JEE Main 2022 (27 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x + k y - 5 z = 1$ and $3 k x - k y + z = 5, k < 3$ and intercepts a unit length on positive x-axis, then the intercept made by the plane $P$ on the y-axis is

Q: Wait, why do we use the dot product of the normals to check for perpendicularity?

Great question! A plane is defined by its normal vector, which points directly out of the surface. If two planes are perpendicular, their normal vectors must also be perpendicular to each other. Since the dot product of two perpendicular vectors is always zero, $\vec{n_1} \cdot \vec{n_2} = 0$ is the most elegant way to capture that geometric relationship algebraically.

Q: Why is there a family of planes? Can't we just find one?

Think of the intersection of two planes as a hinge. Imagine a book spine; the two covers are your planes, and the spine is the line of intersection. You can rotate a third plane around that spine, creating infinitely many planes that pass through that same line. The parameter $\lambda$ acts as the 'angle' or 'position' variable that selects exactly which plane from that infinite set we are talking about.

Q: How do I know which $k$ value to pick when I get $k=1$ and $k=5$?

In JEE problems, constraints are your best friends! The problem explicitly states $k < 3$. This is a filter designed to test your attention to detail. Since $5$ is not less than $3$, we discard it and proceed with $k=1$.

Q: Is there a faster way to find the intercept than writing out the whole equation?

You could potentially use the intercept form of a plane, but because the plane is defined by the intersection of two others, the 'family of planes' method $(P_1 + \lambda P_2 = 0)$ is the most robust and least error-prone approach. It keeps all your variables organized in one place.

Q: What if the denominator becomes zero when I calculate the intercept?

That would imply the plane is parallel to the axis you are investigating! In this specific problem, the math works out cleanly, but always keep an eye on your denominators. If you ever hit a division by zero, it's a physical signal that the plane never crosses that axis (i.e., it's parallel to it).

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. You are dealing with a 'family of planes' passing through the intersection of two given planes. Before you can define this family, you must resolve the unknown parameter

k

using the condition of mutual perpendicularity between the two given planes.

Hint 2

Hint 2: Recall that two planes

A_{1} x + B_{1} y + C_{1} z + D_{1} = 0

and

A_{2} x + B_{2} y + C_{2} z + D_{2} = 0

are perpendicular if and only if the dot product of their normal vectors is zero:

A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2} = 0

. Use this to find

k

, keeping the constraint

k < 3

in mind.

Hint 3

Hint 3: Once you have the specific equations for the two planes, express the family of planes as

(P_{1}) + λ (P_{2}) = 0

. To find the

x

-intercept, set

y = 0

and

z = 0

in this combined equation and equate the resulting

x

value to

1

. Solve for

λ

, then repeat the process for the

y

-intercept by setting

x = 0

and

z = 0

Visualized Solution

The Geometric Setup

Given Planes:
$P_{1} : 2 x + k y - 5 z = 1$
$P_{2} : 3 k x - k y + z = 5$
Condition: $P_{1} ⊥ P_{2}$

Condition for Perpendicular Planes

For perpendicular planes, the dot product of their normal vectors is zero.
$n_{1} \cdot n_{2} = 0$
Normal $n_{1} = (2, k, - 5)$
Normal $n_{2} = (3 k, - k, 1)$

Applying the Dot Product

$n_{1} \cdot n_{2} = (2) (3 k) + (k) (- k) + (- 5) (1) = 0$

Solving for $k$

$6 k - k^{2} - 5 = 0$
Rearranging: $k^{2} - 6 k + 5 = 0$
Factoring: $(k - 1) (k - 5) = 0$
$k = 1$ or $k = 5$

Applying the Constraint

Given constraint: $k < 3$
Therefore, we must choose $k = 1$ .
Updated $P_{1} : 2 x + y - 5 z - 1 = 0$
Updated $P_{2} : 3 x - y + z - 5 = 0$

The Intersection Line and Plane $P$

The two planes intersect along a straight line.
A new plane $P$ passes through this exact intersection line.

Family of Planes

Equation of any plane passing through the intersection of $P_{1}$ and $P_{2}$ :
$P_{1} + λ P_{2} = 0$

Substituting the Plane Equations

$(2 x + y - 5 z - 1) + λ (3 x - y + z - 5) = 0$

Grouping the Variables

Grouping terms of $x, y$ , and $z$ :
$(2 + 3 λ) x + (1 - λ) y + (- 5 + λ) z = 1 + 5 λ$

The $x$ -intercept Condition

The plane $P$ intercepts the positive $x$ -axis at a unit length.
This means the $x$ -intercept is $1$ .
Point on the plane: $(1, 0, 0)$

Applying the $x$ -intercept

To find the $x$ -intercept, set $y = 0$ and $z = 0$ .
$x = \frac{1 + 5 λ}{2 + 3 λ}$
Given $x = 1$ :
$\frac{1 + 5 λ}{2 + 3 λ} = 1$

Solving for $λ$

$1 + 5 λ = 2 + 3 λ$
$5 λ - 3 λ = 2 - 1$
$2 λ = 1 ⟹ λ = \frac{1}{2}$

Finding the $y$ -intercept

We need the intercept made by plane $P$ on the $y$ -axis.
To find the $y$ -intercept, set $x = 0$ and $z = 0$ in the plane equation.
$y = \frac{1 + 5 λ}{1 - λ}$

Final Calculation

Substitute $λ = \frac{1}{2}$ :
$y = \frac{1 + 5 ( \frac{1}{2} )}{1 - \frac{1}{2}}$
$y = \frac{1 + 2.5}{0.5} = \frac{3.5}{0.5}$
$y = 7$

00:00 / 00:00

The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Perpendicularity Constraint

We begin with two planes,

P_{1} : 2 x + k y - 5 z = 1

and

P_{2} : 3 k x - k y + z = 5

. The problem states they are mutually perpendicular. In vector geometry, the normal vector

n = (A, B, C)

defines a plane's orientation.

