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Three Dimensional Geometry: Mathematics | JEE Main

Q: Why is there a factor of -2 in the image formula but -1 in the foot of perpendicular formula?

The foot of perpendicular is at distance along the normal vector, whereas the image is at distance from the initial point along the same normal line.

Let the image of the point

P(2,-1,3)

in the plane

x + 2y - z = 0

Q

. Then the distance of the plane

3x + 2y + z + 29 = 0

from the point

Q

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{24\sqrt{2}}{7}

2\sqrt{14}

3\sqrt{14}

\frac{22\sqrt{2}}{7}

Step-by-step solutionView

Correct answer

The image of $P(2,-1,3)$ in the plane $x+2y-z=0$ is $Q(3,1,2)$ , and the perpendicular distance from $Q$ to the plane $3x+2y+z+29=0$ is $3\sqrt{14}$ .

Option analysis

Why each option works or fails

A ·

\frac{24\sqrt{2}}{7}

Dividing by the norm of the first plane's normal vector $\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}$ or making an arithmetic error when evaluating the perpendicular distance formula. Ensure you divide by the norm of the target plane's normal vector, $\sqrt{3^2+2^2+1^2} = \sqrt{14}$ , not a different plane's normal vector.

B ·

2\sqrt{14}

Calculating the distance from the original point $P(2,-1,3)$ to the second plane instead of using the reflected image point $Q$ . Re-read the question carefully: find the coordinates of the image point $Q$ first, then compute the distance from $Q$ , not from $P$ .

C ·

3\sqrt{14}

None. This is the correct option. Correct. Finding the reflection point gives $Q(3,1,2)$ , and applying the distance formula to $3x+2y+z+29=0$ yields $\frac{|3(3)+2(1)+2+29|}{\sqrt{3^2+2^2+1^2}} = \frac{42}{\sqrt{14}} = 3\sqrt{14}$ .

D ·

\frac{22\sqrt{2}}{7}

Making an algebraic slip when evaluating the numerator or incorrectly simplifying radicals with $\sqrt{14}$ . Verify numerator arithmetic: $9 + 2 + 2 + 29 = 42$ , and $\frac{42}{\sqrt{14}} = \frac{42\sqrt{14}}{14} = 3\sqrt{14}$ .

Reviewed route

Solution

StepWorking

01given

Point $P(2, -1, 3)$ , reflecting plane $\Pi_1: x + 2y - z = 0$ , target plane $\Pi_2: 3x + 2y + z + 29 = 0$ .

02goal

Find the perpendicular distance from the image point $Q$ of $P$ with respect to $\Pi_1$ to the plane $\Pi_2$ .

03approach

Use the standard reflection formula $\frac{\alpha-x_1}{a} = \frac{\beta-y_1}{b} = \frac{\gamma-z_1}{c} = \frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$ to find coordinates of $Q(\alpha, \beta, \gamma)$ , then apply the perpendicular distance formula $D = \frac{|a'\alpha + b'\beta + c'\gamma + d'|}{\sqrt{a'^2+b'^2+c'^2}}$ to plane $\Pi_2$ .

04execute

For point $P(2,-1,3)$ and plane $x+2y-z=0$ : $\frac{\alpha-2}{1} = \frac{\beta+1}{2} = \frac{\gamma-3}{-1} = \frac{-2(1(2) + 2(-1) - 1(3))}{1^2 + 2^2 + (-1)^2} = \frac{-2(-3)}{6} = 1$ . Thus, $\alpha = 2 + 1(1) = 3$ , $\beta = -1 + 2(1) = 1$ , $\gamma = 3 - 1(1) = 2$ . Point $Q$ is $(3, 1, 2)$ .

05execute

Compute the perpendicular distance from $Q(3,1,2)$ to plane $3x + 2y + z + 29 = 0$ : $D = \frac{|3(3) + 2(1) + 1(2) + 29|}{\sqrt{3^2 + 2^2 + 1^2}} = \frac{|9 + 2 + 2 + 29|}{\sqrt{14}} = \frac{42}{\sqrt{14}} = 3\sqrt{14}$ .

✓verify

Verify that the midpoint of $PQ$ is $(\frac{2+3}{2}, \frac{-1+1}{2}, \frac{3+2}{2}) = (2.5, 0, 2.5)$ , which satisfies $x + 2y - z = 2.5 + 0 - 2.5 = 0$ , confirming $Q$ is correctly the mirror image.

