JEE Advanced 2025 — Paper 1 — Question 30

A) $\vec{a}+\vec{b}+\vec{c}+\vec{d}=(\alpha+1) \vec{d}=(\beta+1) \vec{a}$

If $\alpha \neq-1$, then $\vec{d}=\left(\frac{\beta+1}{\alpha+1}\right) \vec{a}$

$\Rightarrow \vec{a}+\vec{b}+\vec{c}=\alpha \vec{d}=\alpha\left(\frac{\beta+1}{\alpha+1}\right) \vec{a}$

$\Rightarrow\left\{1-\alpha\left(\frac{\beta+1}{\alpha+1}\right)\right\} \vec{a}+\vec{b}+\vec{c}=0 \Rightarrow \vec{a}, \vec{b}, \vec{c}$

are coplanar, which is against the given condition,

so, $\alpha=-1$ and hence $\vec{a}+\vec{b}+\vec{c}+\vec{d}=\overrightarrow{0}$

B) $|\vec{a}+\vec{b}|<1 \Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}||\vec{b}| \cos \theta<1$

$\Rightarrow \cos \theta<-\frac{1}{2}$

So, $\frac{2 \pi}{3}<\theta<\pi$

C) $\vec{a} \times(\vec{a} \times \vec{b})=(\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=-\vec{b}$

$a \times\{\vec{a} \times(\vec{a} \times \vec{b})\}=\vec{a} \times(-\vec{b})=-\vec{a} \times \vec{b}$

$a \times[\vec{a} \times\{\vec{a} \times(\vec{a} \times \vec{b})\}=\vec{a} \times(-\vec{a} \times \vec{b})]$ $=(\vec{a} \cdot \vec{a}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{a}=\vec{b}$

D) $\vec{a}+\vec{b}=-\vec{c} \Rightarrow|\vec{a}+\vec{b}|^{2}=|\vec{c}|^{2}=1$

$\Rightarrow \vec{a} \cdot \vec{b}=-\frac{1}{2} \Rightarrow \theta=\frac{2 \pi}{3}$