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Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(7, 10, 11)$ from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is

Q: Why do we introduce the parameter $r$?

The parameter $r$ is your bridge between two worlds. It allows you to express an infinite number of points on a line using a single variable. By setting the coordinates of your path equal to the constraints of the target line, you are essentially asking: 'At what value of $r$ does my path collide with the target?' It turns a geometric problem into a simple algebraic one.

Select Answer:

Progressive Hints

Hint 1

Stop! Do not reach for the perpendicular distance formula. The problem explicitly states 'along the line'. Visualize a path starting at point

P

that moves parallel to the second line until it strikes the target line. You are not looking for the shortest path; you are looking for a specific intersection.

Hint 2

To find this intersection, you need the parametric form of the line. If a line passes through

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

with direction ratios

(a, b, c)

, any point on it can be represented as

(x_{1} + a r, y_{1} + b r, z_{1} + cr)

. Apply this to the path starting at

P (7, 10, 11)

using the direction ratios of the second line.

Hint 3

Now, equate your general point on the path to the constraints of the target line. Remember, the target line is defined by

\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}

. The term

\frac{y - 4}{0}

is not a division by zero error; it is a geometric constraint that forces

y = 4

for every point on that line. Use this to solve for your parameter

r

Visualized Solution

Visualizing the Setup

Given Point $P (7, 10, 11)$
Target Line $L_{1} : \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$

Understanding "Distance Along a Line"

Distance is NOT perpendicular.
It is measured parallel to a given direction line $L_{2}$ .
$L_{2} : \frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$

Direction Ratios of the Path

Direction Ratios (DRs) of $L_{2}$ are $(2, 3, 6)$ .
Our path $L_{3}$ must have the same DRs.

Equation of the Path $L_{3}$

Line $L_{3}$ passes through $P (7, 10, 11)$ .
DRs are $(2, 3, 6)$ .
Equation: $\frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6} = r$

Defining the Intersection Point $Q$

Let $L_{3}$ intersect $L_{1}$ at point $Q$ .
We need the coordinates of $Q$ to find the distance $P Q$ .

General Coordinates of $Q$

From $L_{3}$ , express $x, y, z$ in terms of $r$ :
$x = 2 r + 7$
$y = 3 r + 10$
$z = 6 r + 11$
$Q = (2 r + 7, 3 r + 10, 6 r + 11)$

The Constraint from Line $L_{1}$

Point $Q$ also lies on $L_{1}$ .
Look at $L_{1}$ : $\frac{y - 4}{0}$
This implies $y - 4 = 0 ⟹ y = 4$ for all points on $L_{1}$ .

Solving for Parameter $r$

Equate the $y$ -coordinate of $Q$ to $4$ :
$3 r + 10 = 4$
$3 r = - 6$
$r = - 2$

Exact Coordinates of $Q$

Substitute $r = - 2$ into $Q$ :
$x = 2 (- 2) + 7 = 3$
$y = 3 (- 2) + 10 = 4$
$z = 6 (- 2) + 11 = - 1$
$Q = (3, 4, - 1)$

Setting up the Distance Formula

We have $P (7, 10, 11)$ and $Q (3, 4, - 1)$ .
$P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
$P Q = (7 - 3)^{2} + (10 - 4)^{2} + (11 - (- 1))^{2}$

Calculating the Differences

$P Q = 4^{2} + 6^{2} + 1 2^{2}$
$P Q = 16 + 36 + 144$

Final Distance

$P Q = 196$
$P Q = 14$
The distance is $14$ units.

