Verified Solution • 2025–26 Session moderate QJEEM24Ma3751: JEE Main 2024 Physics - Class 11 JEEM24Ma3751 Solution: Let and be three vectors such that . If , then is equal to: Let a⃗=2i^+5j^−k^,b⃗=2i^−2j^+2k^\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}a=2i^+5j^−k^,b=2i^−2j^+2k^ and c⃗\vec{c}c be three vectors such that (c⃗+i^)×(a⃗+b⃗+i^)=a⃗×(c⃗+i^)(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})(c+i^)×(a+b+i^)=a×(c+i^) . If a⃗⋅c⃗=−29\vec{a} \cdot \vec{c}=-29a⋅c=−29 , then c⃗⋅(−2i^+j^+k^)\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})c⋅(−2i^+j^+k^) is equal to: Multiple Choice Options A 15 B 10 C 5 D 12 Step-by-Step Verified Solution (c⃗+i^)×(a⃗+b⃗+i^+a⃗)=0⇒c⃗+i^=λ(a⃗+b⃗+i^+a⃗)=λ(2a⃗+b⃗+i^)=λ(7i^+8j^)⇒c⃗=(7λ−1)i^+8λj^c⃗⋅a⃗=−29⇒14λ−2+40λ=−29⇒54λ=−27⇒λ=−12\begin{gathered}(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}+\vec{a})=0 \\\Rightarrow \quad \vec{c}+\hat{i}=\lambda(\vec{a}+\vec{b}+\hat{i}+\vec{a}) \\=\lambda(2 \vec{a}+\vec{b}+\hat{i}) \\\quad=\lambda(7 \hat{i}+8 \hat{j}) \\\Rightarrow \quad \vec{c}=(7 \lambda-1) \hat{i}+8 \lambda \hat{j} \\\quad \vec{c} \cdot \vec{a}=-29 \\\Rightarrow \quad 14 \lambda-2+40 \lambda=-29 \\\Rightarrow \quad 54 \lambda=-27 \\\Rightarrow \quad \lambda=-\frac{1}{2}\end{gathered}(c+i^)×(a+b+i^+a)=0⇒c+i^=λ(a+b+i^+a)=λ(2a+b+i^)=λ(7i^+8j^)⇒c=(7λ−1)i^+8λj^c⋅a=−29⇒14λ−2+40λ=−29⇒54λ=−27⇒λ=−21 amp;∴c⃗=(−72−1)i^−4j^=−92i^−4j^amp;c⃗⋅(−2i^+j^+k^)=9−4=5\begin{aligned}& \therefore \quad \vec{c}=\left(\frac{-7}{2}-1\right) \hat{i}-4 \hat{j}=\frac{-9}{2} \hat{i}-4 \hat{j} \\& \vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=9-4=5\end{aligned}amp;∴c=(2−7−1)i^−4j^=2−9i^−4j^amp;c⋅(−2i^+j^+k^)=9−4=5 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