JEE Main 2026 — Vector Algebra Question with Solution

Given $2(\vec{a}\times \vec{b})+3(\vec{b}\times \vec{c})=\vec{0}$

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

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

Since the cross product is zero, the vectors are collinear:

$2\vec{a}-3\vec{c}=\lambda \vec{b}$

$\Rightarrow 3\vec{c}=2\vec{a}-\lambda \vec{b}$

Taking the dot product with $\vec{a}$ on both sides:

$3(\vec{a}\cdot \vec{c})=2|\vec{a}{|}^2-\lambda (\vec{a}\cdot \vec{b})$

We have $\vec{a}=4\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=10\hat{i}+2\hat{j}-\hat{k}$.

$|\vec{a}{|}^2=4^2+(-1{)}^2+3^2=26$

$\vec{a}\cdot \vec{b}=4(10)+(-1)(2)+3(-1)=40-2-3=35$

Given $\vec{a}\cdot \vec{c}=15$, substituting these values:

$3(15)=2(26)-35\lambda$

$45=52-35\lambda$

$35\lambda =7\Rightarrow \lambda =\frac{1}{5}$

Substituting $\lambda$ back into the equation for $\vec{c}$:

$3\vec{c}=2\vec{a}-\frac{1}{5}\vec{b}$

We need to find $\vec{c}\cdot (\hat{i}+\hat{j}-3\hat{k})$. Let $\vec{v}=\hat{i}+\hat{j}-3\hat{k}$.

Taking the dot product with $\vec{v}$ on both sides:

$3(\vec{c}\cdot \vec{v})=2(\vec{a}\cdot \vec{v})-\frac{1}{5}(\vec{b}\cdot \vec{v})$

Calculating $\vec{a}\cdot \vec{v}$ and $\vec{b}\cdot \vec{v}$:

$\vec{a}\cdot \vec{v}=4(1)+(-1)(1)+3(-3)=4-1-9=-6$

$\vec{b}\cdot \vec{v}=10(1)+2(1)+(-1)(-3)=10+2+3=15$

Substituting these values:

$3(\vec{c}\cdot \vec{v})=2(-6)-\frac{1}{5}(15)$

$3(\vec{c}\cdot \vec{v})=-12-3=-15$

$\vec{c}\cdot \vec{v}=-5$

Answer: $-5$