Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
The vector sum of $\vec{a}$ and $\vec{b}$ trisects the angle $\theta$ between them. If $|\vec{a}| = a; |\vec{b}| = b; a > b$, then:
$\theta = 3\cos^{-1}\left(\frac{2b}{a}\right)$
$\theta = 3\cos^{-1}\left(\frac{a}{2b}\right)$
$|\vec{a} + \vec{b}| = \frac{a^2 + b^2}{b}$
$|\vec{a} + \vec{b}| = \frac{a^2 - b^2}{b}$
Step-by-Step Solution
Key Concept: Cross product magnitude combined with orthogonality constraints reduces vector expressions to scalar relationships involving angles.
Using vector addition and cross product properties, express $|\vec{d}|^2$ in terms of magnitudes and dot products. Apply the constraint $\vec{a} \cdot \vec{b} = 0$ to simplify. Calculate $|\vec{d}| \cos \alpha$ using the cross product magnitude formula. From $\cos \alpha = \frac{1}{\sqrt{3}}$, deduce that $\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$ when $\vec{a} \cdot \vec{b} = 0$. Use the parametric geometry to find $\theta = 3\cos^{-1}\left(\frac{a}{2b}\right)$ and compute $|\vec{u} + \vec{v}|$ using the resultant formula.
Correct Answer: 2,4