Q40 / 75 · JEE Main 2026 (23 January Shift 2) · Mathematics
Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b}=2(\vec{a} \times \vec{c})$. If $|\vec{a}|=1,|\vec{b}|=4,|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is equal to
- Correct answerA1
- B2
- C0
- D4
Solution
$\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}) \Rightarrow \vec{a} \times (\vec{b} - 2\vec{c}) = \vec{0}$, so $\vec{b} - 2\vec{c} = k\vec{a}$.
$\vec{b} \cdot \vec{c} = 4 \cdot 2 \cdot \cos 60° = 4$.
Dot with $\vec{c}$: $4 = 2(4) + k(\vec{a} \cdot \vec{c}) \Rightarrow k(\vec{a} \cdot \vec{c}) = -4$.
Dot with $\vec{a}$: $\vec{a} \cdot \vec{b} = 2(\vec{a} \cdot \vec{c}) + k$.
Dot with $\vec{b}$: $16 = 8 + k\vec{a} \cdot \vec{b} = 8 + k\left(-\frac{8}{k} + k\right) = k^2 \Rightarrow k = \pm 4$.
$\vec{a} \cdot \vec{c} = -\frac{4}{k} = \mp 1 \Rightarrow |\vec{a} \cdot \vec{c}| = 1$.
$\vec{b} \cdot \vec{c} = 4 \cdot 2 \cdot \cos 60° = 4$.
Dot with $\vec{c}$: $4 = 2(4) + k(\vec{a} \cdot \vec{c}) \Rightarrow k(\vec{a} \cdot \vec{c}) = -4$.
Dot with $\vec{a}$: $\vec{a} \cdot \vec{b} = 2(\vec{a} \cdot \vec{c}) + k$.
Dot with $\vec{b}$: $16 = 8 + k\vec{a} \cdot \vec{b} = 8 + k\left(-\frac{8}{k} + k\right) = k^2 \Rightarrow k = \pm 4$.
$\vec{a} \cdot \vec{c} = -\frac{4}{k} = \mp 1 \Rightarrow |\vec{a} \cdot \vec{c}| = 1$.
Hence, the correct answer is option: 'A'.