Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade None
Question:
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-coplanar vectors and $\vec{d}$ be a non-zero vector, which is perpendicular to $(\vec{a} + \vec{b} + \vec{c})$. Now if $\vec{d} = \sin x(\vec{a} \times \vec{b}) + \cos y(\vec{b} \times \vec{c}) + 2(\vec{c} \times \vec{a})$, then minimum value of $x^2 + y^2$ is equal to:
$\pi^2$
$0$
$\frac{5\pi^2}{4}$
$\frac{5\pi^2}{4}$
Step-by-Step Solution
Key Concept: Scalar triple product conditions determine unique angle values through constraint equations.
Given $\vec{d} \cdot \vec{a} = \cos y[\vec{a} \vec{b} \vec{c}] = \vec{d}(\vec{b} + \vec{c})$ with $[\vec{a}(\vec{a} + \vec{b} + \vec{c})] = 0$, we have $\cos y = -\frac{\vec{d} \cdot (\vec{a} + \vec{b})}{[\vec{a} \vec{b} \vec{c}]}$. Similarly derive $\sin x = \frac{\vec{d} \cdot (\vec{a} + \vec{b})}{[\vec{a} \vec{b} \vec{c}]}$ and $2 = \frac{\vec{d} \cdot [\vec{a} + \vec{c}]}{[\vec{a} \vec{b} \vec{c}]}$. With $\sin x + \cos y + 2 = 0$, we get $\sin x = -1$ and $\cos y = -1$, giving $x = -\frac{\pi}{2}$ and $y = \pi$.
Correct Answer: 4