JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 19)

Consider three vectors $$\vec{a}, \vec{b}, \vec{c}$$. Let $$|\vec{a}|=2,|\vec{b}|=3$$ and $$\vec{a}=\vec{b} \times \vec{c}$$. If $$\alpha \in\left[0, \frac{\pi}{3}\right]$$ is the angle between the vectors $$\vec{b}$$ and $$\vec{c}$$, then the minimum value of $$27|\vec{c}-\vec{a}|^2$$ is equal to:
124
110
121
105

Explanation

$$\begin{aligned} & \vec{a}=\vec{b} \times \vec{c} \\ & |\vec{a}|=2,|\vec{b}|=3 \end{aligned}$$

$$\vec{a} \cdot \vec{b}=0$$ and $$\vec{a} \cdot \vec{c}=0$$

$$\begin{aligned} & |\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a} \\ & =4+|\vec{c}|^2 \end{aligned}$$

$$\begin{aligned} & |\vec{a}|=|\vec{b} \times \vec{c}|=|\vec{b}| \sin \alpha|\vec{c}| \\ & \Rightarrow \sin \alpha|\vec{c}|=\frac{2}{3} \\ & \Rightarrow \sin ^2 \alpha=\frac{4}{9|\vec{c}|^2} \\ & \Rightarrow|\vec{c}|^2=\frac{4}{9 \sin ^2 \alpha} \\ & \Rightarrow|\vec{c}-\vec{a}|^2=4+\frac{4}{9 \sin ^2 \alpha} \end{aligned}$$

For $$|\vec{c}-\vec{a}|^2$$ to be minimum for $$\alpha \in\left[0, \frac{\pi}{3}\right]$$

$$\begin{gathered} \sin \alpha=\frac{\sqrt{3}}{2} \\ 27|\vec{c}-\vec{a}|^2=27\left[4+\frac{4.4}{9.3}\right] \\ =27\left[\frac{124}{27}\right]=124 \end{gathered}$$

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