Verified Solution • 2025–26 Session moderate

QJEEM19Ma2558: JEE Main 2019 Physics - Class 11 JEEM19Ma2558 Solution: Let and be two vectors. If a vector perpendicular to both the…

Let a→=3i^+2j^+2k^\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat k and b→=i^+2j^−2k^\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k be two vectors. If a vector perpendicular to both the vectors a→+b→\overrightarrow a + \overrightarrow b and a→−b→\overrightarrow a - \overrightarrow b has the magnitude 12 then one such vector is :

Multiple Choice Options

  • A 4(2i^−2j^−k^)4\left( {2\widehat i - 2\widehat j - \widehat k} \right)
  • B 4(−2i^−2j^+k^)4\left( { - 2\widehat i - 2\widehat j + \widehat k} \right)
  • C 4(2i^+2j^+k^)4\left( {2\widehat i + 2\widehat j + \widehat k} \right)
  • D 4(2i^+2j^−k^)4\left( {2\widehat i + 2\widehat j - \widehat k} \right)

Step-by-Step Verified Solution

Required vector is r→\overrightarrow r = λ((a‾+b‾)×(a‾−b‾))\lambda \left( {\left( {\overline a + \overline b } \right) \times \left( {\overline a - \overline b } \right)} \right)

⇒|i^j^k^440204|=λ(16i^−16j^−8k^)

⇒r→=8λ(2i^−2j^−k^)=∣r→∣\Rightarrow \overrightarrow r = 8\lambda \left( {2\widehat i - 2\widehat j - \widehat k} \right) = \left| {\overrightarrow r } \right|

⇒∣8λ∣.3⇒8λ=±4\Rightarrow \left| {8\lambda } \right|.3 \Rightarrow 8\lambda = \pm 4

⇒r→=±4(2i^−2j^−k^)\Rightarrow \overrightarrow r = \pm 4\left( {2\widehat i - 2\widehat j - \widehat k} \right)

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