For two planes to be perpendicular, their normal vectors must satisfy the dot product condition

n_{1} \cdot n_{2} = 0

. Calculating this, we obtain:

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

This simplifies to the quadratic equation

6 k - k^{2} - 5 = 0

, or equivalently,

k^{2} - 6 k + 5 = 0

. Factoring this expression yields

(k - 1) (k - 5) = 0

We have two candidates for

k

1

and

5

. However, the problem imposes the constraint

k < 3

. Consequently, we reject

5

and accept $k = 1$ .

Our planes are now defined as:

P_{1} : 2 x + y - 5 z - 1 = 0

P_{2} : 3 x - y + z - 5 = 0

The Family of Planes

Any plane passing through the line of intersection of

P_{1}

and

P_{2}

can be represented by the linear combination

P_{1} + λ P_{2} = 0

. This allows us to represent the entire family of planes using the parameter

λ

Substituting our specific equations, we get:

(2 x + y - 5 z - 1) + λ (3 x - y + z - 5) = 0

By grouping the terms, the structure of our target plane

P

emerges:

(2 + 3 λ) x + (1 - λ) y + (- 5 + λ) z = 1 + 5 λ

The Intercept and the Final Reveal

The plane

P

makes a unit intercept on the positive

x

-axis. This implies that when

y = 0

and

z = 0

, the value of

x

must be

1

Plugging these coordinates into our equation:

(2 + 3 λ) (1) = 1 + 5 λ

Solving for

λ

2 + 3 λ = 1 + 5 λ \Rightarrow 2 λ = 1 \Rightarrow λ = \frac{1}{2}

With

λ

determined, we find the

y

-intercept by setting

x = 0

and

z = 0

. The equation simplifies to

(1 - λ) y = 1 + 5 λ

Solving for

y

y = \frac{1 + 5 λ}{1 - λ}

Substituting

λ = \frac{1}{2}

y = \frac{1 + 5 ( \frac{1}{2} )}{1 - \frac{1}{2}} = \frac{1 + 2.5}{0.5} = \frac{3.5}{0.5} = 7

The final intercept on the

y

-axis is $7$ .

Common Traps & FAQs

Wait, why do we use the dot product of the normals to check for perpendicularity?

Great question! A plane is defined by its normal vector, which points directly out of the surface. If two planes are perpendicular, their normal vectors must also be perpendicular to each other. Since the dot product of two perpendicular vectors is always zero,

n_{1} \cdot n_{2} = 0

is the most elegant way to capture that geometric relationship algebraically.

Why is there a family of planes? Can't we just find one?

Think of the intersection of two planes as a hinge. Imagine a book spine; the two covers are your planes, and the spine is the line of intersection. You can rotate a third plane around that spine, creating infinitely many planes that pass through that same line. The parameter

λ

acts as the 'angle' or 'position' variable that selects exactly which plane from that infinite set we are talking about.

How do I know which

k

value to pick when I get

k = 1

and

k = 5

In JEE problems, constraints are your best friends! The problem explicitly states

k < 3

. This is a filter designed to test your attention to detail. Since

5

is not less than

3

, we discard it and proceed with

k = 1

Is there a faster way to find the intercept than writing out the whole equation?

You could potentially use the intercept form of a plane, but because the plane is defined by the intersection of two others, the 'family of planes' method

(P_{1} + λ P_{2} = 0)

is the most robust and least error-prone approach. It keeps all your variables organized in one place.

What if the denominator becomes zero when I calculate the intercept?

That would imply the plane is parallel to the axis you are investigating! In this specific problem, the math works out cleanly, but always keep an eye on your denominators. If you ever hit a division by zero, it's a physical signal that the plane never crosses that axis (i.e., it's parallel to it).

Select Answer:

Visualized Solution

The Geometric Setup

Condition for Perpendicular Planes

Applying the Dot Product

Solving for $k$

Applying the Constraint

The Intersection Line and Plane $P$

Family of Planes

Substituting the Plane Equations

Grouping the Variables

The $x$ -intercept Condition

Applying the $x$ -intercept

Solving for $λ$

Finding the $y$ -intercept

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Perpendicularity Constraint

The Family of Planes

The Intercept and the Final Reveal

Similar Questions

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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 through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

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Select Answer:

Visualized Solution

The Geometric Setup

Condition for Perpendicular Planes

Applying the Dot Product

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

Applying the Constraint

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

Family of Planes

Substituting the Plane Equations

Grouping the Variables

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

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

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

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

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Perpendicularity Constraint

The Family of Planes

The Intercept and the Final Reveal

Similar Questions

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Solving for $k$

The Intersection Line and Plane $P$

The $x$ -intercept Condition

Applying the $x$ -intercept

Solving for $λ$

Finding the $y$ -intercept

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 through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

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 :

Equation of a plane at a distance $\frac{2}{21}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2 x + y - 3 z + 4 = 0$ 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