JEE Main 2023 (31 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let the plane $P : 8 x + α_{1} y + α_{2} z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ . If the intercept of $P$ on the $y$ -axis is 1, then the distance between $P$ and $L$ is :

Q: Wait, why do we use the dot product to check for parallelism? Shouldn't it be the cross product?

That is a fantastic question! Remember, the normal vector $\vec{n}$ is perpendicular to the plane itself. If the line is parallel to the plane, it must be perpendicular to the normal vector. When two vectors are perpendicular, their dot product is zero. If we were looking for the line to be *contained* in the plane, we would need both the dot product to be zero AND a point on the line to satisfy the plane equation. Here, the dot product is the perfect tool to ensure the line doesn't 'pierce' the plane.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. The plane

P

has an unknown

y

-intercept, which is given as

1

. This means the point

(0, 1, 0)

must satisfy the equation of the plane. Use this to find the value of

α_{1}

Hint 2

Hint 2: The Arsenal. Since the line

L

is parallel to the plane

P

, the direction vector of the line

d = (2, 3, 5)

must be perpendicular to the normal vector of the plane

n = (8, α_{1}, α_{2})

. Recall that the dot product of two perpendicular vectors is zero.

Hint 3

Hint 3: The Execution. Once you have both

α_{1}

and

α_{2}

, the plane equation is fully defined. To find the distance between the parallel line and the plane, pick any point on the line (for example, by setting the symmetric equations to a parameter) and use the standard point-to-plane distance formula:

d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}

Visualized Solution

Visualizing the Setup

We are given a plane $P : 8 x + α_{1} y + α_{2} z + 12 = 0$ .
A line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ is parallel to $P$ .
Our goal is to find the perpendicular distance between them.

The $y$ -intercept Condition

The problem states the $y$ -intercept of plane $P$ is $1$ .
This means the plane intersects the $y$ -axis at the point $(0, 1, 0)$ .
Any point on the plane must satisfy its equation.

Substituting the Intercept

Let's substitute $(x, y, z) = (0, 1, 0)$ into the plane's equation.
$8 (0) + α_{1} (1) + α_{2} (0) + 12 = 0$

Solving for $α_{1}$

The $x$ and $z$ terms become zero.
We are left with: $α_{1} + 12 = 0$
Therefore, $α_{1} = - 12$ .

Extracting Direction Vectors

From line $L$ , the direction vector is $d = (2, 3, 5)$ .
From plane $P$ , the normal vector is $n = (8, α_{1}, α_{2})$ .
Substituting $α_{1} = - 12$ , we get $n = (8, - 12, α_{2})$ .

Condition for Parallelism

Since the line is parallel to the plane, the line's direction vector $d$ is perpendicular to the plane's normal vector $n$ .
Therefore, their dot product must be zero: $n \cdot d = 0$ .

Setting up the Dot Product

Let's substitute our vectors into the dot product equation.
$(8, - 12, α_{2}) \cdot (2, 3, 5) = 0$
$8 (2) + (- 12) (3) + α_{2} (5) = 0$

Solving for $α_{2}$

$16 - 36 + 5 α_{2} = 0$
$- 20 + 5 α_{2} = 0$
$5 α_{2} = 20 ⟹ α_{2} = 4$

The Complete Plane Equation

Substitute $α_{1} = - 12$ and $α_{2} = 4$ back into $P$ .
$8 x - 12 y + 4 z + 12 = 0$
Dividing the entire equation by $4$ , we get:
$2 x - 3 y + z + 3 = 0$

Distance from Line to Plane

The distance between a parallel line and a plane is constant everywhere.
We can pick any point on the line and find its perpendicular distance to the plane.

Identifying a Point on $L$

Look at the equation of line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ .
A known point on this line is $A (- 2, 3, - 4)$ .

Applying the Distance Formula

The distance $d$ from point $(x_{1}, y_{1}, z_{1})$ to plane $a x + b y + cz + d = 0$ is:
$d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$
Substitute $A (- 2, 3, - 4)$ and plane $2 x - 3 y + z + 3 = 0$ :
$d = \frac{∣2 ( - 2 ) - 3 ( 3 ) + 1 ( - 4 ) + 3∣}{2 ^{2} + ( - 3 ) ^{2} + 1 ^{2}}$

Simplifying Numerator and Denominator

Numerator: $∣ - 4 - 9 - 4 + 3∣ = ∣ - 14∣ = 14$
Denominator: $4 + 9 + 1 = 14$
So, $d = \frac{14}{14}$

Final Calculation

Rationalizing the fraction: $\frac{14}{14} = 14$ .
The perpendicular distance between the plane and the line is $14$ .

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The Sigma Insight: Intersection of a Line and a Plane

Solution Diagram

Analyzing the Setup

The plane

P

is defined by the equation

8 x + α_{1} y + α_{2} z + 12 = 0

. The line

L

is given by the symmetric form:

\frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}

Since the line

L

is parallel to the plane

P

, the distance between them remains constant. Our goal is to determine the constants

α_{1}

and

α_{2}

and calculate the perpendicular distance.

Unlocking the Plane

We are given that the

y

-intercept of the plane is

1

. This implies the plane passes through the point

(0, 1, 0)

Substituting

x = 0

y = 1

, and

z = 0

into the plane equation

8 x + α_{1} y + α_{2} z + 12 = 0

8 (0) + α_{1} (1) + α_{2} (0) + 12 = 0

This simplifies to

α_{1} + 12 = 0

, which yields $α_{1} = - 12$ .

