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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Vector Algebra

Let a =2 i -3 j +k, b =3 i +2 j +5 k and a vector c be such that (...

Let a⃗=2i^−3j^+k,b⃗=3i^+2j^+5k\vec{a}=2 \hat{i}-3 \hat{j}+k, \vec{b}=3 \hat{i}+2 \hat{j}+5 k and a vector c⃗\vec{c} be such that (a→−c→)×b→=−18i^−3j^+12k(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \mathrm{k} and a→⋅c→=3\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3. If b→×c→=d→\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{d}}, then ∣a→⋅d→∣|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}| is equal to :

Options

  1. A1818
  2. B1212
  3. C99
  4. D1515

Correct answer

D. 1515

Step-by-step solution

a→=2i^−3j^+k^,b→=3i^+2j^+5k^\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}
\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|ijk2−31325\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & -3 & 1 \\ 3 & 2 & 5\end{array}\right|
=−17i^−7j^+13k^(a→−c→)×b→=−18i^−3j+12k^⇒(a→×b→)−(c→×b→)=−18i^−3j^+12k^⇒b→×c→=(−18i^−3j^+12k^)−(a→×b→)=(−18i^−3j^+12k^)−(−17i−7j^+13k^)b→×c→=−i^+4j^−k^∴a→⋅d→=a→⋅(b→×c→)=(2i−3j+k)⋅(−i^+4j^−k^)=−2−12−1=−15∴∣a→⋅d→∣=15\begin{aligned} & =-17 \hat{i}-7 \hat{j}+13 \hat{k} \\ & (\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=-18 \hat{\mathrm{i}}-3 \mathrm{j}+12 \hat{k} \\ & \Rightarrow(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})=-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=(-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{k})-(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ & =(-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}})-(-17 \mathrm{i}-7 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=(2 \mathrm{i}-3 \mathrm{j}+\mathrm{k}) \cdot(-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =-2-12-1=-15 \\ & \therefore|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|=15\end{aligned}
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