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Question #40844

Difficulty: 8/10 | 🟢 Free

The Question

Let $P$ be a plane satisfying the equation $|x + 2y - 2z + 2| = 7$ . The normal vector to $P$ drawn from the origin to the plane points into the octant where $x > 0$ , $y > 0$ , and $z < 0$ . Find the coordinates of the image of the point $A(1, -2, 2)$ in the plane $P$ .

Options

A.
$\left( \frac{11}{3}, \frac{10}{3}, -\frac{10}{3} \right)$
B.
$\left( \frac{5}{9}, -\frac{26}{9}, \frac{26}{9} \right)$
C.
$\left( \frac{7}{3}, \frac{2}{3}, -\frac{2}{3} \right)$
D.
$\left( -\frac{5}{3}, -\frac{22}{3}, \frac{22}{3} \right)$

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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

Octant Sign Convention|Image of a point in a plane|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Modulus Trap' combined with the 'Octant Sign Convention'.

The modulus in the plane equation generates two parallel planes.

Students must use the directional signs of the normal vector from the origin to correctly identify the intended plane.

Failing to verify the octant condition leads to selecting the extraneous plane P_2, resulting in the trap answer B.

Additionally, forgetting the multiplier of 2 in the reflection formula yields the foot of the perpendicular (trap answer C).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Octant Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free