The Dance of Vectors

The direction vector of the line

L

d = (2, 3, 5)

. The normal vector to the plane

P

n = (8, - 12, α_{2})

Because the line is parallel to the plane, the direction vector

d

must be perpendicular to the normal vector

n

. Therefore, their dot product must be zero:

n \cdot d = (8) (2) + (- 12) (3) + (α_{2}) (5) = 0

16 - 36 + 5 α_{2} = 0

- 20 + 5 α_{2} = 0 \Rightarrow α_{2} = 4

The equation of the plane is

8 x - 12 y + 4 z + 12 = 0

. Dividing by

4

, we obtain the simplified form:

2 x - 3 y + z + 3 = 0

The Final Leap

To find the distance between the line and the plane, we select a point

A (- 2, 3, - 4)

from the line

L

. We use the perpendicular distance formula from a point

(x_{1}, y_{1}, z_{1})

to a plane

a x + b y + cz + d = 0

d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}

Substituting our values:

d = \frac{∣2 ( - 2 ) - 3 ( 3 ) + 1 ( - 4 ) + 3∣}{2 ^{2} + ( - 3 ) ^{2} + 1 ^{2}}

d = \frac{∣ - 4 - 9 - 4 + 3∣}{4 + 9 + 1} = \frac{∣ - 14∣}{14}

Simplifying the expression, we find the final distance:

$d = 14$

Common Traps & FAQs

Wait, why do we use the dot product to check for parallelism? Shouldn't it be the cross product?

That is a fantastic question! Remember, the normal vector

n

is perpendicular to the plane itself. If the line is parallel to the plane, it must be perpendicular to the normal vector. When two vectors are perpendicular, their dot product is zero. If we were looking for the line to be *contained* in the plane, we would need both the dot product to be zero AND a point on the line to satisfy the plane equation. Here, the dot product is the perfect tool to ensure the line doesn't 'pierce' the plane.

Does it matter which point on the line I choose for the distance formula?

Not at all! Because the line is parallel to the plane, the perpendicular distance from any point on the line to the plane is constant. You can pick the most convenient point—often the one where the numerators of the symmetric form are zero—and the result will be the same.

Why did we divide the plane equation by 4? Is that necessary?

It is not strictly necessary, but it is a great habit to develop! Simplifying the equation

8 x - 12 y + 4 z + 12 = 0

2 x - 3 y + z + 3 = 0

makes the coefficients smaller and easier to work with. It reduces the chance of making a calculation error when you plug the values into the distance formula. Always simplify if you can!

Animated Solution for Mathematics - Three Dimensional Geometry: Let the plane $P : 8 x + α_{1} y + α_{2} z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ . If the intercept of $P$ on the $y$ -axis is 1, then the distance between $P$ and $L$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

The $y$ -intercept Condition

Substituting the Intercept

Solving for $α_{1}$

Extracting Direction Vectors

Condition for Parallelism

Setting up the Dot Product

Solving for $α_{2}$

The Complete Plane Equation

Distance from Line to Plane

Identifying a Point on $L$

Applying the Distance Formula

Simplifying Numerator and Denominator

Final Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Unlocking the Plane

The Dance of Vectors

The Final Leap

Similar Questions

If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ be $P$ . Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$

The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersects at the point $P$ , then the distance of the point $P$ from the plane $z = a$ is :

The distance of the point $(- 1, 9, - 16)$ from the plane $2 x + 3 y - z = 5$ measured parallel to the line $\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}$ is

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

If the foot of the perpendicular from the point $A (- 1, 4, 3)$ on the plane $P : 2 x + m y + n z = 4$ , is $(- 2, \frac{7}{2}, \frac{3}{2})$ , then the distance of the point $A$ from the plane $P$ , measured parallel to a line with direction ratios $3, - 1, - 4$ , is equal to :

The distance of the point $P (3, 4, 4)$ from the point of intersection of the line joining the points $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ and the plane $2 x + y + z = 7$ , is equal to ___

Select Answer:

Visualized Solution

Visualizing the Setup

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

Substituting the Intercept

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α1​

Extracting Direction Vectors

Condition for Parallelism

Setting up the Dot Product

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2​

The Complete Plane Equation

Distance from Line to Plane

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

Applying the Distance Formula

Simplifying Numerator and Denominator

Final Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Unlocking the Plane

The Dance of Vectors

The Final Leap

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1,0,2) from the point of intersection of the line 3x−2​=4y+1​=12z−2​ and the plane x−y+z=16, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foot of the perpendicular from the point (1,2,4) on the line 4x+2​=2y−1​=3z+1​ be P. Then the distance of P from the plane 3x+4y+12z+23=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1,−2,3) from the plane x−y+z=5 measured parallel to the line 2x​=3y​=−6z​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 1x−1​=2y−2​=1z+3​ and 2x−a​=3y+2​=1z−3​ intersects at the point P, then the distance of the point P from the plane z=a is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (−1,9,−16) from the plane 2x+3y−z=5 measured parallel to the line 3x+4​=42−y​=12z−3​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance of the point P(1,−2,1) from the plane x+2y−2z=α, where α>0, is 5, then the foot of the perpendicular from P to the plane is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point P(3,4,4) from the point of intersection of the line joining the points Q(3,−4,−5) and R(2,−3,1) and the plane 2x+y+z=7, is equal to ___

The $y$ -intercept Condition

Solving for $α_{1}$

Solving for $α_{2}$

Identifying a Point on $L$

If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ be $P$ . Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$

The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersects at the point $P$ , then the distance of the point $P$ from the plane $z = a$ is :

The distance of the point $(- 1, 9, - 16)$ from the plane $2 x + 3 y - z = 5$ measured parallel to the line $\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}$ is

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

The distance of the point $P (3, 4, 4)$ from the point of intersection of the line joining the points $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ and the plane $2 x + y + z = 7$ , is equal to ___