JEE Main 2024 (09 Apr Shift 2)

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Animated Solution for Mathematics - Three Dimensional Geometry: The square of the distance of the image of the point $(6, 1, 5)$ in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ , from the origin is

Q: Why do we use the dot product with the direction vector?

The vector $\vec{AM}$ represents the shortest path from point $A$ to the line $L$. In geometry, the shortest distance from a point to a line is always along the perpendicular. Since the line $L$ has a direction vector $\vec{d}$, and $\vec{AM}$ is perpendicular to $L$, the dot product $\vec{AM} \cdot \vec{d}$ must be zero. It is the most elegant way to enforce orthogonality in 3D space!

Q: Can I use the projection formula instead of the parameter method?

Absolutely! You can find the projection of vector $\vec{AP}$ (where $P$ is a point on the line) onto the line's direction vector. However, the parameter method is often preferred in JEE because it directly gives you the coordinates of the foot of the perpendicular $M$, which is exactly what you need to find the image $I$.

Q: Why is the image $I = 2M - A$?

Think of the line $L$ as a mirror. When you reflect a point $A$ across a mirror to get an image $I$, the mirror (the line $L$) must be the perpendicular bisector of the segment $AI$. This means the foot of the perpendicular $M$ is the midpoint of $AI$. By the midpoint formula, $M = \frac{A+I}{2}$. Rearranging this gives $I = 2M - A$. It is pure symmetry!

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Given point $A (6, 1, 5)$ and line $L : \frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ .
We need to find the image of $A$ in the line $L$ .

The Foot of the Perpendicular

Let $M$ be the foot of the perpendicular from $A$ to the line $L$ .
The image $I$ will lie on the extended perpendicular such that $A M = M I$ .

General Point on Line $L$

Equate the line equation to a parameter $λ$ :
$\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4} = λ$
General coordinates of $M$ : $(3 λ + 1, 2 λ, 4 λ + 2)$

Vector $A M$

Position vector of $A = (6, 1, 5)$
Vector $A M = M - A$
$A M = (3 λ + 1 - 6, 2 λ - 1, 4 λ + 2 - 5)$
$A M = (3 λ - 5, 2 λ - 1, 4 λ - 3)$

The Perpendicularity Condition

Direction vector of line $L$ is $d = (3, 2, 4)$ .
Since $A M ⊥ L$ , their dot product must be zero: $A M \cdot d = 0$ .
$3 (3 λ - 5) + 2 (2 λ - 1) + 4 (4 λ - 3) = 0$

Solving for $λ$

Expanding the dot product:
$(9 λ - 15) + (4 λ - 2) + (16 λ - 12) = 0$
Combine like terms: $29 λ - 29 = 0$
$λ = 1$

Exact Coordinates of $M$

Substitute $λ = 1$ into the general coordinates of $M$ :
$M = (3 (1) + 1, 2 (1), 4 (1) + 2)$
$M = (4, 2, 6)$

Locating the Image $I$

Let the image be $I (x_{i}, y_{i}, z_{i})$ .
$M$ is the midpoint of the line segment $A I$ .
Therefore, $M = \frac{A + I}{2}$ , which means $I = 2 M - A$ .

Coordinates of Image $I$

$I = 2 (4, 2, 6) - (6, 1, 5)$
$I = (8, 4, 12) - (6, 1, 5)$
$I = (2, 3, 7)$

Distance from Origin Squared

Origin $O (0, 0, 0)$ and Image $I (2, 3, 7)$ .
Distance squared $O I^{2} = (2 - 0)^{2} + (3 - 0)^{2} + (7 - 0)^{2}$
$O I^{2} = 4 + 9 + 49 = 62$
Final Answer: $62$

00:00 / 00:00

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

Imagine a point

A (6, 1, 5)

and a line

L

defined by the symmetric equations:

\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}

Our goal is to find the reflection

I

of point

A

across line

L

and determine the square of the distance from the origin

O (0, 0, 0)

I

Finding the Foot of the Perpendicular

To find the image

I

, we first identify the point

M

on line

L

that is closest to

A

. This point

M

is the foot of the perpendicular from

A

to the line.

We parameterize the line by setting the ratios equal to

λ

\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4} = λ

Any point

M

on the line can be expressed as

(3 λ + 1, 2 λ, 4 λ + 2)

The Orthogonality Condition

We construct the vector

A M = M - A

. Substituting the coordinates, we get:

A M = (3 λ + 1 - 6, 2 λ - 1, 4 λ + 2 - 5) = (3 λ - 5, 2 λ - 1, 4 λ - 3)

Since

A M

must be perpendicular to the direction vector of the line

d = (3, 2, 4)

, their dot product must be zero:

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

Expanding this equation:

9 λ - 15 + 4 λ - 2 + 16 λ - 12 = 0

Combining like terms results in

29 λ - 29 = 0

, which yields

λ = 1

. Substituting

λ = 1

into our parameterization, the foot of the perpendicular is

M (4, 2, 6)

The Symmetry of Reflection

The point

M

is the midpoint of the segment

A I

. Using the midpoint formula

M = \frac{A + I}{2}

, we solve for

I

I = 2 M - A

I = 2 (4, 2, 6) - (6, 1, 5) = (8, 4, 12) - (6, 1, 5) = (2, 3, 7)

Thus, the coordinates of the reflected point are

I (2, 3, 7)

Final Calculation

The problem requires the square of the distance from the origin

O (0, 0, 0)

I (2, 3, 7)

. Applying the distance formula:

O I^{2} = (2 - 0)^{2} + (3 - 0)^{2} + (7 - 0)^{2}

O I^{2} = 4 + 9 + 49 = 62

The final result for the square of the distance is 62.

