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Question #40877

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{u}_k = \binom{10}{k} \hat{i} + \binom{10}{k+1} \hat{j} + \binom{10}{k+2} \hat{k}$ and $\vec{v}_k = \binom{11}{k+1} \hat{i} + \binom{11}{k+2} \hat{j} + \binom{11}{k+3} \hat{k}$ for any integer $k$ , where $\binom{n}{r} = 0$ if $r < 0$ or $r > n$ . Let $S$ be the set of all integers $k \in [-3, 11]$ such that the points with position vectors $\vec{u}_k$ and $\vec{v}_k$ are collinear with the origin $O$ . Find the number of elements in $S$ .

Options

A.
0
B.
1
C.
2
D.
3

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Collinearity and Coplanarity|General Term Formulation|Closed vs Open Intervals

Why This Question Is Tricky

Common Mistakes

This question tests the transition between discrete binomial coefficients and continuous vector geometry.

The primary trap is the assumption that collinearity strictly requires proportional non-zero components.

Students typically set up the ratio of components, find a mathematical contradiction (20 = 9), and incorrectly conclude there are 0 solutions.

They miss the degenerate boundary cases where the vectors collapse to the zero vector ( $\vec{u}_k = \vec{0}$ or $\vec{u}_{k+1} = \vec{0}$ ), which trivially satisfy the cross-product condition for collinearity.

The inclusion of negative indices (e.g., $k=-3$ ) further tests the rigorous definition of binomial coefficients $\binom{n}{r} = 0$ for $r < 0$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Collinearity and Coplanarity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free