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Question #41653

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the locus of a complex number $z$ in the Argand plane satisfying $|z - 2| = |z - 2i|$ . Let $P$ be the plane in $\mathbb{R}^3$ containing the $z$ -axis and the locus $S$ (where the Argand plane is identified with the $xy$ -plane). Let $L$ be the line of intersection of the planes $2x - y + z = 2$ and $x + 2y - z = 1$ . If $L'$ is the orthogonal projection of the line $L$ onto the plane $P$ , then which of the following points lies on $L'$ ?

Options

A.
$\left(1, 1, \frac{5}{2}\right)$
B.
$\left(0, 0, \frac{5}{2}\right)$
C.
$\left(\frac{3}{2}, \frac{3}{2}, \frac{15}{2}\right)$
D.
$\left(-\frac{1}{2}, \frac{1}{2}, \frac{5}{2}\right)$

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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

Orthogonal projection of a line|Unit Vector Normalization|Locus Problems via Transformation|Symmetric vs Non-symmetric form

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to synthesize complex numbers with 3D geometry.

The first trap is the locus condition: students must correctly identify the locus of $|z - 2| = |z - 2i|$ as the line $x = y$ in the $xy$ -plane and extend it to a 3D plane containing the $z$ -axis, yielding $x - y = 0$ .

A common mistake is a sign error leading to $x + y = 0$ , which is tested by distractor D.

The second trap involves converting a non-symmetric line (intersection of two planes) into symmetric form by finding the cross product of their normals to get the direction vector.

Finally, the student must orthogonally project both a point on the line and its direction vector onto the plane using vector normalization, requiring careful application of the projection formula $\vec{v}' = \vec{v} - (\vec{v} \cdot \hat{n})\hat{n}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Orthogonal projection of a line
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free