JEE Main 2021 (27 Aug Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let $S$ be the mirror image of the point $Q (1, 3, 4)$ with respect to the plane $2 x - y + z + 3 = 0$ and let $R (3, 5, γ)$ be a point of this plane. Then the square of the length of the line segment $S R$ is .

Q: How do I know $R$ is on the plane?

The problem statement explicitly tells us that $R(3, 5, \gamma)$ is a point of the plane. In coordinate geometry, if a point lies on a surface, its coordinates must satisfy the equation of that surface. That is your golden ticket to finding the unknown variable $\gamma$.

Enter Numerical Value:

Visualized Solution

The Mirror Plane and Point $Q$

Given: A plane with equation $2 x - y + z + 3 = 0$ .
Object Point: $Q (1, 3, 4)$ is located in space above the plane.

The Mirror Image $S$

Image Point: $S$ is the exact mirror image of $Q$ with respect to the plane.
The line connecting $Q$ and $S$ is perpendicular to the plane.
The plane bisects the segment $QS$ .

Point $R$ on the Plane

Given: A point $R (3, 5, γ)$ lies exactly on the mirror plane.
We need to find the square of the distance $S R$ .

Condition for a Point on a Plane

Since $R (3, 5, γ)$ lies on the plane, its coordinates must satisfy the plane's equation.
Plane Equation: $2 x - y + z + 3 = 0$

Substituting $R$ into the Plane

Substitute $x = 3$ , $y = 5$ , and $z = γ$ into the equation.
$2 (3) - (5) + (γ) + 3 = 0$

Solving for $γ$

$6 - 5 + γ + 3 = 0$
$1 + γ + 3 = 0$
$4 + γ = 0 ⟹ γ = - 4$
So, $R$ is $(3, 5, - 4)$ .

The Geometric Shortcut

We need $S R^{2}$ . Finding coordinates of $S$ is a long process.
Symmetry Property: Any point on a mirror plane is equidistant from the object and its image.
Therefore, the distance $S R$ is exactly equal to the distance $QR$ .

Equating the Distances

$S R = QR$
Squaring both sides: $S R^{2} = Q R^{2}$
We just need to find the distance between $Q (1, 3, 4)$ and $R (3, 5, - 4)$ .

The Distance Formula

Distance Squared: $d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Points: $Q (1, 3, 4)$ and $R (3, 5, - 4)$

Substituting Coordinates

$Q R^{2} = (3 - 1)^{2} + (5 - 3)^{2} + (- 4 - 4)^{2}$

Evaluating the Differences

$x$ -difference: $3 - 1 = 2$
$y$ -difference: $5 - 3 = 2$
$z$ -difference: $- 4 - 4 = - 8$
$Q R^{2} = (2)^{2} + (2)^{2} + (- 8)^{2}$

Squaring the Terms

$(2)^{2} = 4$
$(2)^{2} = 4$
$(- 8)^{2} = 64$
$Q R^{2} = 4 + 4 + 64$

Final Summation

$Q R^{2} = 8 + 64$
$Q R^{2} = 72$

Concluding the Result

Since $S R^{2} = Q R^{2}$ , we have our final answer.
$S R^{2} = 72$
Pro Tip: Always look for symmetry in 3D geometry problems to save time!

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

We are given a plane defined by the equation

2 x - y + z + 3 = 0

. A point

Q (1, 3, 4)

is reflected across this plane to form an image point

S

Additionally, a point

R (3, 5, γ)

lies on the mirror plane. We are tasked with finding the value of

S R^{2}

The Geometric Insight

The mirror plane acts as the perpendicular bisector of the line segment

QS

. By the definition of a perpendicular bisector, any point

R

lying on the plane is equidistant from the object point

Q

and its image point

S

Therefore, the distance

S R

is equal to the distance

QR

. Consequently, the square of the distances must also be equal:

S R^{2} = Q R^{2}

This realization allows us to bypass the complex calculation of the coordinates of

S

entirely. We only need to determine the value of

γ

and calculate the distance between

Q

and

R

Finding the Unknown Coordinate

Since point

R (3, 5, γ)

lies on the plane

2 x - y + z + 3 = 0

, it must satisfy the plane's equation. We substitute the coordinates of

R

into the equation:

2 (3) - (5) + (γ) + 3 = 0

Simplifying the expression:

6 - 5 + γ + 3 = 0

4 + γ = 0

γ = - 4

Thus, the coordinates of point

R

are

(3, 5, - 4)

Final Calculation

Now that we have

Q (1, 3, 4)

and

R (3, 5, - 4)

, we apply the distance formula to find

Q R^{2}

Q R^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Substituting the known values:

Q R^{2} = (3 - 1)^{2} + (5 - 3)^{2} + (- 4 - 4)^{2}

Q R^{2} = (2)^{2} + (2)^{2} + (- 8)^{2}

Q R^{2} = 4 + 4 + 64

Summing these values, we obtain:

Q R^{2} = 72

Since

S R^{2} = Q R^{2}

, the final answer is 72.

Common Traps & FAQs

Wait, why do we not just find the coordinates of

S

using the line equation and plane intersection?

That is a classic trap! While you *could* find the coordinates of

S

by finding the line passing through

Q

perpendicular to the plane and then finding the intersection, it is a massive time sink. In JEE Advanced, time is your most precious resource. Recognizing the symmetry property

S R = QR

is the difference between a student who finishes the paper and one who runs out of time.

How do I know

R

is on the plane?

