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Question #40445

Difficulty: 8/10 | 🟢 Free

The Question

Let $L$ be the line given by the symmetric equations $\frac{x-5}{1} = \frac{y-3}{1} = \frac{z-4}{2}$ . Let $P(x,y,z)$ be a point on the line $L$ , and let $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ be its position vector. Suppose there exists a real number $\theta$ such that $3|\vec{b}|^2 - (x+y+z)^2 = 3 \sin(2\theta) + 3 \sqrt{3} \cos(2\theta)$ If the point $P$ also lies on the plane $\Pi: 2x - y + kz = 15$ , then the value of $k$ is:

Options

A.
1
B.
2
C.
3
D.
4

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

Monotonicity and Boundedness|Lagrange Identity|Line-Plane intersection logic

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of 3D coordinate geometry and algebraic boundedness.

The primary trap lies in the equation $3|\vec{b}|^2 - (x+y+z)^2 = f(\theta)$ .

Students who fail to recognize the LHS as the expanded form of Lagrange's Identity ( $|\vec{a} \times \vec{b}|^2$ ) will face a grueling algebraic expansion.

Furthermore, the critical Advanced-level trap is ignoring the range of the functions.

By expressing the LHS as a quadratic in the line's parameter $s$ (minimum value 6) and the RHS as a single trigonometric function (maximum value 6), the problem forces a unique intersection point.

Students who blindly attempt to solve for $s$ without applying the boundedness principle will assume infinite solutions or extraneous intersection points, completely missing the geometric constraint.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Monotonicity and Boundedness
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free