Hints that build this answer step by step

What formula gives the coordinates of the image $(x_2, y_2, z_2)$ of a point $(x_1, y_1, z_1)$ reflected in the plane $ax + by + cz + d = 0$ ?
$\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{z_2 - z_1}{c} = -2\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$
Using the reflection formula for $P(2,-1,3)$ and the plane $x + 2y - z = 0$ , what are the coordinates of $Q$ ?
$Q(3, 1, 2)$
What is the perpendicular distance from $Q(3, 1, 2)$ to the plane $3x + 2y + z + 29 = 0$ ?
$3\sqrt{14}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is there a factor of -2 in the image formula but -1 in the foot of perpendicular formula?

The foot of perpendicular is at distance $d$ along the normal vector, whereas the image is at distance $2d$ from the initial point along the same normal line.

Answer

The image of $P(2,-1,3)$ in the plane $x+2y-z=0$ is $Q(3,1,2)$ , and the perpendicular distance from $Q$ to the plane $3x+2y+z+29=0$ is $3\sqrt{14}$ .

Why each option works or fails

A: $\frac{24\sqrt{2}}{7}$ - Dividing by the norm of the first plane's normal vector $\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}$ or making an arithmetic error when evaluating the perpendicular distance formula. Ensure you divide by the norm of the target plane's normal vector, $\sqrt{3^2+2^2+1^2} = \sqrt{14}$ , not a different plane's normal vector.
B: $2\sqrt{14}$ - Calculating the distance from the original point $P(2,-1,3)$ to the second plane instead of using the reflected image point $Q$ . Re-read the question carefully: find the coordinates of the image point $Q$ first, then compute the distance from $Q$ , not from $P$ .
C · correct: $3\sqrt{14}$ - None. This is the correct option. Correct. Finding the reflection point gives $Q(3,1,2)$ , and applying the distance formula to $3x+2y+z+29=0$ yields $\frac{|3(3)+2(1)+2+29|}{\sqrt{3^2+2^2+1^2}} = \frac{42}{\sqrt{14}} = 3\sqrt{14}$ .
D: $\frac{22\sqrt{2}}{7}$ - Making an algebraic slip when evaluating the numerator or incorrectly simplifying radicals with $\sqrt{14}$ . Verify numerator arithmetic: $9 + 2 + 2 + 29 = 42$ , and $\frac{42}{\sqrt{14}} = \frac{42\sqrt{14}}{14} = 3\sqrt{14}$ .

Step-by-step solution

given: Point $P(2, -1, 3)$ , reflecting plane $\frac{24\sqrt{2}}{7}$ , target plane $\Pi_2: 3x + 2y + z + 29 = 0$ .
goal: Find the perpendicular distance from the image point $Q$ of $P$ with respect to $\Pi_1$ to the plane $\Pi_2$ .
approach: Use the standard reflection formula $\frac{\alpha-x_1}{a} = \frac{\beta-y_1}{b} = \frac{\gamma-z_1}{c} = \frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$ to find coordinates of $Q(\alpha, \beta, \gamma)$ , then apply the perpendicular distance formula $\sqrt{3^2+2^2+1^2} = \sqrt{14}$ to plane $\Pi_2$ .
execute: For point $P(2,-1,3)$ and plane $x+2y-z=0$ : $\frac{\alpha-2}{1} = \frac{\beta+1}{2} = \frac{\gamma-3}{-1} = \frac{-2(1(2) + 2(-1) - 1(3))}{1^2 + 2^2 + (-1)^2} = \frac{-2(-3)}{6} = 1$ . Thus, $\alpha = 2 + 1(1) = 3$ , $\beta = -1 + 2(1) = 1$ , $\gamma = 3 - 1(1) = 2$ . Point $Q$ is $(3, 1, 2)$ .
execute: Compute the perpendicular distance from $Q(3,1,2)$ to plane $3x + 2y + z + 29 = 0$ : $D = \frac{|3(3) + 2(1) + 1(2) + 29|}{\sqrt{3^2 + 2^2 + 1^2}} = \frac{|9 + 2 + 2 + 29|}{\sqrt{14}} = \frac{42}{\sqrt{14}} = 3\sqrt{14}$ .
verify: Verify that the midpoint of $PQ$ is $(\frac{2+3}{2}, \frac{-1+1}{2}, \frac{3+2}{2}) = (2.5, 0, 2.5)$ , which satisfies $x + 2y - z = 2.5 + 0 - 2.5 = 0$ , confirming $Q$ is correctly the mirror image.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The image of $P(2,-1,3)$ in the plane $x+2y-z=0$ is $Q(3,1,2)$ , and the perpendicular distance from $Q$ to the plane $3x+2y+z+29=0$ is $3\sqrt{14}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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