JEE Main 2017

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Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ , having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is:

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Progressive Hints

Hint 1

Imagine the plane as a sheet of paper floating in space. To define its orientation, you need a 'normal' vector—a needle sticking straight out of it. Since the plane contains two lines, the normal must be perpendicular to both. How do you find a vector perpendicular to two others?

Hint 2

You have the direction vectors of the two lines, say

b_{1}

and

b_{2}

. The cross product

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

will give you the direction of the normal. Remember, the cross product is the key to unlocking the orientation of the plane.

Hint 3

Once you have the normal vector

n = (a, b, c)

and a point

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

on the plane, the equation is

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

. After finding the plane equation

a x + b y + cz + d = 0

, use the distance formula

d = \frac{∣ a x _{0} + b y _{0} + c z _{0} + d ∣}{a ^{2} + b ^{2} + c ^{2}}

Visualized Solution

Visualizing the Setup

Given point: $P (1, 3, - 7)$
Plane passes through: $A (1, - 1, - 1)$
Normal $n$ is $⊥$ to $L_{1} : \frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$
Normal $n$ is $⊥$ to $L_{2} : \frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$

Direction Vectors of $L_{1}$ and $L_{2}$

Line $L_{1} : \frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$
Direction vector $b_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}$
Line $L_{2} : \frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$
Direction vector $b_{2} = 2 \hat{i} - \hat{j} - \hat{k}$

Normal Vector via Cross Product

Since $n ⊥ b_{1}$ and $n ⊥ b_{2}$ , we use the cross product.
$n = b_{1} \times b_{2}$
$n = \hat{i} 12 \hat{j} - 2 - 1 \hat{k} 3 - 1$

Expanding the Determinant

$n = \hat{i} ((- 2) (- 1) - (3) (- 1)) - \hat{j} ((1) (- 1) - (3) (2)) + \hat{k} ((1) (- 1) - (- 2) (2))$
$n = \hat{i} (2 + 3) - \hat{j} (- 1 - 6) + \hat{k} (- 1 + 4)$
$n = 5 \hat{i} + 7 \hat{j} + 3 \hat{k}$

Setting up the Plane Equation

General equation: $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$
Normal vector $(a, b, c) = (5, 7, 3)$
Point on plane $(x_{1}, y_{1}, z_{1}) = (1, - 1, - 1)$
Substitute: $5 (x - 1) + 7 (y - (- 1)) + 3 (z - (- 1)) = 0$

Simplifying the Plane Equation

Expand: $5 x - 5 + 7 y + 7 + 3 z + 3 = 0$
Combine constants: $- 5 + 7 + 3 = 5$
Final Plane Equation: $5 x + 7 y + 3 z + 5 = 0$

Perpendicular Distance Formula

Distance $d$ from point $(x_{0}, y_{0}, z_{0})$ to plane $a x + b y + cz + d = 0$ is:
$d = \frac{∣ a x _{0} + b y _{0} + c z _{0} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$
Point $P (1, 3, - 7)$
Plane: $5 x + 7 y + 3 z + 5 = 0$

Calculating the Numerator

Substitute $P (1, 3, - 7)$ into $∣5 x + 7 y + 3 z + 5∣$
Numerator $= ∣5 (1) + 7 (3) + 3 (- 7) + 5∣$
Numerator $= ∣5 + 21 - 21 + 5∣$
Numerator $= ∣10∣ = 10$

Calculating the Denominator

Denominator $= a^{2} + b^{2} + c^{2}$
Denominator $= 5^{2} + 7^{2} + 3^{2}$
Denominator $= 25 + 49 + 9$
Denominator $= 83$

Final Distance

Distance $d = \frac{10}{83}$
Comparing with options, the correct choice is Option 2.

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

To find the shortest distance from the point

P (1, 3, - 7)

to the plane, we must first determine the equation of the plane. We are given a point

A (1, - 1, - 1)

that lies on the plane.

The plane's orientation is defined by its normal vector

n

, which is perpendicular to two given lines with direction vectors

b_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}

and

b_{2} = 2 \hat{i} - \hat{j} - \hat{k}

Determining the Normal Vector

Since the normal vector

n

is perpendicular to both

b_{1}

and

b_{2}

, it is parallel to their cross product. We calculate this using the determinant:

n = \hat{i} 12 \hat{j} - 2 - 1 \hat{k} 3 - 1

Expanding the determinant, we obtain:

n = \hat{i} ((- 2) (- 1) - (3) (- 1)) - \hat{j} ((1) (- 1) - (3) (2)) + \hat{k} ((1) (- 1) - (- 2) (2))

n = 5 \hat{i} + 7 \hat{j} + 3 \hat{k}

Defining the Plane Equation

With the normal vector

n = (5, 7, 3)

and the point

A (1, - 1, - 1)

, we use the point-normal form

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

5 (x - 1) + 7 (y + 1) + 3 (z + 1) = 0

Expanding and simplifying this expression yields the standard form of the plane:

5 x + 7 y + 3 z + 5 = 0

Calculating the Shortest Distance

The shortest distance

d

from a point

(x_{0}, y_{0}, z_{0})

to a plane

a x + b y + cz + d = 0

is given by the formula:

d = \frac{∣ a x _{0} + b y _{0} + c z _{0} + d ∣}{a ^{2} + b ^{2} + c ^{2}}

Substituting

P (1, 3, - 7)

and the plane coefficients into the formula:

d = \frac{∣5 ( 1 ) + 7 ( 3 ) + 3 ( - 7 ) + 5∣}{5 ^{2} + 7 ^{2} + 3 ^{2}}

d = \frac{∣5 + 21 - 21 + 5∣}{25 + 49 + 9} = \frac{10}{83}

The shortest distance from the point to the plane is $\frac{10}{83}$ .

Common Traps & FAQs

Wait, why do we use the cross product of the direction vectors? Can't we just use the dot product?

The dot product gives a scalar, which tells us about angles, but the cross product gives a vector perpendicular to the plane defined by the two lines. Since our normal vector must be perpendicular to both lines, the cross product is the only tool that gives us that specific direction.

Why do we use the point

(1, - 1, - 1)

instead of the points

(1, - 2, 4)

(2, - 1, - 7)

from the lines?

Great question! The problem states the plane passes through the point

(1, - 1, - 1)

. While the lines exist in space, the plane is specifically anchored to that point. Using a point from one of the lines would only be correct if that line was guaranteed to lie on the plane, which isn't stated here.

Why do we take the absolute value in the distance formula?

Distance is a physical length, and length cannot be negative. The expression

a x_{0} + b y_{0} + c z_{0} + d

can be negative depending on which side of the plane the point lies, so the absolute value ensures we get the magnitude of that separation.

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Visualized Solution

Visualizing the Setup

Direction Vectors of $L_{1}$ and $L_{2}$

Normal Vector via Cross Product

Expanding the Determinant

Setting up the Plane Equation

Simplifying the Plane Equation

Perpendicular Distance Formula

Calculating the Numerator

Calculating the Denominator

Final Distance

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Determining the Normal Vector

Defining the Plane Equation

Calculating the Shortest Distance

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Setup

Direction Vectors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Normal Vector via Cross Product

Expanding the Determinant

Setting up the Plane Equation

Simplifying the Plane Equation

Perpendicular Distance Formula

Calculating the Numerator

Calculating the Denominator

Final Distance

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Determining the Normal Vector

Defining the Plane Equation

Calculating the Shortest Distance

Similar Questions

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Direction Vectors of $L_{1}$ and $L_{2}$

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Let a plane $P$ pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

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 :