Let vec a =4hat i -hat j +3hat k , vec b =10hat i +2hat j -hat k and a vector vec c be such that 2(vec a ×vec b )+3(vec b ×vec c )=vec 0 . If vec a ×vec c =15 then vec c ×(hat i +hat j -3hat k ) is equal to:

Mathematics · Vector Algebra

JEE Main 2026 — 8 April, Evening Shift — Question 31

Let $\vec{a}=4\hat{i}-\hat{j}+3\hat{k}$ , $\vec{b}=10\hat{i}+2\hat{j}-\hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a}\times\vec{b})+3(\vec{b}\times\vec{c})=\vec{0}$ . If $\vec{a}\cdot\vec{c}=15$ then $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$ is equal to:

Option A:
$-6$
Option B:
$-5$
Correct
Option C:
$-4$
Option D:
$-3$

Answer: B

Step-by-step solution

$2(\vec{a} \times \vec{b})-3(\vec{c} \times \vec{b})=\overrightarrow{0}$ $(2 \vec{a}-3 \vec{c}) \times \vec{b}=\overrightarrow{0}$ $\vec{b} \|(2 \vec{a}-3 \vec{c})$ $2 \vec{a}-3 \vec{c}=\lambda \vec{b}$

\overrightarrow{\mathrm{c}}=\frac{2 \overrightarrow{\mathrm{a}}-\lambda \overrightarrow{\mathrm{b}}}{3} \end{gathered}$$ $\overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{c}}=15$ $\left(\frac{2 \vec{a}-\lambda \vec{b}}{3}\right) \cdot \vec{a}=15$ $\frac{2|\mathrm{a}|^{2}-\lambda \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}}{3}=15$ $2(26)-\lambda(40-2-3)=45$ $\lambda=\frac{1}{5}$ $\Rightarrow \vec{c} \cdot(\hat{i}+\hat{j}-3 \hat{k})=\frac{\left(2(4 \hat{i}-\hat{j}+3 k)-\frac{1}{5}(10 \hat{i}+2 \hat{j}-k) \cdot(\hat{i}+\hat{j}-3 \hat{k})\right)}{3}$ $=\frac{2(4-1-9)-\frac{1}{5}(10+2+3)}{3}=-5$

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Scalar or Dot Product of Two Vectors