Wait, why does the dot product of $\vec{PM}$ and $\vec{d}$ have to be zero?

That is the definition of perpendicularity! If two vectors are perpendicular, the angle $\theta$ between them is $90^\circ$. Since the dot product is defined as $\vec{A} \cdot \vec{B} = \|A\| \|B\| \cos(\theta)$, and $\cos(90^\circ) = 0$, the dot product must be zero. It is the mathematical heartbeat of orthogonality.

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Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $P (4, 6, - 2)$ from the line passing through the point $(- 3, 2, 3)$ and parallel to a line with direction ratios $3, 3, - 1$ is equal to :

Q: Why do we use $\lambda$ to represent a point on the line? Can't we just use the distance formula directly?

Using $\lambda$ is the most elegant way to 'anchor' a point onto the line. Because the line extends infinitely, we don't know which specific point $M$ is the foot of the perpendicular. By defining $M$ in terms of $\lambda$, we transform a 3D geometry problem into a simple 1D algebraic equation. It is the bridge between the geometry of space and the algebra of lines.

Q: Is there a faster way to do this using the cross product?

Yes! You can use the formula $d = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}$, where $A$ is a point on the line and $\vec{d}$ is the direction vector. While this avoids finding $\lambda$, it requires calculating a cross product and a magnitude, which can be prone to arithmetic errors. Both methods are valid, but the $\lambda$ method is often more intuitive for visualizing the foot of the perpendicular.

Q: What if I get a negative value for $\lambda$? Did I do something wrong?

Not at all! $\lambda$ is just a parameter. A negative $\lambda$ simply means the foot of the perpendicular $M$ lies 'behind' the reference point $A$ on the line. The math remains perfectly valid regardless of the sign of $\lambda$.

Select Answer:

Visualized Solution

Visualizing the Geometry

Given point: $P (4, 6, - 2)$
Line $L$ passes through: $A (- 3, 2, 3)$
Direction Ratios of $L$ : $d = (3, 3, - 1)$

The Foot of the Perpendicular

Let $M$ be the foot of the perpendicular from $P$ to $L$ .
We need to find the length of $P M$ .
$P M ⊥ L$

Equation of Line $L$

Equation of line passing through $(x_{1}, y_{1}, z_{1})$ with DRs $(a, b, c)$ :
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = λ$
$\frac{x + 3}{3} = \frac{y - 2}{3} = \frac{z - 3}{- 1} = λ$

General Point $M$

Any point on the line can be expressed in terms of $λ$ .
$x = 3 λ - 3$
$y = 3 λ + 2$
$z = - λ + 3$
$M = (3 λ - 3, 3 λ + 2, - λ + 3)$

Vector $P M$

Vector $P M = Position vector of M - Position vector of P$
$P M = (3 λ - 3 - 4) \hat{i} + (3 λ + 2 - 6) \hat{j} + (- λ + 3 - (- 2)) \hat{k}$
$P M = (3 λ - 7) \hat{i} + (3 λ - 4) \hat{j} + (- λ + 5) \hat{k}$

The Perpendicularity Condition

Since $P M$ is perpendicular to line $L$ , the dot product of their direction vectors is zero.
$P M \cdot d = 0$
Direction vector of line: $d = (3, 3, - 1)$

Applying the Dot Product

$P M = (3 λ - 7, 3 λ - 4, - λ + 5)$
$d = (3, 3, - 1)$
$(3 λ - 7) (3) + (3 λ - 4) (3) + (- λ + 5) (- 1) = 0$

Solving for $λ$

Expand: $9 λ - 21 + 9 λ - 12 + λ - 5 = 0$
Combine $λ$ terms: $9 λ + 9 λ + λ = 19 λ$
Combine constants: $- 21 - 12 - 5 = - 38$
$19 λ - 38 = 0 ⟹ 19 λ = 38 ⟹ λ = 2$

Exact Coordinates of $M$

Substitute $λ = 2$ into $M (3 λ - 3, 3 λ + 2, - λ + 3)$
$x = 3 (2) - 3 = 6 - 3 = 3$
$y = 3 (2) + 2 = 6 + 2 = 8$
$z = - (2) + 3 = 1$
$M = (3, 8, 1)$

Calculating Distance $P M$

Distance formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
$P (4, 6, - 2)$ and $M (3, 8, 1)$
$P M = (3 - 4)^{2} + (8 - 6)^{2} + (1 - (- 2))^{2}$
$P M = (- 1)^{2} + (2)^{2} + (3)^{2}$

Final Answer

$P M = 1 + 4 + 9$
$P M = 14$
The perpendicular distance from point $P$ to the line is $14$ .

