Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Let $\vec{a} = a\vec{i} + b\vec{j} + c\vec{k}$ and $\vec{\beta} = b\vec{i} + \vec{c}\vec{j} + a\vec{k}$, where $a, b, c \in \mathbb{R}$. If '$\theta$' be the angle between $\vec{a}$ and $\vec{\beta}$ then:
θ ∈ (0, π/2)
θ ∈ [0, 2π/3]
θ ∈ (2π/3, π]
None of these
Step-by-Step Solution
Key Concept: Bounds on dot products come from algebraic inequalities applied to sum of squares and perfect squares.
Using the dot product formula $\vec{a} \cdot \vec{b} = ab + bc + ca = \sqrt{a^2 + b^2 + c^2}\sqrt{b^2 + c^2 + a^2}\cos\theta$, we get $\cos\theta = \frac{ab + bc + ca}{a^2 + b^2 + c^2}$. From $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$, we deduce $ab + bc + ca \leq a^2 + b^2 + c^2$, so $\cos\theta \leq 1$. Also $(a+b+c)^2 \geq 0$ gives $ab + bc + ca \geq -\frac{1}{2}(a^2+b^2+c^2)$, hence $\cos\theta \geq -\frac{1}{2}$, placing $\theta \in [0, 2\pi/3]$.
Correct Answer: 2