Before You Begin

Question #41741

Difficulty: 8/10 | 🟢 Free

The Question

Consider the family of lines $L_t$ defined by the intersection of the planes $P_1: tx + y + 2z = 0$ and $P_2: x + ty + 2z = 0$ , where $t \in \mathbb{R} \setminus \{-1, 1\}$ . Let $(l_t, m_t, n_t)$ be the direction cosines of $L_t$ such that $n_t > 0$ . Find the set of all values of $t$ for which $l_t + m_t + n_t > 0$ .

Options

A.
$(-\infty, -1) \cup (3, \infty)$
B.
$(3, \infty)$
C.
$(-\infty, 3) \setminus \{-1, 1\}$
D.
$(-1, 1) \cup (3, \infty)$

Check Your Answer

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

Symmetric vs Non-symmetric form|Direction Cosines constraint|Wavy Curve (Method of Intervals)|Indeterminate Forms

Why This Question Is Tricky

Common Mistakes

The trap in this problem is failing to account for the absolute value when computing the magnitude of the direction vector, specifically $\sqrt{(t-1)^2} = |t-1|$ .

Students often blindly cancel $(1-t)$ with $(t-1)$ without considering the sign, leading to an incorrect constant sign for the direction cosines.

Furthermore, enforcing the constraint $n_t > 0$ requires multiplying the normalized vector by the sign of its z-component, $\text{sgn}(t^2-1)$ .

This creates a piecewise inequality that must be solved using the Method of Intervals (Wavy Curve), trapping those who ignore the indeterminate nature of the normalization at $t=1$ and the sign flip at $t=-1$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Symmetric vs Non-symmetric form
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free