JEE Main 2021 (26 February Shift 1)

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Animated Solution for Mathematics - Three Dimensional Geometry: Let $(λ, 2, 1)$ be a point on the plane which passes through the point $(4, - 2, 2)$ . If the plane is perpendicular to the line joining the point $(- 2, - 21, 29)$ and $(- 1, - 16, 23)$ , then $(\frac{λ}{11})^{2} - \frac{4 λ}{11} - 4$ is equal to

Q: I keep getting the signs wrong in the plane equation. How do I avoid this?

It is a common trap! Always write the formula $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$ clearly before substituting. When you plug in negative coordinates, use parentheses, like $y - (-2)$, to ensure you don't accidentally turn a subtraction into an addition. Slowing down for that one step will save you from a lot of frustration.

Q: Can I use the dot product to find the plane equation?

Absolutely! The equation of a plane can also be written as $\vec{r} \cdot \vec{n} = \vec{A} \cdot \vec{n}$, where $\vec{r} = (x, y, z)$ and $\vec{A}$ is the position vector of the point on the plane. Both methods are mathematically identical. The point-normal form is just a more expanded, algebraic way of looking at the same dot product relationship.

Enter Numerical Value:

Visualized Solution

Visualizing the Plane

Let the required plane be $π$ .
The plane passes through the given point $A (4, - 2, 2)$ .

The Perpendicular Line

A line passes through points $B (- 2, - 21, 29)$ and $C (- 1, - 16, 23)$ .
The plane $π$ is strictly perpendicular to this line.

Identifying the Normal Vector

The normal vector $n$ of a plane is perpendicular to the plane.
Since line $B C$ is perpendicular to the plane, $n$ is parallel to $B C$ .

Calculating Vector $B C$

$n = B C = (x_{C} - x_{B}) \hat{i} + (y_{C} - y_{B}) \hat{j} + (z_{C} - z_{B}) \hat{k}$
$n = (- 1 - (- 2)) \hat{i} + (- 16 - (- 21)) \hat{j} + (23 - 29) \hat{k}$
$n = \hat{i} + 5 \hat{j} - 6 \hat{k}$

Equation of a Plane

The general equation of a plane passing through $(x_{1}, y_{1}, z_{1})$ with normal direction ratios $(a, b, c)$ is:
$a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$

Substituting Known Values

Point $A (4, - 2, 2) ⟹ x_{1} = 4, y_{1} = - 2, z_{1} = 2$
Normal $n = (1, 5, - 6) ⟹ a = 1, b = 5, c = - 6$
$1 (x - 4) + 5 (y - (- 2)) - 6 (z - 2) = 0$

Expanding the Equation

Expand the brackets:
$x - 4 + 5 (y + 2) - 6 z + 12 = 0$
$x - 4 + 5 y + 10 - 6 z + 12 = 0$

Final Plane Equation

Group the variables and constants:
$x + 5 y - 6 z + (- 4 + 10 + 12) = 0$
$x + 5 y - 6 z + 18 = 0$

Introducing Point P

The problem states that point $P (λ, 2, 1)$ lies on this plane.

Satisfying the Plane Equation

Since $P (λ, 2, 1)$ is on the plane, it must satisfy $x + 5 y - 6 z + 18 = 0$ .
Substitute $x = λ, y = 2, z = 1$ :
$λ + 5 (2) - 6 (1) + 18 = 0$

Solving for $λ$

$λ + 10 - 6 + 18 = 0$
$λ + 22 = 0$
$λ = - 22$

The Target Expression

We need to evaluate the expression:
$(\frac{λ}{11})^{2} - \frac{4 λ}{11} - 4$

Substituting $λ$

Substitute $λ = - 22$ :
$(\frac{- 22}{11})^{2} - \frac{4 ( - 22 )}{11} - 4$
Notice that $\frac{- 22}{11} = - 2$

Final Calculation

Replace $\frac{λ}{11}$ with $- 2$ :
$(- 2)^{2} - 4 (- 2) - 4$
$4 + 8 - 4$
$= 8$

Conclusion

Key Takeaway: The direction ratios of a line perpendicular to a plane are exactly the direction ratios of the plane's normal.
Final Answer: $8$

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. Before you lies a flat, infinite surface—a plane, which we shall call

π

This plane is anchored to a specific point

A (4, - 2, 2)

. This point serves as our gateway and anchor in the coordinate system.

The Perpendicular Connection

The problem introduces a line formed by joining two points:

B (- 2, - 21, 29)

and

C (- 1, - 16, 23)

. We are given that our plane

π

is strictly perpendicular to this line.

If a line is perpendicular to a plane, the line's direction vector is parallel to the plane's normal vector

n

. The normal vector dictates the tilt and orientation of the plane.

