Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = x_1\hat{i} + x_2\hat{j} + x_3\hat{k}$, where $x_1, x_2, x_3 \in \{-3, -2, -1, 0, 1, 2\}$. Number of possible vectors $\vec{b}$ such that $\vec{a}$ and $\vec{b}$ are mutually perpendicular, is $p$ then $\frac{p}{5} = _______.
Step-by-Step Solution
Key Concept: Transform the dot product constraint into a linear Diophantine equation, then count solutions using generating functions or combinatorial methods.
Given $\vec{a} \cdot \vec{b} = 0$ where $\vec{a} = (x_1, x_2, x_3)$ and $\vec{b} = (1, 1, 1)$, we need $x_1 + x_2 + x_3 = 0$. The number of integral solutions is found by computing the coefficient of $x^9$ in $(1 - x)^3(1 - x^{-1})^3 = (1 - x^3)^3(1 - x)^{-3}$, which equals $\binom{11}{9}\binom{3}{3} - 3\binom{8}{9}\binom{3}{3} = 25$ using the multinomial expansion.
Correct Answer: I need to find the number of vectors $\vec{b} = x_1\hat{i} + x_2\hat{j} + x_3\hat{k}$ where $x_1, x_2, x_3 \in \{-3, -2, -1, 0, 1, 2\}$