Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are four non-coplanar unit vectors, $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular, such that $\vec{d}$ makes equal angles with all the three vectors $\vec{a}, \vec{b}, \vec{c}$, then:
$\vec{a} + \vec{b} = \vec{b} + \vec{c} = \vec{c} + \vec{d}$
$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$
$[\vec{d} \vec{a} \vec{b}] = [\vec{d} \vec{b} \vec{c}] = [\vec{d} \vec{c} \vec{a}]$
$[\vec{d} \vec{a} \vec{b}] = [\vec{d} \vec{b} \vec{c}] = [\vec{d} \vec{c} \vec{a}]$
Step-by-Step Solution
Key Concept: A line through the centroid with equal direction cosines satisfies equal scalar triple product conditions.
Since $\vec{d}$ makes equal angles with $\vec{a}, \vec{b}, \vec{c}$, it passes through the centroid with $\vec{d} = \frac{\mu(\vec{a} + \vec{b} + \vec{c})}{3}$ for some $\mu$. From the scalar triple product identity $[\vec{a} \vec{b} \vec{c}]\vec{d} = [\vec{d} \vec{b} \vec{c}]\vec{a} + [\vec{a} \vec{d} \vec{c}]\vec{b} + [\vec{a} \vec{b} \vec{d}]\vec{c}$ and the condition that components are equal, we conclude $[\vec{d} \vec{c}] = [\vec{d} \vec{a}] = [\vec{d} \vec{b}]$.
Correct Answer: 4