JEE Advanced 2003

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: (i) Find the equation of the plane passing through the points $(2, 1, 0), (5, 0, 1)$ and $(4, 1, 1)$ . (ii) If $P$ is the point $(2, 1, 6)$ then find the point $Q$ such that $P Q$ is perpendicular to the plane in (i) and the midpoint of $P Q$ lies on it.

Q: Is there a direct formula for the image of a point? Can I just memorize that?

You could memorize the formula $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$, but in JEE Advanced, memorization is a trap. Understanding the derivation via the normal vector and midpoint ensures you can solve variations of the problem, such as when the plane is given in different forms or when you are asked for the foot of the perpendicular instead of the reflection.

Progressive Hints

Hint 1

Visualize the three points as vertices of a triangle floating in space. The plane is the unique flat surface containing this triangle. The determinant method is essentially a compact way of checking if the volume of the parallelepiped formed by vectors from one point to the others is zero, which forces them to be coplanar.

Hint 2

Remember that the coefficients of

x, y,

and

z

in the plane equation

a x + b y + cz = d

represent the components of the normal vector

n = (a, b, c)

. This vector is perpendicular to every line lying on the plane, which is the key to connecting the plane to the line segment

P Q

Hint 3

For the reflection

Q

of point

P

, use the property that the line segment

P Q

is parallel to the normal vector

n

. Express

Q

in terms of a parameter

λ

Q = P + λ n

, then force the midpoint of

P Q

to satisfy the plane equation to solve for

λ

Visualized Solution

Given Points on a Plane

Given points: $A (2, 1, 0)$ , $B (5, 0, 1)$ , and $C (4, 1, 1)$
Objective: Find the equation of the plane passing through $A, B,$ and $C$

Equation of a Plane (3 Points)

The equation of a plane passing through $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ , and $(x_{3}, y_{3}, z_{3})$ is given by a determinant:
$x - x_{1} x_{2} - x_{1} x_{3} - x_{1} y - y_{1} y_{2} - y_{1} y_{3} - y_{1} z - z_{1} z_{2} - z_{1} z_{3} - z_{1} = 0$

Substituting the Coordinates

Let $(x_{1}, y_{1}, z_{1}) = A (2, 1, 0)$
Substitute $A, B, C$ into the determinant:
$x - 2 5 - 2 4 - 2 y - 1 0 - 1 1 - 1 z - 0 1 - 0 1 - 0 = 0$

Simplifying the Determinant

Simplify the numerical values in the second and third rows:
$x - 2 32 y - 1 - 1 0 z 11 = 0$

Expanding the Determinant

Expanding along the first row:
$(x - 2) [(- 1) (1) - (0) (1)] - (y - 1) [(3) (1) - (2) (1)] + z [(3) (0) - (- 1) (2)] = 0$
$(x - 2) (- 1) - (y - 1) (1) + z (2) = 0$

Final Plane Equation

Distribute and simplify:
$- x + 2 - y + 1 + 2 z = 0$
$- x - y + 2 z + 3 = 0$
Equation of the plane: $x + y - 2 z = 3$

Introducing Point P and Image Q

Given point $P (2, 1, 6)$
Let $Q (α, β, γ)$ be the image of $P$ with respect to the plane.
The line segment $P Q$ is perpendicular to the plane.

Direction of Line PQ

The normal vector to the plane $x + y - 2 z = 3$ is $n = (1, 1, - 2)$ .
Since $P Q$ is perpendicular to the plane, it is parallel to $n$ .
Direction ratios of line $P Q$ are $(1, 1, - 2)$ .

Equation of Line PQ

Equation of line $P Q$ passing through $P (2, 1, 6)$ with direction $(1, 1, - 2)$ :
$\frac{x - 2}{1} = \frac{y - 1}{1} = \frac{z - 6}{- 2} = λ$

General Coordinates of Q

Express the coordinates of any point on the line in terms of $λ$ :
$x = λ + 2$
$y = λ + 1$
$z = - 2 λ + 6$
Let $Q = (λ + 2, λ + 1, - 2 λ + 6)$

Finding the Midpoint M

The midpoint $M$ of $P (2, 1, 6)$ and $Q (λ + 2, λ + 1, - 2 λ + 6)$ lies on the plane.
$M = (\frac{λ + 4}{2}, \frac{λ + 2}{2}, \frac{12 - 2 λ}{2})$

M Lies on the Plane

Substitute the coordinates of $M$ into the plane equation $x + y - 2 z = 3$ :
$(\frac{λ + 4}{2}) + (\frac{λ + 2}{2}) - 2 (\frac{12 - 2 λ}{2}) = 3$

Solving for $λ$

Multiply the entire equation by $2$ to clear denominators:
$(λ + 4) + (λ + 2) - 2 (12 - 2 λ) = 6$
$2 λ + 6 - 24 + 4 λ = 6$
$6 λ - 18 = 6 ⟹ 6 λ = 24 ⟹ λ = 4$

Calculating Coordinates of Q

Substitute $λ = 4$ back into the coordinates of $Q$ :
$α = 4 + 2 = 6$
$β = 4 + 1 = 5$
$γ = - 2 (4) + 6 = - 2$
Image Point Q: $(6, 5, - 2)$

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

To define the plane passing through points

A (2, 1, 0)

