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Question #40713

Difficulty: 8/10 | 🟢 Free

The Question

For each $k \in \{1, 2, 3\}$ , let $E_k$ be the quadratic equation $x^2 - \frac{k}{2}x + \frac{k^2-2}{4} = 0$ . Let $S$ be the set of all valid real vectors $\vec{v}_k = l_k \hat{i} + m_k \hat{j} + n_k \hat{k}$ such that $l_k, m_k, n_k$ are the direction cosines of a line in 3D space, where $l_k$ and $m_k$ are the two roots of $E_k$ (with $l_k > m_k$ ) and $n_k > 0$ . Let $W$ be the vector space spanned by the elements of $S$ over the field of real numbers. What is the dimension of $W$ ?

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

Quadratic Roots in Sequences|Linear Independence vs Dependence|Direction Cosines constraint

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of quadratic roots, 3D geometry constraints, and linear algebra.

The primary trap lies in the mechanical verification of the direction cosine constraint ( $l^2+m^2+n^2=1$ ).

A student might calculate $n_k^2 = 1 - (l_k^2+m_k^2) = 1 - (k^2/4 - (k^2-2)/2) = k^2/4 \ge 0$ and falsely conclude that real vectors exist for all $k=1, 2, 3$ .

Proceeding blindly, they will construct a determinant to check linear independence, notice $l_k + m_k = k/2 = n_k$ , find the determinant is zero, and incorrectly deduce the dimension is 2.

The 'Lethal Collision' occurs because they fail to check the discriminant of the quadratic ( $D_k \ge 0$ ); for $k=2, 3$ , the roots are complex, rendering the vectors imaginary.

Thus, only one real vector exists, making the true dimension 1.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Quadratic Roots in Sequences
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free