Verified Solution • 2025–26 Session moderate

QJEEM22Ma037c: JEE Main 2022 Physics - Class 11 JEEM22Ma037c Solution: Let and . If the projection of on the vector is 30, then is…

Let a→=αi^+j^−k^\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\hat{j}-\hat{k} and \overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}, \alpha>0 . If the projection of a→×b→\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} on the vector −i^+2j^−2k^-\hat{i}+2 \hat{j}-2 \hat{k} is 30, then α\alpha is equal to :

Multiple Choice Options

  • A 152\frac{15}{2}
  • B 8
  • C 132\frac{13}{2}
  • D 7

Step-by-Step Verified Solution

Given : a→=(α,1,−1)\overrightarrow a = (\alpha ,1, - 1) and b→=(2,1,−α)\overrightarrow b = (2,1, - \alpha )

c→=a→×b→=|i^j^k^α1−121−α|

=(−α+1)i^+(α2−2)j^+(α−2)k^= ( - \alpha + 1)\widehat i + ({\alpha ^2} - 2)\widehat j + (\alpha - 2)\widehat k

Projection of c→\overrightarrow c on d→=−i^+2j^−2k^\overrightarrow d = - \widehat i + 2\widehat j - 2\widehat k

=∣c→ . d→∣d∣∣=30= \left| {\overrightarrow c \,.\,{{\overrightarrow d } \over {|d|}}} \right| = 30 {Given}

⇒ =∣α−1−4+2α2−2α+41+4+4∣=30\Rightarrow \, = \left| {{{\alpha - 1 - 4 + 2{\alpha ^2} - 2\alpha + 4} \over {\sqrt {1 + 4 + 4} }}} \right| = 30

On solving α=−132\alpha = {{ - 13} \over 2} (Rejected as \alpha > 0 )

and α=7\alpha = 7

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