B (5, 0, 1)

, and

C (4, 1, 1)

, we utilize the property that the volume of the parallelepiped formed by vectors originating from

A

must be zero for the points to be coplanar. We set up the following determinant:

x - 2 5 - 2 4 - 2 y - 1 0 - 1 1 - 1 z - 0 1 - 0 1 - 0 = 0

Simplifying the entries within the determinant, we obtain:

x - 2 32 y - 1 - 1 0 z 11 = 0

The Master Equation

Expanding this determinant along the first row, we perform the calculation:

(x - 2) [(- 1) (1) - (0) (1)] - (y - 1) [(3) (1) - (2) (1)] + z [(3) (0) - (- 1) (2)] = 0

This simplifies to the linear expression:

- (x - 2) - (y - 1) + 2 z = 0

Rearranging the terms, we arrive at the equation of the plane:

- x + 2 - y + 1 + 2 z = 0 \Rightarrow x + y - 2z = 3

The Mirror in the Void

We now seek the reflection

Q

of point

P (2, 1, 6)

across the plane. The line

P Q

must be perpendicular to the plane, meaning its direction ratios are defined by the normal vector

n = (1, 1, - 2)

The equation of the line

P Q

passing through

P (2, 1, 6)

is:

\frac{x - 2}{1} = \frac{y - 1}{1} = \frac{z - 6}{- 2} = λ

Any point on this line can be expressed in terms of the parameter

λ

(λ + 2, λ + 1, - 2 λ + 6)

The Final Convergence

The midpoint

M

of the segment

P Q

must lie on the plane. The coordinates of

M

are:

M = (\frac{λ + 4}{2}, \frac{λ + 2}{2}, \frac{12 - 2 λ}{2})

Substituting these coordinates into the plane equation

x + y - 2 z = 3

(\frac{λ + 4}{2}) + (\frac{λ + 2}{2}) - 2 (\frac{12 - 2 λ}{2}) = 3

Multiplying by

2

to clear the denominators, we get:

(λ + 4) + (λ + 2) - 2 (12 - 2 λ) = 6

Solving for

λ

2 λ + 6 - 24 + 4 λ = 6 \Rightarrow 6 λ - 18 = 6 \Rightarrow 6 λ = 24 \Rightarrow λ = 4

Substituting

λ = 4

back into the expression for

Q

, we find

x = 4 + 2 = 6

y = 4 + 1 = 5

, and

z = - 2 (4) + 6 = - 2

. The reflection of point

P

across the plane is $Q (6, 5, - 2)$ .

Common Traps & FAQs

Why do we use the determinant method instead of finding two vectors and taking their cross product?

Both are perfectly valid! The determinant is just a compact, elegant way of writing the scalar triple product. It is often preferred in exams because it handles the coplanarity condition in one single, organized step, reducing the chance of algebraic errors that can occur when calculating cross products separately.

Why must the midpoint of

P Q

lie on the plane?

By the geometric definition of a reflection, the plane acts as the perpendicular bisector of the segment

P Q

. This means the plane cuts the segment

P Q

exactly in half. Therefore, the point where the line

P Q

intersects the plane is, by definition, the midpoint of the segment

P Q

Is there a direct formula for the image of a point? Can I just memorize that?

You could memorize the formula

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = - 2 \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}

, but in JEE Advanced, memorization is a trap. Understanding the derivation via the normal vector and midpoint ensures you can solve variations of the problem, such as when the plane is given in different forms or when you are asked for the foot of the perpendicular instead of the reflection.

Visualized Solution

Given Points on a Plane

Equation of a Plane (3 Points)

Substituting the Coordinates

Simplifying the Determinant

Expanding the Determinant

Final Plane Equation

Introducing Point P and Image Q

Direction of Line PQ

Equation of Line PQ

General Coordinates of Q

Finding the Midpoint M

M Lies on the Plane

Solving for $λ$

Calculating Coordinates of Q

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Equation

The Mirror in the Void

The Final Convergence

Similar Questions

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is:

The equation of the plane passing through the point $(1, 2, - 3)$ and perpendicular to the planes $3 x + y - 2 z = 5$ and $2 x - 5 y - z = 7$ , is:

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

Let $P$ be the plane, passing through the point $(1, - 1, - 5)$ and perpendicular to the line joining the points $(4, 1, - 3)$ and $(2, 4, 3)$ . Then the distance of $P$ from the point $(3, - 2, 2)$ is

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

A plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$ is

Visualized Solution

Given Points on a Plane

Equation of a Plane (3 Points)

Substituting the Coordinates

Simplifying the Determinant

Expanding the Determinant

Final Plane Equation

Introducing Point P and Image Q

Direction of Line PQ

Equation of Line PQ

General Coordinates of Q

Finding the Midpoint M

M Lies on the Plane

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Calculating Coordinates of Q

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Equation

The Mirror in the Void

The Final Convergence

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the plane containing the straight line 2x​=3y​=4z​ and perpendicular to the plane containing the straight lines 3x​=4y​=2z​ and 4x​=2y​=3z​ is

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Solving for $λ$

The equation of the plane passing through the point $(1, 2, - 3)$ and perpendicular to the planes $3 x + y - 2 z = 5$ and $2 x - 5 y - z = 7$ , is:

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

A plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$ is