Verified Solution • 2025–26 Session moderate

QJEEM22Maad3a: JEE Main 2022 Physics - Class 11 JEEM22Maad3a Solution: Let and let be a vector such that and . Then the projection of…

Let a⃗=i^−j^+2k^\vec{a}=\hat{i}-\hat{j}+2 \hat{k} and let b⃗\vec{b} be a vector such that a⃗×b⃗=2i^−k^\vec{a} \times \vec{b}=2 \hat{i}-\hat{k} and a⃗⋅b⃗=3\vec{a} \cdot \vec{b}=3 . Then the projection of b⃗\vec{b} on the vector a⃗−b⃗\vec{a}-\vec{b} is :

Multiple Choice Options

  • A 221\frac{2}{\sqrt{21}}
  • B 2372 \sqrt{\frac{3}{7}}
  • C 2373\frac{2}{3} \sqrt{\frac{7}{3}}
  • D 23\frac{2}{3}

Step-by-Step Verified Solution

a→=i^−j^+2k^\overrightarrow a = \widehat i - \widehat j + 2\widehat k

a→×b→=2i^−k^\overrightarrow a \times \overrightarrow b = 2\widehat i - \widehat k

a→ . b→=3\overrightarrow a \,.\,\overrightarrow b = 3

∣a→×b→∣2+∣a→ . b→∣2=∣a→∣2 . ∣b→∣2|\overrightarrow a \times \overrightarrow b {|^2} + |\overrightarrow a \,.\,\overrightarrow b {|^2} = |\overrightarrow a {|^2}\,.\,|\overrightarrow b {|^2}

⇒5+9=6∣b→∣2\Rightarrow 5 + 9 = 6|\overrightarrow b {|^2}

⇒∣b∣2=73\Rightarrow |b {|^2} = {7 \over 3}

∣a→−b→∣=∣a→∣2+∣b→∣2−2a→ . b→=73|\overrightarrow a - \overrightarrow b | = \sqrt {|\overrightarrow a {|^2} + |\overrightarrow b {|^2} - 2\overrightarrow a \,.\,\overrightarrow b } = \sqrt {{7 \over 3}}

projection of b→\overrightarrow b on a→−b→=b→ . (a→−b→)∣a→−b→∣\overrightarrow a - \overrightarrow b = {{\overrightarrow b \,.\,(\overrightarrow a - \overrightarrow b )} \over {|\overrightarrow a - \overrightarrow b |}}

=b→ . a→−∣b→∣2∣a→−b→∣=3−7373= {{\overrightarrow b \,.\,\overrightarrow a - |\overrightarrow b {|^2}} \over {|\overrightarrow a - \overrightarrow b |}} = {{3 - {7 \over 3}} \over {\sqrt {{7 \over 3}} }}

=221= {2 \over {\sqrt {21} }}

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