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

We are given a point

P (7, 10, 11)

and a line

L_{1}

defined by:

\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}

We must find the distance from

P

L_{1}

along a path parallel to

L_{2}

, where

L_{2}

is defined by:

\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}

The direction ratios of

L_{2}

are

(2, 3, 6)

. Since our path

L_{3}

passes through

P (7, 10, 11)

and is parallel to

L_{2}

, its equation is:

\frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6} = r

Finding the Intersection Point $Q$

Any point

Q

on the path

L_{3}

can be expressed in terms of the parameter

r

as:

x = 2 r + 7, y = 3 r + 10, z = 6 r + 11

Since

Q

must also lie on

L_{1}

, it must satisfy the constraints of

L_{1}

. The term

\frac{y - 4}{0}

in the equation of

L_{1}

implies that the

y

-coordinate is constant:

y - 4 = 0 ⟹ y = 4

By substituting the expression for

y

from our path

L_{3}

into this constraint, we solve for

r

3 r + 10 = 4 ⟹ 3 r = - 6 ⟹ r = - 2

Calculating the Coordinates and Distance

Using

r = - 2

, we find the specific coordinates of the intersection point

Q

x = 2 (- 2) + 7 = 3

y = 3 (- 2) + 10 = 4

z = 6 (- 2) + 11 = - 1

Thus, the point of intersection is

Q (3, 4, - 1)

. We now calculate the distance

P Q

using the 3D distance formula:

P Q = (7 - 3)^{2} + (10 - 4)^{2} + (11 - (- 1))^{2}

P Q = 4^{2} + 6^{2} + 1 2^{2}

P Q = 16 + 36 + 144 = 196

The final distance is $14$ units.

Common Traps & FAQs

Why can't I just use the shortest distance formula

\frac{∣ ( P - A ) \times b ∣}{∣ b ∣}

That formula calculates the perpendicular distance, which is the shortest path. But this problem imposes a constraint: you must travel 'along' a specific direction. Think of it like walking through a forest; you aren't taking the shortest path to the river, you are walking along a specific compass bearing until you hit the bank.

What does

\frac{y - 4}{0}

actually mean in the equation of the line?

It is a common point of confusion! In 3D geometry, a denominator of zero in the symmetric form of a line equation implies that the numerator must be zero. It essentially locks that coordinate. So,

\frac{y - 4}{0}

tells us that

y - 4 = 0

, or

y = 4

. It means the line lies entirely on the plane

y = 4

Why do we introduce the parameter

r

The parameter

r

is your bridge between two worlds. It allows you to express an infinite number of points on a line using a single variable. By setting the coordinates of your path equal to the constraints of the target line, you are essentially asking: 'At what value of

r

does my path collide with the target?' It turns a geometric problem into a simple algebraic one.

Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(7, 10, 11)$ from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Understanding "Distance Along a Line"

Direction Ratios of the Path

Equation of the Path $L_{3}$

Defining the Intersection Point $Q$

General Coordinates of $Q$

The Constraint from Line $L_{1}$

Solving for Parameter $r$

Exact Coordinates of $Q$

Setting up the Distance Formula

Calculating the Differences

Final Distance

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Intersection Point $Q$

Calculating the Coordinates and Distance

Similar Questions

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

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 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 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 :

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

Let $d$ be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point $(7, 8, 9)$ . Then $d^{2} + 6$ is equal to :

Let $P$ be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of $P$ from the line $4 x = 2 y = z$ is

Let $L$ be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$ , whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line $L$ is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to :

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

Select Answer:

Visualized Solution

Visualizing the Setup

Understanding "Distance Along a Line"

Direction Ratios of the Path

Equation of the Path .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

Defining the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

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

The Constraint from Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

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

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

Setting up the Distance Formula

Calculating the Differences

Final Distance

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculating the Coordinates and Distance

Similar Questions

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let d be the distance of the point of intersection of the lines 3x+6​=2y​=1z+1​ and 4x−7​=3y−9​=2z−4​ from the point (7,8,9). Then d2+6 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be the point of intersection of the lines 1x−2​=5y−4​=1z−2​ and 2x−3​=3y−2​=2z−3​. Then, the shortest distance of P from the line 4x=2y=z 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 :

Equation of the Path $L_{3}$

Defining the Intersection Point $Q$

General Coordinates of $Q$

The Constraint from Line $L_{1}$

Solving for Parameter $r$

Exact Coordinates of $Q$

Finding the Intersection Point $Q$

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 :

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 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 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 :

Let $d$ be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point $(7, 8, 9)$ . Then $d^{2} + 6$ is equal to :

Let $P$ be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of $P$ from the line $4 x = 2 y = z$ is

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