Verified Solution • 2025–26 Session moderate

QJEEM21Ma2589: JEE Main 2021 Physics - Class 11 JEEM21Ma2589 Solution: Let = + 2 3 and 3 + 5 . If = , . = 3 and = 1, R, then the value …

Let a→\overrightarrow a = i^\widehat i + 2 j^\widehat j −- 3 k^\widehat k and b→=2i^\overrightarrow b = 2\widehat i −- 3 j^\widehat j + 5 k^\widehat k . If r→\overrightarrow r ×\times a→\overrightarrow a = b→\overrightarrow b ×\times r→\overrightarrow r ,

r→\overrightarrow r . (αi^+2j^+k^)\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right) = 3 and r→ . (2i^+5j^−αk^)\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right) = −- 1, α\alpha ∈\in R, then the

value of α\alpha + ∣r→∣2{\left| {\overrightarrow r } \right|^2} is equal to :

Multiple Choice Options

  • A 13
  • B 11
  • C 9
  • D 15

Step-by-Step Verified Solution

Given r→\overrightarrow r ×\times a→\overrightarrow a = b→\overrightarrow b ×\times r→\overrightarrow r

⇒\Rightarrow r→×a→=−r→×b→\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b

r→×(a→+b→)=0\overrightarrow r \times (\overrightarrow a + \overrightarrow b ) = 0

r→∣∣(a→+b→)\overrightarrow r ||(\overrightarrow a + \overrightarrow b )

r→=λ(a→+b→)\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b )

(a→+b→=3i^−j^+2k^)(\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k)

∵\because r→ . (2i^+5j^−αk^)=−1\overrightarrow r \,.\,(2\widehat i + 5\widehat j - \alpha \widehat k) = - 1

λ[3i^−j^+2k^] . [2i^+5j^−αk^]=−1\lambda \left[ {3\widehat i - \widehat j + 2\widehat k} \right]\,.\,\left[ {2\widehat i + 5\widehat j - \alpha \widehat k} \right] = - 1

⇒λ(6−5−2α)=−1\Rightarrow \lambda (6 - 5 - 2\alpha ) = - 1

λ(1−2α)=−1\lambda (1 - 2\alpha ) = - 1 .... (1)

r→ . (αi^+2j^+k^)=3\overrightarrow r \,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3

λ(3i^−j^+2k^) . (αi^+2j^+k^)=3\lambda (3\widehat i - \widehat j + 2\widehat k)\,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3

⇒λ[3α−2+2]=3⇒λα=1\Rightarrow \lambda [3\alpha - 2 + 2] = 3 \Rightarrow \lambda \alpha = 1 .... (2)

From (1) & (2)

λ[1−2λ]=−1\lambda \left[ {1 - {2 \over \lambda }} \right] = - 1

λ−2=−1⇒λ=1 α=1\lambda - 2 = - 1 \Rightarrow \lambda = 1\,\alpha = 1

r→=3i^−j^+2k^\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k

α+∣r→∣2=1+14=15\alpha + |\overrightarrow r {|^2} = 1 + 14 = 15

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