We calculate the direction vector

n = B C

as follows:

n = (x_{C} - x_{B}) \hat{i} + (y_{C} - y_{B}) \hat{j} + (z_{C} - z_{B}) \hat{k}

Substituting the given coordinates:

n = (- 1 - (- 2)) \hat{i} + (- 16 - (- 21)) \hat{j} + (23 - 29) \hat{k}

This simplifies to:

n = 1 \hat{i} + 5 \hat{j} - 6 \hat{k}

Constructing the Equation

With the normal vector

n = (1, 5, - 6)

and the point

A (4, - 2, 2)

on the plane, we use the standard point-normal form:

a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0

Substituting our values:

1 (x - 4) + 5 (y - (- 2)) - 6 (z - 2) = 0

Expanding this expression:

x - 4 + 5 (y + 2) - 6 (z - 2) = 0

x - 4 + 5 y + 10 - 6 z + 12 = 0

Grouping the constants, we find the final equation of the plane:

x + 5 y - 6 z + 18 = 0

The Intersection of Point and Plane

The problem states that a point

P (λ, 2, 1)

lies on this plane. Therefore, the coordinates of

P

must satisfy the equation of the plane.

Substituting

x = λ

y = 2

, and

z = 1

λ + 5 (2) - 6 (1) + 18 = 0

λ + 10 - 6 + 18 = 0

λ + 22 = 0 \Rightarrow λ = - 22

Final Calculation

We must evaluate the expression:

(\frac{λ}{11})^{2} - \frac{4 λ}{11} - 4

Given

λ = - 22

, the term

\frac{λ}{11}

becomes:

\frac{- 22}{11} = - 2

Substituting this into the expression:

(- 2)^{2} - 4 (- 2) - 4

4 + 8 - 4 = 8

The final answer is 8.

Common Traps & FAQs

Wait, why is the normal vector of the plane the same as the direction vector of the line?

That is a fantastic question! Think of it this way: the normal vector

n

is defined as a vector perpendicular to the entire surface of the plane. If a line is also perpendicular to that same plane, it must be parallel to the normal vector. Since they share the same direction, their direction ratios are proportional, and we can use the line's direction vector directly as our normal vector.

I keep getting the signs wrong in the plane equation. How do I avoid this?

It is a common trap! Always write the formula

a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0

clearly before substituting. When you plug in negative coordinates, use parentheses, like

y - (- 2)

, to ensure you don't accidentally turn a subtraction into an addition. Slowing down for that one step will save you from a lot of frustration.

Can I use the dot product to find the plane equation?

Absolutely! The equation of a plane can also be written as

r \cdot n = A \cdot n

, where

r = (x, y, z)

and

A

is the position vector of the point on the plane. Both methods are mathematically identical. The point-normal form is just a more expanded, algebraic way of looking at the same dot product relationship.

Enter Numerical Value:

Visualized Solution

Visualizing the Plane

The Perpendicular Line

Identifying the Normal Vector

Calculating Vector $B C$

Equation of a Plane

Substituting Known Values

Expanding the Equation

Final Plane Equation

Introducing Point P

Satisfying the Plane Equation

Solving for $λ$

The Target Expression

Substituting $λ$

Final Calculation

Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Perpendicular Connection

Constructing the Equation

The Intersection of Point and Plane

Final Calculation

Similar Questions

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

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 $P$ 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 length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - 2 \hat{k})$ 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:

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

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

The distance of the point $(7, - 3, - 4)$ from the plane passing through the points $(2, - 3, 1)$ , $(- 1, 1, - 2)$ and $(3, - 4, 2)$ is :

Enter Numerical Value:

Visualized Solution

Visualizing the Plane

The Perpendicular Line

Identifying the Normal Vector

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

Equation of a Plane

Substituting Known Values

Expanding the Equation

Final Plane Equation

Introducing Point P

Satisfying the Plane Equation

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

The Target Expression

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

Final Calculation

Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Perpendicular Connection

Constructing the Equation

The Intersection of Point and Plane

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the perpendicular drawn from the point (2,1,4) to the plane containing the lines r=(i^+j^​)+λ(i^+2j^​−k^) and r=(i^+j^​)+μ(−i^+j^​−2k^) is :

.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:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane passing through the point (4, -1, 2) and parallel to the lines 3x+2​=−1y−2​=2z+1​ and 1x−2​=2y−3​=3z−4​ also passes through the point :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (7,−3,−4) from the plane passing through the points (2,−3,1), (−1,1,−2) and (3,−4,2) is :

Calculating Vector $B C$

Solving for $λ$

Substituting $λ$

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 length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - 2 \hat{k})$ 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:

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The distance of the point $(7, - 3, - 4)$ from the plane passing through the points $(2, - 3, 1)$ , $(- 1, 1, - 2)$ and $(3, - 4, 2)$ is :