JEE Main 2022 (25 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let $P$ be the plane passing through the intersection of the planes $r \cdot (\hat{i} + 3 \hat{j} - \hat{k}) = 5$ and $r \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 3$ , and the point $(2, 1, - 2)$ . Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2 \hat{j} + 4 \hat{k}$ and $5 \hat{i} - \hat{j} + 2 \hat{k}$ respectively. Then the points

Q: How do I know if the points are on the same side or opposite sides?

It is all about the sign of the expression $ax + by + cz + d$. If you plug in the coordinates of a point and get a positive value, it is on one side. If you get a negative value, it is on the other. If the signs are different, the plane must lie between the two points, meaning they are on opposite sides.

Select Answer:

Visualized Solution

Visualizing the Intersection

Plane $P_{1} : r \cdot (\hat{i} + 3 \hat{j} - \hat{k}) = 5 \Rightarrow x + 3 y - z - 5 = 0$
Plane $P_{2} : r \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 3 \Rightarrow 2 x - y + z - 3 = 0$
Target Point: $(2, 1, - 2)$

The Family of Planes Concept

Equation of family of planes: $P_{1} + λ P_{2} = 0$
$(x + 3 y - z - 5) + λ (2 x - y + z - 3) = 0$

Applying the Point Constraint

Substitute $(x, y, z) = (2, 1, - 2)$ into the family equation:
$(2 + 3 (1) - (- 2) - 5) + λ (2 (2) - 1 + (- 2) - 3) = 0$

Solving for $λ$

$(2 + 3 + 2 - 5) + λ (4 - 1 - 2 - 3) = 0$
$2 + λ (- 2) = 0 \Rightarrow 2 λ = 2$
$λ = 1$

Final Equation of Plane $P$

Substitute $λ = 1$ into $(x + 3 y - z - 5) + λ (2 x - y + z - 3) = 0$ :
$(x + 2 x) + (3 y - y) + (- z + z) - (5 + 3) = 0$
Final Equation of Plane $P : 3 x + 2 y - 8 = 0$

The Side of Plane Test

Define function $f (x, y, z) = 3 x + 2 y - 8$
If $f (X) \cdot f (Y) < 0$ , points are on opposite sides.
If $f (X) \cdot f (Y) > 0$ , points are on the same side.

Evaluating Point $X$

Point $X$ position vector: $\hat{i} - 2 \hat{j} + 4 \hat{k} \Rightarrow X (1, - 2, 4)$
$f (X) = 3 (1) + 2 (- 2) - 8$
$f (X) = 3 - 4 - 8 = - 9$ (Negative)

Evaluating Point $Y$

Point $Y$ position vector: $5 \hat{i} - \hat{j} + 2 \hat{k} \Rightarrow Y (5, - 1, 2)$
$f (Y) = 3 (5) + 2 (- 1) - 8$
$f (Y) = 15 - 2 - 8 = 5$ (Positive)

Final Conclusion

$f (X) = - 9 < 0$
$f (Y) = 5 > 0$
Since $f (X)$ and $f (Y)$ have opposite signs, $X$ and $Y$ are on opposite sides of $P$ .
Correct Option: (3)

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

We begin by translating the given vector equations of the planes into Cartesian coordinates. The first plane,

r \cdot (\hat{i} + 3 \hat{j} - \hat{k}) = 5

, is expressed as:

x + 3 y - z - 5 = 0

The second plane,

r \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 3

, is expressed as:

2 x - y + z - 3 = 0

The Master Equation

To find the plane passing through the line of intersection of these two planes, we utilize the Family of Planes concept. Any such plane can be represented by the equation:

(x + 3 y - z - 5) + λ (2 x - y + z - 3) = 0

We are given that this plane must pass through the point

(2, 1, - 2)

. Substituting these coordinates into our master equation, we obtain:

(2 + 3 (1) - (- 2) - 5) + λ (2 (2) - 1 + (- 2) - 3) = 0

Solving for the Plane

Simplifying the expression above, we get:

2 + λ (- 2) = 0

This yields the value

λ = 1

. Substituting

λ = 1

back into the family equation, we combine the terms:

(x + 2 x) + (3 y - y) + (- z + z) - (5 + 3) = 0

This simplifies to the final equation of the plane:

3 x + 2 y - 8 = 0

The Side of the Plane Test

To determine the relative positions of points

X (1, - 2, 4)

and

Y (5, - 1, 2)

, we define the function

f (x, y, z) = 3 x + 2 y - 8

For point

X (1, - 2, 4)

f (X) = 3 (1) + 2 (- 2) - 8 = 3 - 4 - 8 = - 9

For point

Y (5, - 1, 2)

f (Y) = 3 (5) + 2 (- 1) - 8 = 15 - 2 - 8 = 5

Since

f (X)

is negative and

f (Y)

is positive, the points

X

and

Y

lie on opposite sides of the plane.

Common Traps & FAQs

Wait, why do we use the family of planes equation

P_{1} + λ P_{2} = 0

Think of the line of intersection as the binding of a book. The two given planes are two pages meeting at that binding. Any other plane passing through that same line is just another page in the same book. The parameter

λ

allows us to rotate through all possible planes that share that specific line of intersection.

Does it matter if I write

P_{1} + λ P_{2} = 0

P_{2} + λ P_{1} = 0

Mathematically, it does not matter! You will simply get a different value for

λ

, but the final equation of the plane will be identical. The family of planes is defined by the set of all linear combinations, so as long as you account for the constraint, the result is unique.

How do I know if the points are on the same side or opposite sides?

It is all about the sign of the expression

a x + b y + cz + d

. If you plug in the coordinates of a point and get a positive value, it is on one side. If you get a negative value, it is on the other. If the signs are different, the plane must lie between the two points, meaning they are on opposite sides.

Select Answer:

Visualized Solution

Visualizing the Intersection

The Family of Planes Concept

Applying the Point Constraint

Solving for $λ$

Final Equation of Plane $P$

The Side of Plane Test

Evaluating Point $X$

Evaluating Point $Y$

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Equation

Solving for the Plane

The Side of the Plane Test

Similar Questions

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

The vector equation of the plane passing through the intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (\hat{i} - 2 \hat{j}) = - 2$ , and the point $(1, 0, 2)$ 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 :

The equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis is:

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points ?

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

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + 4 \hat{k}) = 16$ and $r \cdot (- \hat{i} + \hat{j} + \hat{k}) = 6$ . Then which of the following points does NOT lie on $P$ ?

Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ and $P 2 : 2 x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2 λ, λ, - λ)$ from the plane $P 2$ is

Select Answer:

Visualized Solution

Visualizing the Intersection

The Family of Planes Concept

Applying the Point Constraint

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

Final Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

The Side of Plane Test

Evaluating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Evaluating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Equation

Solving for the Plane

The Side of the Plane Test

Similar Questions

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vector equation of the plane passing through the intersection of the planes r⋅(i^+j^​+k^)=1 and r⋅(i^−2j^​)=−2, and the point (1,0,2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vector equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x−y+z=0 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane passing through the line of intersection of the planes r⋅(i^+j^​+k^)=1 and r⋅(2i^+3j^​−k^)+4=0 and parallel to the x-axis is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane containing the line 2x−3​=−1y+2​=3z−1​ and also containing its projection on the plane 2x+3y−z=5, contains which one of the following points ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the plane passing through the point (−1,0,−2) and perpendicular to each of the planes 2x+y−z=2 and x−y−z=3 be ax+by+cz+8=0. Then the value of a+b+c is equal to:

Solving for $λ$

Final Equation of Plane $P$

Evaluating Point $X$

Evaluating Point $Y$

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

The vector equation of the plane passing through the intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (\hat{i} - 2 \hat{j}) = - 2$ , and the point $(1, 0, 2)$ 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 :

The equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis is:

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points ?

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to: