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Question #41618

Difficulty: 8/10 | 🟢 Free

The Question

A line $L$ passes through the origin $O$ and has direction cosines $(l, m, n)$ proportional to $(t^2-4, 4t, t^2+4)$ for a real parameter $t \notin \{0, 2\}$ . It is given that the angle $\gamma$ made by $L$ with the positive z-axis is acute. The line $L$ intersects the plane $\Pi: x - y + z = 2\sqrt{2}$ at the point $Q$ . If the sum of the $x$ and $y$ coordinates of $Q$ is at most $-\sqrt{2}$ , then the complete set of values for $t$ is:

Options

A.
$[-2, 0) \cup [1, 2)$
B.
$[-2, 1]$
C.
$(-\infty, -2] \cup (0, 1] \cup (2, \infty)$
D.
$[-2, 0) \cup (0, 1]$

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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

Line-Plane intersection logic|Direction Cosines constraint|Dot Product Projection Sign|Wavy Curve (Method of Intervals)

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of a parametrically defined line with a plane, requiring the normalization of direction ratios to direction cosines.

The core trap lies in the final inequality $\frac{t^2+4t-4}{t(t-2)} \le -1$ .

Students often cross-multiply the denominator $t(t-2)$ without considering its sign, incorrectly simplifying the inequality to $t^2+4t-4 \le -t^2+2t \implies (t+2)(t-1) \le 0$ , which yields the erroneous closed interval $[-2, 1]$ .

The correct Wavy Curve (Method of Intervals) approach requires bringing all terms to one side and finding the critical points of both numerator and denominator, revealing the sign flips at the asymptotes $t=0$ and $t=2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Line-Plane intersection logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free