Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade None
Question:
The position vectors of the vertices $A, B, C$ of a triangle are $\vec{a}, \vec{b}$ and $\vec{c}$ respectively, where $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{a}$ and $\vec{b}$ are non-collinear vectors. If $\vec{g}$, the position vector of the centroid of the triangle $ABC$, makes equal angles $\alpha$ with the vectors $\vec{a}, \vec{b}$ and $\vec{c}$, then:
$|\vec{a}| = |\vec{b}|$
$|\vec{a}| \neq |\vec{b}|$
the value of $\alpha$ is $\cos^{-1}\frac{1}{\sqrt{3}}$ if $\vec{a} \cdot \vec{b} = 0$
the value of $\alpha$ is $\cos^{-1}\sqrt{\frac{2}{5}}$ if $\vec{a} \cdot \vec{b} = 0$
Step-by-Step Solution
Key Concept: Equal angles condition forces $|\vec{a}| = |\vec{b}|$, and when these vectors are orthogonal, the angle with the centroid is $\cos^{-1}\frac{1}{\sqrt{3}}$.
The centroid is $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$. Since $\vec{g}$ makes equal angles $\alpha$ with $\vec{a}$, $\vec{b}$, and $\vec{c}$, we have $\cos\alpha = \frac{\vec{g}\cdot\vec{a}}{|\vec{g}||\vec{a}|} = \frac{\vec{g}\cdot\vec{b}}{|\vec{g}||\vec{b}|} = \frac{\vec{g}\cdot\vec{c}}{|\vec{g}||\vec{c}|}$. Computing $\vec{g}\cdot\vec{a} = \frac{1}{3}(|\vec{a}|^2 + \vec{a}\cdot\vec{b})$ and $\vec{g}\cdot\vec{b} = \frac{1}{3}(\vec{a}\cdot\vec{b} + |\vec{b}|^2)$. For equal angles: $\frac{|\vec{a}|^2 + \vec{a}\cdot\vec{b}}{|\vec{a}|} = \frac{\vec{a}\cdot\vec{b} + |\vec{b}|^2}{|\vec{b}|}$, which gives $|\vec{a}| = |\vec{b}|$. When $\vec{a}\cdot\vec{b} = 0$: $\vec{g}\cdot\vec{a} = \frac{|\vec{a}|^2}{3}$ and $|\vec{g}|^2 = \frac{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{a}|^2|\vec{b}|^2}{9} = \frac{2|\vec{a}|^4 + |\vec{a}|^4}{9} = \frac{|\vec{a}|^4}{3}$, so $\cos\alpha = \frac{1}{\sqrt{3}}$.
Correct Answer: 1,3