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 [JEE Main 2026, 23 Jan (Shift 2)]

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Correct answer: b) $1$. Given: $ \vec{a} \times \vec{b}-2(\vec{a} \times \vec{c})=0 $ $\Rightarrow \vec{a} \times(\vec{b}-2 \vec{c})=0$ $\Rightarrow \vec{b}-2 \vec{c}=\lambda \vec{a} \ldots . .(i) $ $\Rightarrow |\lambda \vec{a}|^2=|\vec{b}-2 \vec{c}|^2$ $\Rightarrow \lambda^2|\vec{a}|^2=|\vec{b}|^2+4 |\vec{c}|^2-4 \vec{b} \cdot \vec{c} $ $\Rightarrow \lambda^2=16+16-4.4 \cdot 2 \cdot \frac{1}{2} $ $\Rightarrow \lambda^2=16 $ $\Rightarrow \lambda= \pm 4 $ $ \because \vec{b}-2 \vec{c}= \pm 4 \vec{a}$ Dot with $\vec{c} $ $\vec{b} \cdot \vec{c}-2|\vec{c}|^2= \pm 4(\vec{a} \cdot \vec{c})$ $\Rightarrow 4 \cdot 2 \cdot \frac{1}{2}-2 \cdot 4= \pm 4(\vec{a} \cdot \vec{c}) $ $\Rightarrow |\vec{a} \cdot \vec{c}|=1$

Let \(\vec{a}, \vec{b}, \vec{c}\) be three vectors such that \(\vec{a} \times \vec{b}=2(\vec{a} \times \vec{c})\). If \(…

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