Common Traps & FAQs

Why do we use the dot product with the direction vector?

The vector

A M

represents the shortest path from point

A

to the line

L

. In geometry, the shortest distance from a point to a line is always along the perpendicular. Since the line

L

has a direction vector

d

, and

A M

is perpendicular to

L

, the dot product

A M \cdot d

must be zero. It is the most elegant way to enforce orthogonality in 3D space!

Can I use the projection formula instead of the parameter method?

Absolutely! You can find the projection of vector

A P

(where

P

is a point on the line) onto the line's direction vector. However, the parameter method is often preferred in JEE because it directly gives you the coordinates of the foot of the perpendicular

M

, which is exactly what you need to find the image

I

Why is the image

I = 2 M - A

Think of the line

L

as a mirror. When you reflect a point

A

across a mirror to get an image

I

, the mirror (the line

L

) must be the perpendicular bisector of the segment

A I

. This means the foot of the perpendicular

M

is the midpoint of

A I

. By the midpoint formula,

M = \frac{A + I}{2}

. Rearranging this gives

I = 2 M - A

. It is pure symmetry!

Animated Solution for Mathematics - Three Dimensional Geometry: The square of the distance of the image of the point $(6, 1, 5)$ in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ , from the origin is

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Foot of the Perpendicular

General Point on Line $L$

Vector $A M$

The Perpendicularity Condition

Solving for $λ$

Exact Coordinates of $M$

Locating the Image $I$

Coordinates of Image $I$

Distance from Origin Squared

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Foot of the Perpendicular

The Orthogonality Condition

The Symmetry of Reflection

Final Calculation

Similar Questions

Let $Q (a, b, c)$ be the image of the point $P (3, 2, 1)$ in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ . Then the distance of $Q$ from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is

Let the line $L$ be the projection of the line $\frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 4}{2}$ in the plane $x - 2 y - z = 3$ . If $d$ is the distance of the point $(0, 0, 6)$ from $L$ , then $d^{2}$ is equal to .

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is:

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$ . The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line $L$ along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to

Let P be the image of the point $Q (7, - 2, 5)$ in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R (5, p, q)$ be a point on L. Then the square of the area of $△ P QR$ is

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :

If the length of the perpendicular drawn from the point $P (a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $26$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point $P$ in this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to :

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Foot of the Perpendicular

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

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

The Perpendicularity Condition

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

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

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

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

Distance from Origin Squared

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Foot of the Perpendicular

The Orthogonality Condition

The Symmetry of Reflection

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Q(a,b,c) be the image of the point P(3,2,1) in the line 1x−1​=2y​=1z−1​. Then the distance of Q from the line 3x−9​=2y−9​=−2z−5​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the line L be the projection of the line 2x−1​=1y−3​=2z−4​ in the plane x−2y−z=3. If d is the distance of the point (0,0,6) from L, then d2 is equal to .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point P(1,2,a) in the line 3x−6​=2y−7​=27−z​ is Q(5,b,c), then a2+b2+c2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The square of the distance of the point (715​,732​,7) from the line 3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^​+7k^ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the line 2x−2​=3y−6​=4z−3​ from the point (1,4,0) along the line 1x​=2y−2​=3z+3​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be the image of the point Q(7,−2,5) in the line L:2x−1​=3y+1​=4z​ and R(5,p,q) be a point on L. Then the square of the area of △PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of line 3y−2z−1=0=3x−z+4 from the point (2,−1,6) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance, of the point (7,−2,11) from the line 1x−6​=0y−4​=3z−8​ along the line 2x−5​=−3y−1​=6z−5​, is :

General Point on Line $L$

Vector $A M$

Solving for $λ$

Exact Coordinates of $M$

Locating the Image $I$

Coordinates of Image $I$

Let $Q (a, b, c)$ be the image of the point $P (3, 2, 1)$ in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ . Then the distance of $Q$ from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is

Let the line $L$ be the projection of the line $\frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 4}{2}$ in the plane $x - 2 y - z = 3$ . If $d$ is the distance of the point $(0, 0, 6)$ from $L$ , then $d^{2}$ is equal to .

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is:

Let P be the image of the point $Q (7, - 2, 5)$ in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R (5, p, q)$ be a point on L. Then the square of the area of $△ P QR$ is

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :