Verified Solution • 2025–26 Session moderate

QJEEM24Ma880f: JEE Main 2024 Physics - Class 11 JEEM24Ma880f Solution: Let a unit vector make angles and with the vectors and…

Let a unit vector u^=xi^+yj^+zk^\hat{u}=x \hat{i}+y \hat{j}+z \hat{k} make angles π2,π3\frac{\pi}{2}, \frac{\pi}{3} and 2π3\frac{2 \pi}{3} with the vectors 12i^+12k^,12j^+12k^\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 12i^+12j^\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j} respectively. If v⃗=12i^+12j^+12k^\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k} then ∣u^−v⃗∣2|\hat{u}-\vec{v}|^2 is equal to

Multiple Choice Options

  • A 112\frac{11}{2}
  • B 52\frac{5}{2}
  • C 7
  • D 9

Step-by-Step Verified Solution

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

amp;p→1=12i^+12k^,p→2=12j^+12k^amp;p→3=12i^+12j^\begin{aligned}& \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}\end{aligned}

Now angle between u^\hat{\mathrm{u}} and p→1=π2\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}

u^⋅p→1=0⇒x2+z2=0\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0

⇒x+z=0\Rightarrow \mathrm{x}+\mathrm{z}=0 ...... (i)

Angle between u^\hat{\mathrm{u}} and p→2=π3\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}

u^⋅p→2=∣u^∣⋅∣p→2∣cos⁡π3\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}

⇒y2+z2=12⇒y+z=12\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \Rightarrow y+z=\frac{1}{\sqrt{2}} ...... (ii)

Angle between u^\hat{\mathrm{u}} and p→3=2π3\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}

u^⋅p→3=∣u^∣⋅∣p→3∣cos⁡2π3\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3}

⇒x2+42=−12⇒x+y=−12\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}} ..... (iii)

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

x=−12y=0z=12x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}

amp; Thus u^−v→=−12i^+12k^−12i^−12j^−12k^amp;u^−v→=−22i^−12j^amp;∴∣u^−v→∣2=(42+12)2=52\begin{aligned}& \text { Thus } \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\& \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \\& \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2}\end{aligned}

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