JEE Advanced 2022

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let $S$ be the reflection of a point $Q$ with respect to the plane given by $r = - (t + p) \hat{i} + t \hat{j} + (1 + p) \hat{k}$ where $t, p$ are real parameters and $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10 \hat{i} + 15 \hat{j} + 20 \hat{k}$ and $α \hat{i} + β \hat{j} + γ \hat{k}$ respectively, then which of the following is/are TRUE ?

Q: How do I know which vectors to cross product?

In the form $\vec{r} = \vec{a} + t\vec{v}_1 + p\vec{v}_2$, the vectors $\vec{v}_1$ and $\vec{v}_2$ are parallel to the plane. Any two non-parallel vectors on a plane define its orientation. Their cross product $\vec{v}_1 \times \vec{v}_2$ is guaranteed to be perpendicular to the plane, giving you the normal vector $\vec{n}$.

Q: Why is the constant in the reflection formula $-2$?

The reflection point $S$ is such that the plane is the perpendicular bisector of the segment $QS$. If the distance from $Q$ to the plane is $d$, the distance from the plane to $S$ is also $d$, but in the opposite direction. The factor $-2$ accounts for this displacement across the plane.

Select Answer:

* Multiple Correct

Visualized Solution

Visualizing the Setup

Given:
Point $Q (10, 15, 20)$
Plane: $r = - (t + p) \hat{i} + t \hat{j} + (1 + p) \hat{k}$
Let $S (α, β, γ)$ be the reflection of $Q$ .

Rearranging the Plane Equation

Expand the position vector $r$ :
$r = - t \hat{i} - p \hat{i} + t \hat{j} + \hat{k} + p \hat{k}$
Group by parameters $t$ and $p$ :
$r = \hat{k} + t (- \hat{i} + \hat{j}) + p (- \hat{i} + \hat{k})$

Identifying Vectors on the Plane

Compare with standard form: $r = a + t v_{1} + p v_{2}$
Position vector of a point on the plane: $a = \hat{k} ⟹ (0, 0, 1)$
First parallel vector: $v_{1} = - \hat{i} + \hat{j}$
Second parallel vector: $v_{2} = - \hat{i} + \hat{k}$

Finding the Normal Vector $n$

The normal vector $n$ is perpendicular to the plane.
It is the cross product of the two parallel vectors:
$n = v_{1} \times v_{2}$
$n = (- \hat{i} + \hat{j}) \times (- \hat{i} + \hat{k})$

Evaluating the Cross Product

Expand the cross product:
$n = (- \hat{i} \times - \hat{i}) + (- \hat{i} \times \hat{k}) + (\hat{j} \times - \hat{i}) + (\hat{j} \times \hat{k})$
Using cross product rules:
$n = 0 + \hat{j} + \hat{k} + \hat{i}$
$n = \hat{i} + \hat{j} + \hat{k}$

Cartesian Equation of the Plane

Point-normal form: $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$
Normal direction ratios: $(a, b, c) = (1, 1, 1)$
Point on plane: $(x_{1}, y_{1}, z_{1}) = (0, 0, 1)$
$1 (x - 0) + 1 (y - 0) + 1 (z - 1) = 0$
$x + y + z = 1$

Visualizing the Reflection $S$

Let $S (α, β, γ)$ be the reflection of $Q (10, 15, 20)$ .
The line segment $QS$ is perpendicular to the plane.
The plane bisects $QS$ at point $M$ .

The 3D Reflection Formula

For a point $(x_{1}, y_{1}, z_{1})$ and plane $a x + b y + cz + d = 0$ :
The reflection $(α, β, γ)$ is given by:
$\frac{α - x _{1}}{a} = \frac{β - y _{1}}{b} = \frac{γ - z _{1}}{c} = - 2 \frac{a x _{1} + b y _{1} + c z _{1} + d}{a ^{2} + b ^{2} + c ^{2}}$

Substituting the Values

Point $Q$ : $(x_{1}, y_{1}, z_{1}) = (10, 15, 20)$
Plane: $x + y + z - 1 = 0 ⟹ (a, b, c, d) = (1, 1, 1, - 1)$
$\frac{α - 10}{1} = \frac{β - 15}{1} = \frac{γ - 20}{1} = - 2 \frac{1 ( 10 ) + 1 ( 15 ) + 1 ( 20 ) - 1}{1 ^{2} + 1 ^{2} + 1 ^{2}}$

Evaluating the Right-Hand Side

Numerator: $10 + 15 + 20 - 1 = 44$
Denominator: $1^{2} + 1^{2} + 1^{2} = 3$
RHS $= - 2 (\frac{44}{3}) = - \frac{88}{3}$

Solving for $α$

Equating the first term to the RHS:
$α - 10 = - \frac{88}{3}$
$α = 10 - \frac{88}{3}$
$α = \frac{30 - 88}{3} = - \frac{58}{3}$

Solving for $β$ and $γ$

For $β$ : $β - 15 = - \frac{88}{3} ⟹ β = 15 - \frac{88}{3} = \frac{45 - 88}{3} = - \frac{43}{3}$
For $γ$ : $γ - 20 = - \frac{88}{3} ⟹ γ = 20 - \frac{88}{3} = \frac{60 - 88}{3} = - \frac{28}{3}$
So, $S \equiv (- \frac{58}{3}, - \frac{43}{3}, - \frac{28}{3})$

