat{ k }-\frac{1}{\sqrt{2}} \hat{ i }-\frac{1}{\sqrt{2}} \hat{ j }-\frac{1}{\sqrt{2}} \hat{ k }$
$\hat{u}-\overrightarrow{v}=\frac{-2}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{j}$
$\therefore|\hat{u}-\overrightarrow{v}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2}$

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