Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point (1

Q: Wait, why can't we just use the perpendicular distance formula from a point to a line?

That is a fantastic question! The perpendicular distance formula gives the shortest distance, which is only valid if the path of measurement is perpendicular to the line. Here, the problem explicitly defines a specific direction for the path. We must follow that path until it hits the line, which is why we use the parametric intersection method instead.

Q: How do I know if the path equation is correctly standardized?

A line equation is in standard symmetric form when the coefficients of $x$, $y$, and $z$ are all $1$. If you see coefficients like $3x$ or $2y$, you must divide the entire numerator and denominator by that coefficient to isolate the variables. For example, $\frac{3x-11}{2}$ becomes $\frac{x-11/3}{2/3}$. This reveals the true direction ratios.

Q: What if the parameter $\lambda$ turns out to be zero?

If $\lambda = 0$, it simply means that the point $A$ itself lies on the line $L$. In that case, the distance would be zero. It is a perfectly valid result, so don't be alarmed if you encounter it in similar problems!

Question

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 $\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$ is equal to

Options:

(A) 6

(B) 5

(C) 4

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a three-dimensional coordinate system containing an infinite highway, line $L$, passing through points $P(1, 2, 3)$ and $Q(2, 3, 5)$. You are positioned at point $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$.

The objective is to travel from $A$ to $L$ along a specific trajectory defined by the equation:

$$\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$$

### Pinning Down the Highway

To define the line $L$, we first determine its direction vector $\vec{d_L}$ using points $P$ and $Q$:

$$\vec{d_L} = \langle 2-1, 3-2, 5-3 \rangle = \langle 1, 1, 2 \rangle$$

Using point $P(1, 2, 3)$, the symmetric equation of the highway $L$ is:

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

### Decoding the Path

The given trajectory equation contains coefficients that must be standardized to reveal its true direction. By dividing the numerators and denominators by $3$, we rewrite the path as:

$$\frac{x - 11/3}{2/3} = \frac{y - 11/3}{1/3} = \frac{z - 19/3}{2/3}$$

The direction vector of this path, $\vec{d_{L'}}$, is $\langle 2/3, 1/3, 2/3 \rangle$, which is parallel to the vector $\langle 2, 1, 2 \rangle$.

### The Intersection

We assume the path intersects the highway $L$ at point $B$. Since $B$ lies on the path starting from $A$, we parameterize it using $\lambda$:

$$x = \frac{11}{3} + 2\lambda, \quad y = \frac{11}{3} + \lambda, \quad z = \frac{19}{3} + 2\lambda$$

Point $B$ must also satisfy the equation of the highway $L$. Substituting these coordinates into the equation of $L$:

$$\frac{(\frac{11}{3} + 2\lambda) - 1}{1} = \frac{(\frac{11}{3} + \lambda) - 2}{1}$$

Simplifying this expression yields:

$$2\lambda + \frac{8}{3} = \lambda + \frac{5}{3}$$

Solving for the parameter, we find $\lambda = -1$.

### Final Calculation

The distance between $A$ and $B$ is the magnitude of the vector $\vec{AB}$. Given $\vec{AB} = B - A = \langle 2\lambda, \lambda, 2\lambda \rangle$, the distance $d$ is:

$$d = \sqrt{(2\lambda)^2 + (\lambda)^2 + (2\lambda)^2} = \sqrt{9\lambda^2} = 3|\lambda|$$

Substituting $\lambda = -1$, we obtain the final result:

**$d = 3|-1| = 3$ units.**

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Equation of Line $L$

Direction of the Path

Parameterizing Point $B$

Intersection Logic

Solving for $λ$

Calculating Distance $A B$

Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Pinning Down the Highway

Decoding the Path

The Intersection

Final Calculation

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 :

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 $(7, 10, 11)$ 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

If the distances of the point $(1, 2, a)$ from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and $L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ 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 :