Verified Solution • 2025–26 Session moderate

QJEEM24Ma6f20: JEE Main 2024 Physics - Class 11 JEEM24Ma6f20 Solution: Let and be a vector such that . If , then is equal to __________.

Let a→=i^−3j^+7k^,b→=2i^−j^+k^\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k} and c→\overrightarrow{\mathrm{c}} be a vector such that (a→+2b→)×c→=3(c→×a→)(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}) . If a⃗⋅c⃗=130\vec{a} \cdot \vec{c}=130 , then b⃗⋅c⃗\vec{b} \cdot \vec{c} is equal to __________.

Step-by-Step Verified Solution

(a⃗+2b⃗)×c⃗=3(c⃗×a⃗)(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})

⇒amp;b⃗×c⃗+2(a⃗×c⃗)=0amp;(b⃗+2a⃗)×c⃗=0amp;c⃗=λ(b⃗+2a⃗)amp;c⃗⋅a⃗=130⇒λ=1amp;c⃗=4i^−7j^+15k^amp;b⃗.c⃗=30\begin{aligned}\Rightarrow \quad & \vec{b} \times \vec{c}+2(\vec{a} \times \vec{c})=0 \\& (\vec{b}+2 \vec{a}) \times \vec{c}=0 \\& \vec{c}=\lambda(\vec{b}+2 \vec{a}) \\& \vec{c} \cdot \vec{a}=130 \Rightarrow \lambda=1 \\& \vec{c}=4 \hat{i}-7 \hat{j}+15 \hat{k} \\& \vec{b} . \vec{c}=30\end{aligned}

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