Verified Solution • 2025–26 Session moderate QJEEM24Mafdb4: JEE Main 2024 Physics - Class 11 JEEM24Mafdb4 Solution: Let and be three vectors. If a vectors satisfies and , then is… Let a⃗=3i^+j^−2k^,b⃗=4i^+j^+7k^\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=4 \hat{i}+\hat{j}+7 \hat{k}a=3i^+j^−2k^,b=4i^+j^+7k^ and c⃗=i^−3j^+4k^\vec{c}=\hat{i}-3 \hat{j}+4 \hat{k}c=i^−3j^+4k^ be three vectors. If a vectors p⃗\vec{p}p satisfies p⃗×b⃗=c⃗×b⃗\vec{p} \times \vec{b}=\vec{c} \times \vec{b}p×b=c×b and p⃗⋅a⃗=0\vec{p} \cdot \vec{a}=0p⋅a=0 , then p⃗⋅(i^−j^−k^)\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})p⋅(i^−j^−k^) is equal to Multiple Choice Options A 36 B 28 C 32 D 24 Step-by-Step Verified Solution amp;p→×b→−c→×b→=0→amp;(p→−c→)×b→=0→amp;p→−c→=λb→⇒p→=c→+λb→\begin{aligned}& \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\& (\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\& \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{b}}\end{aligned}amp;p×b−c×b=0amp;(p−c)×b=0amp;p−c=λb⇒p=c+λb Now, p→⋅a→=0\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0p⋅a=0 (given) amp; So, c→⋅a→+λa→⋅b→=0amp;(3−3−8)+λ(12+1−14)=0amp;λ=−8amp;p→=c→−8b→amp;p→=−31i^−11j^−52k^amp; So, p→⋅(i^−j^−k^)amp;=−31+11+52amp;=32\begin{aligned}& \text { So, } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \\& (3-3-8)+\lambda(12+1-14)=0 \\& \lambda=-8 \\& \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}-8 \overrightarrow{\mathrm{b}} \\& \overrightarrow{\mathrm{p}}=-31 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-52 \hat{\mathrm{k}} \\& \text { So, } \overrightarrow{\mathrm{p}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\& =-31+11+52 \\& =32\end{aligned}amp; So, c⋅a+λa⋅b=0amp;(3−3−8)+λ(12+1−14)=0amp;λ=−8amp;p=c−8bamp;p=−31i^−11j^−52k^amp; So, p⋅(i^−j^−k^)amp;=−31+11+52amp;=32 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