The problem statement explicitly tells us that

R (3, 5, γ)

is a point of the plane. In coordinate geometry, if a point lies on a surface, its coordinates must satisfy the equation of that surface. That is your golden ticket to finding the unknown variable

γ

Is the distance

S R

always equal to

QR

Yes, absolutely. By the definition of a mirror reflection, the plane acts as the perpendicular bisector of the segment connecting the object and its image. Therefore, any point on that plane is equidistant from the object and the image. This is a powerful geometric invariant you should keep in your toolkit.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $S$ be the mirror image of the point $Q (1, 3, 4)$ with respect to the plane $2 x - y + z + 3 = 0$ and let $R (3, 5, γ)$ be a point of this plane. Then the square of the length of the line segment $S R$ is .

Enter Numerical Value:

Visualized Solution

The Mirror Plane and Point $Q$

The Mirror Image $S$

Point $R$ on the Plane

Condition for a Point on a Plane

Substituting $R$ into the Plane

Solving for $γ$

The Geometric Shortcut

Equating the Distances

The Distance Formula

Substituting Coordinates

Evaluating the Differences

Squaring the Terms

Final Summation

Concluding the Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Geometric Insight

Finding the Unknown Coordinate

Final Calculation

Similar Questions

Let $α, β, γ, δ$ be real numbers such that $α^{2} + β^{2} + γ^{2} \neq = 0$ and $α + γ = 1$ . Suppose the point $(3, 2, - 1)$ is the mirror image of the point $(1, 0, - 1)$ with respect to the plane $α x + β y + γ z = δ$ . Then which of the following statements is/are TRUE?

If the mirror image of the point $(2, 4, 7)$ in the plane $3 x - y + 4 z = 2$ is $(a, b, c)$ , the $2 a + b + 2 c$ is equal to :

A plane $E$ is perpendicular to the two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $32$ , then $(P Q)^{2}$ is equal to

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3 x - 4 y + 12 z + 19 = 0$ be $(a - 6, β, γ)$ . If $a + b + c = 5$ , then $7 β - 9 γ$ is equal to ____.

Let the image of the point $P (1, 2, 3)$ in the plane $2 x - y + z = 9$ be $Q$ . If the coordinates of the point $R$ are $(6, 10, 7)$ , then the square of the area of the triangle $P QR$ is _______.

If $Q (0, - 1, - 3)$ is the image of the point $P$ in the plane $3 x - y + 4 z = 2$ and $R$ is the point $(3, - 1, - 2)$ , then the area (in sq. units) of $Δ P QR$ is :

Let the image of the point $P (2, - 1, 3)$ in the plane $x + 2 y - z = 0$ be $Q$ . Then the distance of the plane $3 x + 2 y + z + 29 = 0$ from the point $Q$ is

Let a plane $P$ pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to

Let the plane $2 x + 3 y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3 y + 5 z = 8$ . If the mirror image of the point $(2, - \frac{1}{2}, 2)$ in the rotated plane is $B (a, b, c)$ , then :

Enter Numerical Value:

Visualized Solution

The Mirror Plane and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

The Mirror Image .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R on the Plane

Condition for a Point on a Plane

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R into the Plane

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

The Geometric Shortcut

Equating the Distances

The Distance Formula

Substituting Coordinates

Evaluating the Differences

Squaring the Terms

Final Summation

Concluding the Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Geometric Insight

Finding the Unknown Coordinate

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mirror image of the point (2,4,7) in the plane 3x−y+4z=2 is (a,b,c), the 2a+b+2c is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the mirror image of the point (a,b,c) with respect to the plane 3x−4y+12z+19=0 be (a−6,β,γ). If a+b+c=5, then 7β−9γ is equal to ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the image of the point P(1,2,3) in the plane 2x−y+z=9 be Q. If the coordinates of the point R are (6,10,7), then the square of the area of the triangle PQR is _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Q(0,−1,−3) is the image of the point P in the plane 3x−y+4z=2 and R is the point (3,−1,−2), then the area (in sq. units) of ΔPQR is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the image of the point P(2,−1,3) in the plane x+2y−z=0 be Q. Then the distance of the plane 3x+2y+z+29=0 from the point Q is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a plane P pass through the point (3,7,−7) and contain the line, −3x−2​=2y−3​=1z+2​. If distance of the plane P from the origin is d, then d2 is equal to

The Mirror Plane and Point $Q$

The Mirror Image $S$

Point $R$ on the Plane

Substituting $R$ into the Plane

Solving for $γ$

If the mirror image of the point $(2, 4, 7)$ in the plane $3 x - y + 4 z = 2$ is $(a, b, c)$ , the $2 a + b + 2 c$ is equal to :

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3 x - 4 y + 12 z + 19 = 0$ be $(a - 6, β, γ)$ . If $a + b + c = 5$ , then $7 β - 9 γ$ is equal to ____.

Let the image of the point $P (1, 2, 3)$ in the plane $2 x - y + z = 9$ be $Q$ . If the coordinates of the point $R$ are $(6, 10, 7)$ , then the square of the area of the triangle $P QR$ is _______.

If $Q (0, - 1, - 3)$ is the image of the point $P$ in the plane $3 x - y + 4 z = 2$ and $R$ is the point $(3, - 1, - 2)$ , then the area (in sq. units) of $Δ P QR$ is :

Let the image of the point $P (2, - 1, 3)$ in the plane $x + 2 y - z = 0$ be $Q$ . Then the distance of the plane $3 x + 2 y + z + 29 = 0$ from the point $Q$ is

Let a plane $P$ pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to