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Question #40853

Difficulty: 8/10 | 🟢 Free

The Question

Let $M = \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 5 \\ 2 & 3 & 5 \end{pmatrix}$ . The position vectors of the vertices $A, B,$ and $C$ of a tetrahedron $ABCD$ are given by the first, second, and third column vectors of $M$ , respectively. Let $M'$ be the matrix obtained by interchanging the first and second columns of $M$ . If the position vector of vertex $D$ is given by $(\det(M), \text{Tr}(M), \det(M'))$ , find the distance between the centroid of the tetrahedron $ABCD$ and the vertex $D$ .

Options

A.
$\sqrt{83}$
B.
$\sqrt{38}$
C.
$\sqrt{65}$
D.
$\sqrt{74}$

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

Distance Formula in 3D|Centroid of a Tetrahedron|Row/Column Exchange Parity|Matrix Trace Properties

Why This Question Is Tricky

Common Mistakes

This question integrates Linear Algebra with 3D Coordinate Geometry.

The primary trap lies in the 'Row/Column Exchange Parity' property of determinants.

Students often incorrectly assume $\det(M') = \det(M)$ or miscalculate the trace, leading to an incorrect vertex $D$ (e.g., $D=(6,8,6)$ yields a distance of $\sqrt{38}$ ).

Furthermore, it tests the precise application of the tetrahedron centroid formula, ensuring students divide by 4 rather than 3, a common cognitive slip when transitioning from 2D triangles to 3D tetrahedrons.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Distance Formula in 3D
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free