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Question #41836

Difficulty: 8/10 | 🟢 Free

The Question

A fair die is rolled three times, yielding outcomes $a, b, c \in \{1, 2, 3, 4, 5, 6\}$ . Three vectors are defined as: $\vec{u} = (a^2-5a+6)\hat{i} + (b-4)\hat{j} + (c-5)\hat{k}$ $\vec{v} = (4-b)\hat{i} + (a^2-5a+6)\hat{j} - a\hat{k}$ $\vec{w} = (5-c)\hat{i} + a\hat{j} + (a^2-5a+6)\hat{k}$ What is the probability that the vectors $\vec{u}, \vec{v},$ and $\vec{w}$ are coplanar?

Options

A.
$\frac{1}{6}$
B.
$\frac{1}{3}$
C.
$\frac{1}{2}$
D.
$\frac{1}{4}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Equiprobable Outcomes|Scalar Triple Product Volume Logic|Skew-Symmetric Determinant

Why This Question Is Tricky

Common Mistakes

This question tests the geometric interpretation of the scalar triple product alongside the algebraic properties of skew-symmetric matrices.

The trap lies in brute-forcing the $3 \times 3$ determinant, which becomes algebraically overwhelming.

The intended logical path requires recognizing that the component matrix can be decomposed into a scalar matrix $xI$ and a skew-symmetric matrix $S$ .

Students must realize that $\det(xI + S) = x(x^2 + \sum S_{ij}^2)$ , meaning the volume is zero if and only if the scalar part $x$ is zero (since the sum of squares is strictly positive).

This elegantly reduces a complex 3-variable probability problem into a simple quadratic condition on a single die roll.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Equiprobable Outcomes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free