00:00 / 00:00

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room at point

P (4, 6, - 2)

. A long, straight wire passes through point

A (- 3, 2, 3)

with direction ratios

(3, 3, - 1)

Our goal is to find the shortest distance from

P

to this wire, which corresponds to the perpendicular distance from a point to a line. We define the foot of the perpendicular as point

M

Defining the Path

To find the distance, we represent the line in parametric form. Using the parameter

λ

, we write the equation of the line as:

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

λ

varies, we slide along the line. Thus, any general point

M

on the line can be expressed as:

M (3 λ - 3, 3 λ + 2, - λ + 3)

The Perpendicularity Condition

Consider the vector

P M

connecting your position

P

to the point

M

. If

M

is the foot of the perpendicular,

P M

must be orthogonal to the line's direction vector

d = (3, 3, - 1)

Mathematically, this requires the dot product

P M \cdot d = 0

. First, we calculate

P M

by subtracting the coordinates of

P

from

M

P M = (3 λ - 7) \hat{i} + (3 λ - 4) \hat{j} + (- λ + 5) \hat{k}

Solving for the Unknown

We apply the condition

P M \cdot d = 0

by multiplying the corresponding components:

(3 λ - 7) (3) + (3 λ - 4) (3) + (- λ + 5) (- 1) = 0

Expanding this expression yields:

9 λ - 21 + 9 λ - 12 + λ - 5 = 0

Combining like terms results in

19 λ - 38 = 0

, which simplifies to the elegant result $λ = 2$ .

Final Calculation

With

λ = 2

, we locate the exact coordinates of

M

by substituting the value back into our general point expression:

M = (3 (2) - 3, 3 (2) + 2, - (2) + 3) = (3, 8, 1)

The shortest distance is the magnitude of the vector

P M

between

P (4, 6, - 2)

and

M (3, 8, 1)

. Using the distance formula:

P M = (3 - 4)^{2} + (8 - 6)^{2} + (1 - (- 2))^{2}

P M = (- 1)^{2} + (2)^{2} + (3)^{2} = 1 + 4 + 9 = 14

The shortest distance from the point to the line is $14$ .

Common Traps & FAQs

Why do we use

λ

to represent a point on the line? Can't we just use the distance formula directly?

Using

λ

is the most elegant way to 'anchor' a point onto the line. Because the line extends infinitely, we don't know which specific point

M

is the foot of the perpendicular. By defining

M

in terms of

λ

, we transform a 3D geometry problem into a simple 1D algebraic equation. It is the bridge between the geometry of space and the algebra of lines.

Is there a faster way to do this using the cross product?

Yes! You can use the formula

d = \frac{∥ A P \times d ∥}{∥ d ∥}

, where

A

is a point on the line and

d

is the direction vector. While this avoids finding

λ

, it requires calculating a cross product and a magnitude, which can be prone to arithmetic errors. Both methods are valid, but the

λ

method is often more intuitive for visualizing the foot of the perpendicular.

Wait, why does the dot product of

P M

and

d

have to be zero?

That is the definition of perpendicularity! If two vectors are perpendicular, the angle

θ

between them is

9 0^{\circ}

. Since the dot product is defined as

A \cdot B = ∥ A ∥∥ B ∥ cos (θ)

, and

cos (9 0^{\circ}) = 0

, the dot product must be zero. It is the mathematical heartbeat of orthogonality.

What if I get a negative value for

λ

? Did I do something wrong?

Not at all!

λ

is just a parameter. A negative

λ

simply means the foot of the perpendicular

M

lies 'behind' the reference point

A

on the line. The math remains perfectly valid regardless of the sign of

λ

Why do we need the direction ratios

(3, 3, - 1)

The direction ratios define the 'orientation' of the line in 3D space. Without them, we would know where the line starts, but not where it points. They act as the 'compass' for the line, allowing us to define the vector

d

which is essential for both the parametric equation and the perpendicularity condition.

Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $P (4, 6, - 2)$ from the line passing through the point $(- 3, 2, 3)$ and parallel to a line with direction ratios $3, 3, - 1$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Geometry

The Foot of the Perpendicular

Equation of Line $L$

General Point $M$

Vector $P M$

The Perpendicularity Condition

Applying the Dot Product

Solving for $λ$

Exact Coordinates of $M$

Calculating Distance $P M$

Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Defining the Path

The Perpendicularity Condition

Solving for the Unknown

Final Calculation

Similar Questions

The distance of the point having position vector $- \hat{i} + 2 \hat{j} + 6 \hat{k}$ from the straight line passing through the point $(2, 3, - 4)$ and parallel to the vector, $6 \hat{i} + 3 \hat{j} - 4 \hat{k}$ 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 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 the line passing through the points $(- 1, 2, 1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x+2}{3}= rac{y-3}{2}= rac{z-4}{1}$ at the point P. Then the distance of P from the point $Q (4, - 5, 1)$ 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 :

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 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 .

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

Select Answer:

Visualized Solution

Visualizing the Geometry

The Foot of the Perpendicular

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

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

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

The Perpendicularity Condition

Applying the Dot Product

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

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

Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Defining the Path

The Perpendicularity Condition

Solving for the Unknown

Final Calculation

Similar Questions

.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 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; } 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 :

.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; } The perpendicular distance, of the line 2x−1​=−1y+2​=2z+3​ from the point P(2,−10,1), is :

Equation of Line $L$

General Point $M$

Vector $P M$

Solving for $λ$

Exact Coordinates of $M$

Calculating Distance $P M$

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 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:

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 :

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 .

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :