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Three Dimensional Geometry: Mathematics | JEE Main

Let the image of the point

\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)

in the plane

\mathrm{x}-2\mathrm{y}+\mathrm{z}-2=0

\mathrm{P}

. If the distance of the point

\mathrm{Q}(6,-2, \alpha), \alpha>0

, from

\mathrm{P}

13

, then

\alpha

is equal to

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Step-by-step solutionView

Correct answer

The image point $P$ is $(2, 1, 3)$ , and using the distance formula with $Q(6, -2, \alpha)$ and $\alpha > 0$ gives $\alpha = 15$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given point $A\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ , plane $\Pi: x - 2y + z - 2 = 0$ , point $Q(6, -2, \alpha)$ with $\alpha > 0$ , and distance $PQ = 13$ where $P$ is the reflection of $A$ in $\Pi$ .

02goal

Find the value of the positive parameter $\alpha$ .

03approach

Use the standard reflection formula for a point $(x_1, y_1, z_1)$ across a plane $ax + by + cz + d = 0$ : $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$ After finding $P(x, y, z)$ , apply the 3D distance formula $PQ^2 = 13^2$ and solve for $\alpha > 0$ .

04execute

Calculate the scaling factor $k$ for the image: $k = \frac{-2\left(1\cdot\frac{5}{3} - 2\cdot\frac{5}{3} + 1\cdot\frac{8}{3} - 2\right)}{1^2 + (-2)^2 + 1^2} = \frac{-2\left(\frac{5 - 10 + 8 - 6}{3}\right)}{6} = \frac{-2\left(-\frac{3}{3}\right)}{6} = \frac{2}{6} = \frac{1}{3}$ Now find coordinates of $P(x, y, z)$ : $x = \frac{5}{3} + 1\left(\frac{1}{3}\right) = 2$ $y = \frac{5}{3} - 2\left(\frac{1}{3}\right) = 1$ $z = \frac{8}{3} + 1\left(\frac{1}{3}\right) = 3$ So $P = (2, 1, 3)$ .

05execute

Apply the distance formula between $P(2, 1, 3)$ and $Q(6, -2, \alpha)$ : $PQ^2 = (6 - 2)^2 + (-2 - 1)^2 + (\alpha - 3)^2 = 13^2$ $4^2 + (-3)^2 + (\alpha - 3)^2 = 169$ $16 + 9 + (\alpha - 3)^2 = 169$ $(\alpha - 3)^2 = 169 - 25 = 144$ Since $\alpha > 0$ : $\alpha - 3 = 12 \implies \alpha = 15$

✓verify

Check: for $P(2,1,3)$ , midpoint of $A\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ and $P(2,1,3)$ is $M\left(\frac{11}{6}, \frac{8}{6}, \frac{17}{6}\right)$ . Substitute $M$ into plane: $\frac{11}{6} - 2\left(\frac{8}{6}\right) + \frac{17}{6} - 2 = \frac{11 - 16 + 17 - 12}{6} = 0$ , which lies on the plane. Distance $PQ = \sqrt{(6-2)^2 + (-2-1)^2 + (15-3)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$ . All constraints verified.

Hints that build this answer step by step

What is the standard formula for finding the image $(x_1, y_1, z_1)$ of a point $(x_0, y_0, z_0)$ in the plane $ax + by + cz + d = 0$ ?
$\frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = \frac{z_1 - z_0}{c} = \frac{-2(ax_0 + by_0 + cz_0 + d)}{a^2 + b^2 + c^2}$
Applying the image formula to $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ and the plane $x - 2y + z - 2 = 0$ , what are the coordinates of $P$ ?
$P(2, 1, 3)$
Given the distance $PQ = 13$ with $P(2, 1, 3)$ and $Q(6, -2, \alpha)$ where $\alpha > 0$ , what is the value of $\alpha$ ?
$\alpha = 15$

✓ Source and academic review

Question type: Numerical
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 the factor -2 in the image formula instead of -1?

The foot of the perpendicular from a point to a plane is the midpoint between the point and its image. Therefore, the vector to the image is double the displacement vector to the foot. This gives the factor of -2 instead of -1.

Answer

The image point $P$ is $(2, 1, 3)$ , and using the distance formula with $Q(6, -2, \alpha)$ and $\alpha > 0$ gives $\alpha = 15$ .

Why each option works or fails

Step-by-step solution

given: Given point $A\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ , plane $\Pi: x - 2y + z - 2 = 0$ , point $Q(6, -2, \alpha)$ with $\alpha > 0$ , and distance $PQ = 13$ where $P$ is the reflection of $A$ in $\Pi$ .
goal: Find the value of the positive parameter $\alpha$ .
approach: Use the standard reflection formula for a point $(x_1, y_1, z_1)$ across a plane $ax + by + cz + d = 0$ : $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$ After finding $P(x, y, z)$ , apply the 3D distance formula $PQ^2 = 13^2$ and solve for $\alpha > 0$ .
execute: Calculate the scaling factor $k$ for the image: $k = \frac{-2\left(1\cdot\frac{5}{3} - 2\cdot\frac{5}{3} + 1\cdot\frac{8}{3} - 2\right)}{1^2 + (-2)^2 + 1^2} = \frac{-2\left(\frac{5 - 10 + 8 - 6}{3}\right)}{6} = \frac{-2\left(-\frac{3}{3}\right)}{6} = \frac{2}{6} = \frac{1}{3}$ Now find coordinates of $P(x, y, z)$ : $x = \frac{5}{3} + 1\left(\frac{1}{3}\right) = 2$ $y = \frac{5}{3} - 2\left(\frac{1}{3}\right) = 1$ $z = \frac{8}{3} + 1\left(\frac{1}{3}\right) = 3$ So $P = (2, 1, 3)$ .
execute: Apply the distance formula between $P(2, 1, 3)$ and $Q(6, -2, \alpha)$ : $PQ^2 = (6 - 2)^2 + (-2 - 1)^2 + (\alpha - 3)^2 = 13^2$ $4^2 + (-3)^2 + (\alpha - 3)^2 = 169$ $16 + 9 + (\alpha - 3)^2 = 169$ $(\alpha - 3)^2 = 169 - 25 = 144$ Since $\alpha > 0$ : $\alpha - 3 = 12 \implies \alpha = 15$
verify: Check: for $P(2,1,3)$ , midpoint of $A\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ and $P(2,1,3)$ is $M\left(\frac{11}{6}, \frac{8}{6}, \frac{17}{6}\right)$ . Substitute $M$ into plane: $\frac{11}{6} - 2\left(\frac{8}{6}\right) + \frac{17}{6} - 2 = \frac{11 - 16 + 17 - 12}{6} = 0$ , which lies on the plane. Distance $PQ = \sqrt{(6-2)^2 + (-2-1)^2 + (15-3)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$ . All constraints verified.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The image point $P$ is $(2, 1, 3)$ , and using the distance formula with $Q(6, -2, \alpha)$ and $\alpha > 0$ gives $\alpha = 15$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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