The shortest distance between the lines (x-2)/3 = (y+1)/2 = (z-6)/2 and (x-6)/3 = (1-y)/2 = (z+8)/0

Q: Wait, why can't I just use the distance formula between two points on the lines?

That is a common trap! The distance between two arbitrary points on the lines depends on which points you choose. The 'shortest distance' is a unique geometric property that only occurs along the common perpendicular. You must use the vector projection method to isolate this specific distance.

Q: Why did we have to change $1-y$ to $y-1$ in the second line?

Great catch. The standard symmetric form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. Notice the coefficients of $x, y,$ and $z$ must be positive one. If you have $1-y$, the coefficient of $y$ is $-1$. By rewriting it as $\frac{y-1}{-2}$, you ensure the direction vector is correctly extracted.

Q: What if the cross product $\vec{b}_1 \times \vec{b}_2$ turns out to be zero?

If the cross product of the direction vectors is the zero vector, it means the lines are parallel. In that case, the formula for skew lines would fail (division by zero). For parallel lines, you would instead find the distance from a point on one line to the other line using a different approach.

Question

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 ______.

Sigma Percentile · Accepted Answer

### The Geometry of the Void: Mastering Skew Lines

Imagine you are standing in a vast, three-dimensional void. Two lines are streaking through this space like contrails of high-speed jets.

They are not parallel, meaning they will never run side-by-side, yet they are also not intersecting. They are 'skew' lines—a concept that often feels abstract until you realize it is the most common state of lines in 3D space.

Today, we are going to bridge the gap between these two lines. We are going to find the shortest distance between them, a journey that requires us to be precise, vigilant, and mathematically elegant.

### The Trap of the Standard Form

Before we dive into the vectors, we must address a classic JEE trap. Look at the second line:

$$\frac{x-6}{3} = \frac{1-y}{2} = \frac{z+8}{0}$$

Many students rush through this, grabbing the direction vector as $(3, 2, 0)$. But stop! Look at the $y$-term. It is $1-y$, not $y-1$.

In the standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, the coefficient of the variable must be positive one. By factoring out a negative sign, we rewrite the term as $\frac{y-1}{-2}$.

Now, and only now, is our direction vector $\vec{b}_2 = 3\hat{i} - 2\hat{j} + 0\hat{k}$ correct. Precision is the hallmark of an engineer.

### The Vector Bridge

To find the shortest distance $d$, we need a bridge. Geometrically, this bridge is a line segment perpendicular to both lines.

We find this by taking the cross product of the two direction vectors, $\vec{b}_1 = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b}_2 = 3\hat{i} - 2\hat{j} + 0\hat{k}$. The cross product $\vec{b}_1 	imes \vec{b}_2$ gives us the normal vector, the direction of our bridge.

We set up the determinant:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 2 \ 3 & -2 & 0 \end{vmatrix}$$

Expanding this, we get $4\hat{i} + 6\hat{j} - 12\hat{k}$. This vector is the backbone of our calculation.

### The Final Projection

Now, we connect the two lines using a displacement vector between a point on $L_1$ (let us use $\vec{a}_1 = 2\hat{i} - \hat{j} + 6\hat{k}$) and a point on $L_2$ (let us use $\vec{a}_2 = 6\hat{i} + \hat{j} - 8\hat{k}$).

The vector connecting them is $\vec{a}_2 - \vec{a}_1 = 4\hat{i} + 2\hat{j} - 14\hat{k}$. The shortest distance is simply the projection of this connecting vector onto our normal vector.

We calculate the dot product of $(\vec{a}_2 - \vec{a}_1)$ and our normal vector $(4\hat{i} + 6\hat{j} - 12\hat{k})$:

$$| (4)(4) + (2)(6) + (-14)(-12) | = | 16 + 12 + 168 | = 196$$

Dividing this by the magnitude of the normal vector, which is $\sqrt{4^2 + 6^2 + (-12)^2} = \sqrt{196} = 14$, we arrive at our final answer:

$$d = \frac{196}{14} = 14$$

The math is clean, the logic is sound, and the distance is found. You have successfully navigated the void. The final answer is **14**.

Animated Solution for Mathematics - Three Dimensional Geometry: 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 ______.

Enter Numerical Value:

Visualized Solution

Visualizing Skew Lines

Extracting Data from $L_{1}$

The Standard Form Trap in $L_{2}$

The Shortest Distance Formula

Connecting the Points: $a_{2} - a_{1}$

Setting up the Cross Product

Evaluating the Cross Product

Magnitude of the Normal Vector

The Dot Product (Numerator)

Evaluating the Dot Product

Final Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of the Void

The Trap of the Standard Form

The Vector Bridge

The Final Projection

Similar Questions

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 $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

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 :

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

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

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______.

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 $x + 1 = 2 y = - 12 z$ and $x = y + 2 = 6 z - 6$ is

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ and be two lines. Then which of the following points lies on the line of the shortest distance between L₁ and $L_{2}$ ?