Checking the Given Options

Option A: $3 (α + β) = 3 (- \frac{58}{3} - \frac{43}{3}) = - 101$ (TRUE)
Option B: $3 (β + γ) = 3 (- \frac{43}{3} - \frac{28}{3}) = - 71$ (TRUE)
Option C: $3 (γ + α) = 3 (- \frac{28}{3} - \frac{58}{3}) = - 86$ (TRUE)
Option D: $3 (α + β + γ) = - 129 \neq = - 121$ (FALSE)

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, empty 3D space. You have a point

Q

(10, 15, 20)

, and you are tasked with finding its mirror image

S

across a mysterious plane. The plane is given by

r = - (t + p) \hat{i} + t \hat{j} + (1 + p) \hat{k}

At first glance, this looks like a jumble of parameters. However, we can rewrite the equation as:

r = \hat{k} + t (- \hat{i} + \hat{j}) + p (- \hat{i} + \hat{k})

This is the standard form

r = a + t v_{1} + p v_{2}

. Here,

a = (0, 0, 1)

is a point on the plane, while

v_{1} = (- 1, 1, 0)

and

v_{2} = (- 1, 0, 1)

are vectors parallel to the plane.

The Power of the Normal Vector

To find the Cartesian equation of the plane, we need the normal vector

n

, which defines the plane's orientation. We find it by taking the cross product of our two parallel vectors:

n = v_{1} \times v_{2}

Performing the cross product, we obtain:

n = \hat{i} - 1 - 1 \hat{j} 10 \hat{k} 01 = \hat{i} + \hat{j} + \hat{k}

This is a beautifully symmetric result. With the normal vector

(1, 1, 1)

and the point

(0, 0, 1)

, the Cartesian equation becomes

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

, which simplifies to:

x + y + z = 1

The Reflection Journey

Now, we want to find the reflection

S (α, β, γ)

Q (10, 15, 20)

. The line segment

QS

is perpendicular to the plane, and the plane bisects

QS

at its midpoint.

We use the standard reflection formula:

\frac{α - 10}{1} = \frac{β - 15}{1} = \frac{γ - 20}{1} = - 2 \frac{1 ( 10 ) + 1 ( 15 ) + 1 ( 20 ) - 1}{1 ^{2} + 1 ^{2} + 1 ^{2}}

The right-hand side simplifies as follows:

- 2 \times \frac{44}{3} = - \frac{88}{3}

Final Calculation

Now, we solve for each coordinate individually:

α = 10 - \frac{88}{3} = - \frac{58}{3}

β = 15 - \frac{88}{3} = - \frac{43}{3}

γ = 20 - \frac{88}{3} = - \frac{28}{3}

We have found the reflection point $S (- \frac{58}{3}, - \frac{43}{3}, - \frac{28}{3})$ . This problem serves as a masterclass in transforming a complex vector representation into a simple, elegant geometric reality.

Common Traps & FAQs

Why can't we just use the parametric form directly to find the reflection?

The reflection formula requires the normal vector of the plane. The parametric form gives us the plane's orientation via two parallel vectors, but it doesn't give the normal directly. We must convert it to Cartesian form

a x + b y + cz + d = 0

to use the standard reflection formula.

How do I know which vectors to cross product?

In the form

r = a + t v_{1} + p v_{2}

, the vectors

v_{1}

and

v_{2}

are parallel to the plane. Any two non-parallel vectors on a plane define its orientation. Their cross product

v_{1} \times v_{2}

is guaranteed to be perpendicular to the plane, giving you the normal vector

n

Why is the constant in the reflection formula

- 2

The reflection point

S

is such that the plane is the perpendicular bisector of the segment

QS

. If the distance from

Q

to the plane is

d

, the distance from the plane to

S

is also

d

, but in the opposite direction. The factor

- 2

accounts for this displacement across the plane.

Select Answer:

Visualized Solution

Visualizing the Setup

Rearranging the Plane Equation

Identifying Vectors on the Plane

Finding the Normal Vector $n$

Evaluating the Cross Product

Cartesian Equation of the Plane

Visualizing the Reflection $S$

The 3D Reflection Formula

Substituting the Values

Evaluating the Right-Hand Side

Solving for $α$

Solving for $β$ and $γ$

Checking the Given Options

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Normal Vector

The Reflection Journey

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?

Let a unit vector $O P$ make angle $α, β, γ$ with the positive directions of the co-ordinate axes $O X, O Y, O Z$ respectively, where $β \in (0, π /2)$ . $O P$ is perpendicular to the plane through points $(1, 2, 3)$ , $(2, 3, 4)$ and $(1, 5, 7)$ , then which one of the following is true ?

Let $Q$ be the mirror image of the point $P (1, 2, 1)$ with respect to the plane $x + 2 y + 2 z = 16$ . Let $T$ be a plane passing through the point $Q$ and contains the line $r = - \hat{k} + λ (\hat{i} + \hat{j} + 2 \hat{k}), λ \in R$ . Then, which of the following points lies on $T$ ?

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 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 :

Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then $3 (α + β + γ)$ equals

List-I

List-II

Match the following :

List-I

List-II

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 .

Select Answer:

Visualized Solution

Visualizing the Setup

Rearranging the Plane Equation

Identifying Vectors on the Plane

Finding the Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Evaluating the Cross Product

Cartesian Equation of the Plane

Visualizing the Reflection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The 3D Reflection Formula

Substituting the Values

Evaluating the Right-Hand Side

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

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

Checking the Given Options

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Normal Vector

The Reflection Journey

Final Calculation

Similar Questions

.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 ____.

List-I

List-II

Match the following :

List-I

List-II

Finding the Normal Vector $n$

Visualizing the Reflection $S$

Solving for $α$

Solving for $β$ and $γ$

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 ____.