Verified Solution • 2025–26 Session moderate QJEEM22Ma8e5b: JEE Main 2022 Mathematics - Class 12 JEEM22Ma8e5b Solution: Let , R. If is a vector such that and , then is equal to… Let b→=i^+j^+λk^\overrightarrow b = \widehat i + \widehat j + \lambda \widehat kb=i+j+λk , λ\lambdaλ ∈\in∈ R. If a→\overrightarrow aa is a vector such that a→×b→=13i^−j^−4k^\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat ka×b=13i−j−4k and a→ . b→+21=0\overrightarrow a \,.\,\overrightarrow b + 21 = 0a.b+21=0 , then (b→−a→). (k^−j^)+(b→+a→). (i^−k^)\left( {\overrightarrow b - \overrightarrow a } \right).\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right).\,\left( {\widehat i - \widehat k} \right)(b−a).(k−j)+(b+a).(i−k) is equal to _____________. Step-by-Step Verified Solution Let a→=xi^=yj^+zk^\overrightarrow a = x\widehat i = y\widehat j + z\widehat ka=xi=yj+zk So, |i^j^k^xyz11λ|=i^(λy−z)+j^(z−λx)+k^(x−y) ⇒λy−z=13, z−λx=−1, x−y=−4\Rightarrow \lambda y - z = 13,\,z - \lambda x = - 1,\,x - y = - 4⇒λy−z=13,z−λx=−1,x−y=−4 and x+y+λz=−21x + y + \lambda z = - 21x+y+λz=−21 ⇒\Rightarrow⇒ Clearly, λ=3\lambda = 3λ=3 , x=−2x = - 2x=−2 , y=2y = 2y=2 and z=−7z = - 7z=−7 So, b→−a→=3i^−j^+10k^\overrightarrow b - \overrightarrow a = 3\widehat i - \widehat j + 10\widehat kb−a=3i−j+10k and b→+a→=−i^+3j^−4k^\overrightarrow b + \overrightarrow a = - \widehat i + 3\widehat j - 4\widehat kb+a=−i+3j−4k ⇒(b→−a→) . (k^−j^)+(b→+a→) . (i^−k^)=11+3=14\Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right)\,.\,\left( {\widehat i - \widehat k} \right) = 11 + 3 = 14⇒(b−a).(k−j)+(b+a).(i−k)=11+3=14 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring