Verified Solution • 2025–26 Session moderate QJEEM23Ma2f93: JEE Main 2023 Physics - Class 11 JEEM23Ma2f93 Solution: If and . Then is equal to : If a→=i^+2k^,b→=i^+j^+k^,c→=7i^−3j^+4k^,r→×b→+b→×c→=0→\overrightarrow a = \widehat i + 2\widehat k,\overrightarrow b = \widehat i + \widehat j + \widehat k,\overrightarrow c = 7\widehat i - 3\widehat j + 4\widehat k,\overrightarrow r \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0a=i+2k,b=i+j+k,c=7i−3j+4k,r×b+b×c=0 and r→ . a→=0\overrightarrow r \,.\,\overrightarrow a = 0r.a=0 . Then r→ . c→\overrightarrow r \,.\,\overrightarrow cr.c is equal to : Multiple Choice Options A 36 B 30 C 34 D 32 Step-by-Step Verified Solution (r→−c→)×b→=0(\overrightarrow r - \overrightarrow c ) \times \overrightarrow b = 0(r−c)×b=0 r→=λb→+c→\overrightarrow r = \lambda \overrightarrow b + \overrightarrow cr=λb+c ⇒λb→ . a→+c→ . a→=0\Rightarrow \lambda \overrightarrow b \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a = 0⇒λb.a+c.a=0 ⇒λ(3)+(7+8)=0\Rightarrow \lambda (3) + (7 + 8) = 0⇒λ(3)+(7+8)=0 ⇒λ=−5\Rightarrow \lambda = - 5⇒λ=−5 r→=5b→+c→\overrightarrow r = 5\overrightarrow b + \overrightarrow cr=5b+c =−5i^−5j^−5k^+(7i^+3j^+4k^)= - 5\widehat i - 5\widehat j - 5\widehat k + (7\widehat i + 3\widehat j + 4\widehat k)=−5i−5j−5k+(7i+3j+4k) =2i^−8j^−k^= 2\widehat i - 8\widehat j - \widehat k=2i−8j−k ∴\therefore∴ r→ . c→=17+24−4=34\overrightarrow r \,.\,\overrightarrow c = 17 + 24 - 4 = 34r.c=17+24−4=34 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