The shortest distance between the lines (x-3)/3 = (y-8)/-1 = (z-3)/1 and (x+3)/-3 = (y+7)/2 = (z-6)/

Q: Wait, why do we call these 'skew lines'? Can't we just find the point where they intersect?

That is a fantastic question! In 2D, any two non-parallel lines will eventually intersect. But in 3D space, lines have more freedom. Skew lines are special because they are not parallel, yet they exist in different planes, meaning they will never meet. If you try to solve for an intersection point by equating the coordinates, you will find that no such point exists. That is exactly why we need the concept of the 'shortest distance'—it is the only way to measure the gap between them.

Q: Does the order of the cross product $\vec{p} \times \vec{q}$ matter? What if I do $\vec{q} \times \vec{p}$?

Great observation! The cross product $\vec{q} \times \vec{p}$ is just the negative of $\vec{p} \times \vec{q}$. However, in our shortest distance formula, we take the absolute value of the dot product in the numerator and the magnitude of the normal vector in the denominator. Since the absolute value and the magnitude will remain the same regardless of the sign, the final result for the distance will be identical. So, you are safe either way!

Q: Why is the shortest distance formula a projection?

Think of the vector $\vec{AB}$ as a bridge connecting the two lines. This bridge is tilted. The shortest distance is the component of this bridge that lies perfectly along the direction of the common perpendicular. Mathematically, the projection of any vector $\vec{V}$ onto a vector $\vec{n}$ is given by $\frac{|\vec{V} \cdot \vec{n}|}{|\vec{n}|}$. Here, $\vec{AB}$ is our bridge, and $\vec{n}$ is the common perpendicular, so we are simply projecting the bridge onto the perpendicular direction.

Question

The shortest distance between the lines $\frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$ is :

Options:

(A) $2\sqrt{30}$

(B) $\frac{7}{2}\sqrt{30}$

(C) $3$

(D) $3\sqrt{30}$

Sigma Percentile · Accepted Answer

### The Geometry of Skew Lines: A 3D Odyssey

Imagine standing in a vast, three-dimensional space. You see two airplanes flying at different altitudes, their paths crossing in your field of vision, yet they never collide.

They are moving in different directions, at different heights. In the language of mathematics, these are skew lines. They are not parallel, and they do not intersect.

Today, we are going to calculate the shortest distance between two such lines, a problem that feels daunting but is actually a beautiful exercise in vector algebra.

### Phase 1: The Detective Work

Our journey begins by extracting the DNA of these lines. We are given:

$$L_1: \frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1}$$

$$L_2: \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$$

To work with these, we need a point on each line and their direction vectors. For $L_1$, comparing it to the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, we identify point $A = (3, 8, 3)$ and direction vector $\vec{p} = 3\hat{i} - \hat{j} + \hat{k}$.

For $L_2$, we must be vigilant with signs! The equation $\frac{x+3}{-3}$ implies $x - (-3)$, so our point $B$ is $(-3, -7, 6)$, and the direction vector is $\vec{q} = -3\hat{i} + 2\hat{j} + 4\hat{k}$. This is the foundation of our solution.

### Phase 2: The Bridge and the Perpendicular

Now, we need a bridge between these two lines. We define the vector $\vec{AB} = \vec{B} - \vec{A}$.

Calculating this, we get $\vec{AB} = (-3-3)\hat{i} + (-7-8)\hat{j} + (6-3)\hat{k}$, which simplifies to $\vec{AB} = -6\hat{i} - 15\hat{j} + 3\hat{k}$.

This vector connects our two lines, but it is not the shortest path. The shortest path must be perpendicular to both lines. To find this direction, we use the cross product: $\vec{n} = \vec{p} 	imes \vec{q}$.

Setting up the determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ -3 & 2 & 4 \end{vmatrix}$$

We expand it to find $\vec{n} = \hat{i}(-4 - 2) - \hat{j}(12 - (-3)) + \hat{k}(6 - 3)$, resulting in $\vec{n} = -6\hat{i} - 15\hat{j} + 3\hat{k}$.

### Phase 3: The Grand Finale

In a delightful twist, we notice that our normal vector $\vec{n}$ is identical to our bridge vector $\vec{AB}$. This simplifies our life significantly!

The shortest distance $SD$ is the projection of $\vec{AB}$ onto $\vec{n}$, given by:

$$SD = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|}$$

Since $\vec{AB} = \vec{n}$, the dot product $\vec{AB} \cdot \vec{n}$ is simply the magnitude squared, $|\vec{n}|^2$. Calculating the magnitude:

$$|\vec{n}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270}$$

The dot product is also $270$. Thus:

$$SD = \frac{270}{\sqrt{270}} = \sqrt{270}$$

Simplifying this, we get $\sqrt{9 	imes 30} = 3\sqrt{30}$. We have successfully navigated the 3D space to find the shortest distance.

The final result is **$3\sqrt{30}$ units**. Remember, math is not just about numbers; it is about visualizing the invisible connections in the world around us.

Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is :

Select Answer:

Visualized Solution

Visualizing Skew Lines

Extracting Data from Line $L_{1}$

Extracting Data from Line $L_{2}$

Finding Vector $A B$

The Common Perpendicular Vector

Setting up the Cross Product

Calculating the Cross Product

Magnitude of the Perpendicular Vector

The Dot Product Calculation

Final Shortest Distance Formula

Final Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Skew Lines

Phase 1

Phase 2

Phase 3

Similar Questions

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is:

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______.

The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}$ and $x + y + z + 1 = 0, 2 x - y + z + 3 = 0$ is:

The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = z$ and $\frac{7 - x}{2} = y - 2 = z - 6$ is

The shortest distance between lines $L_{1}$ and $L_{2}$ , where $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$ and $L_{2}$ is the line passing through the points $A (- 4, 4, 3), B (- 1, 6, 3)$ and perpendicular to the line $\frac{x - 3}{- 2} = \frac{y}{3} = \frac{z - 1}{1}$ , is

Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 3x−3​=−1y−8​=1z−3​ and −3x+3​=2y+7​=4z−6​ is :