Jee Main 2024 29 01 2024 Shift 2 - Question8
Question 8
Unit vector $\vec{u}=x \hat{i}+y \hat{j}+z \hat{k}$ makes angles
$\frac{\pi}{2}, \frac{\pi}{3}, \frac{2 \pi}{3}$ with $\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}\right),\left(\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}\right)$,
$\left(\frac{\hat{i}}{\sqrt{2}}+\frac{\hat{j}}{\sqrt{2}}\right)$ respectively and
$\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$ find $|\vec{u}-\vec{v}|$.
(1) $\sqrt{\frac{5}{2}}$
(2) $\sqrt{\frac{7}{2}}$
(3) $\sqrt{\frac{2}{5}}$
(4) $\sqrt{\frac{2}{7}}$
Show Answer
Answer: (1)
Solution:
$\frac{x}{\sqrt{2}}+\frac{z}{\sqrt{2}}=0$
$\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2}$
$\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{-1}{2}$
$\Rightarrow y=0, z=\frac{1}{\sqrt{2}}, x=\frac{-1}{\sqrt{2}}$
$\vec{v}-\vec{u}=\sqrt{2} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$
$|\vec{v}-\vec{u}|=\sqrt{2+\frac{1}{2}}$
$=\sqrt{\frac{5}{2}}$
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