The shortest distance between the lines (x-3)/4 = (y+7)/-11 = (z-1)/5 and (x-5)/3 = (y-9)/-6 = (z+2)

Q: Wait, why can't we just find the intersection point and say the distance is zero?

That is a fantastic question! In 2D, lines almost always intersect. But in 3D, lines can be 'skew.' Skew lines are like two highways on different levels of a flyover—they never touch, and they aren't parallel. Because they don't intersect, the shortest distance is not zero; it is the length of the bridge connecting them at their closest point.

Q: Does the order of $\vec{a}_1$ and $\vec{a}_2$ matter when calculating $\vec{a}_2 - \vec{a}_1$?

Not at all! Whether you calculate $\vec{a}_2 - \vec{a}_1$ or $\vec{a}_1 - \vec{a}_2$, the resulting vector will just be the negative of the other. Since the formula for the shortest distance uses the absolute value of the dot product, the negative sign will be cancelled out. The distance remains the same.

Q: Why do we use the cross product $\vec{b}_1 \times \vec{b}_2$?

The cross product is our geometric 'perpendicularity machine.' By definition, the cross product of two vectors produces a third vector that is orthogonal (perpendicular) to both. Since we need a line segment that is perpendicular to both skew lines, the cross product of their direction vectors gives us the exact direction we need for that segment.

Question

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:

Options:

(A) $\frac{178}{\sqrt{563}}$

(B) $\frac{187}{\sqrt{563}}$

(C) $\frac{185}{\sqrt{563}}$

(D) $\frac{179}{\sqrt{563}}$

Sigma Percentile · Accepted Answer

### The Geometry of the Infinite: Conquering Skew Lines

Imagine you are standing in a vast, three-dimensional void. You see two infinite lines stretching out into the distance. They are not parallel, yet they never cross.

These are skew lines, and they represent one of the most elegant challenges in 3D geometry. Today, we are going to find the shortest distance between them using the vector tools that define our universe.

### Phase 1: Extracting the DNA of the Lines

Every line in 3D space has a unique signature: a point it passes through and a direction it follows.

For our first line, $L_1: \frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5}$, we extract a point $\vec{a}_1 = (3, -7, 1)$ and a direction vector $\vec{b}_1 = (4, -11, 5)$.

For our second line, $L_2: \frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1}$, we find a point $\vec{a}_2 = (5, 9, -2)$ and a direction vector $\vec{b}_2 = (3, -6, 1)$.

Notice how we carefully flip the signs in the numerators to find the coordinates of the points. This is the foundation of our journey.

### Phase 2: The Bridge Between Worlds

To find the shortest distance, we must bridge the gap between these two lines. We start by creating a vector that connects a point on $L_1$ to a point on $L_2$:

$$\vec{a}_2 - \vec{a}_1 = (5-3, 9-(-7), -2-1) = (2, 16, -3)$$

This vector represents the displacement between our two starting points. Now, we need the direction of the shortest path, which must be perpendicular to both lines.

We invoke the power of the cross product: $\vec{n} = \vec{b}_1 	imes \vec{b}_2$. We set up our determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -11 & 5 \ 3 & -6 & 1 \end{vmatrix}$$

Expanding this, we calculate the components:
*   $\hat{i}$ component: $(-11)(1) - (5)(-6) = -11 + 30 = 19$
*   $\hat{j}$ component: $-( (4)(1) - (5)(3) ) = -(4 - 15) = 11$
*   $\hat{k}$ component: $(4)(-6) - (-11)(3) = -24 + 33 = 9$

Thus, our normal vector is $\vec{n} = 19\hat{i} + 11\hat{j} + 9\hat{k}$.

### Phase 3: The Final Projection

The shortest distance is the projection of our connecting vector $\vec{a}_2 - \vec{a}_1$ onto the normal vector $\vec{n}$. The formula is:

$$S.D. = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot \vec{n}|}{|\vec{n}|}$$

First, let's find the magnitude of the normal vector:

$$|\vec{n}| = \sqrt{19^2 + 11^2 + 9^2} = \sqrt{361 + 121 + 81} = \sqrt{563}$$

Next, we calculate the dot product:

$$|(\vec{a}_2 - \vec{a}_1) \cdot \vec{n}| = |(2)(19) + (16)(11) + (-3)(9)| = |38 + 176 - 27| = 187$$

Putting it all together, the shortest distance is:

$$\mathbf{S.D. = \frac{187}{\sqrt{563}}}$$

This is the precise, calculated gap between two infinite paths in space. You have successfully navigated the 3D landscape and emerged with the correct answer.

Animated Solution for Mathematics - Three Dimensional Geometry: 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:

Select Answer:

Visualized Solution

Visualizing the Skew Lines

Extracting Parameters for Line $1$

Extracting Parameters for Line $2$

Vector Joining the Two Points

The Common Perpendicular Direction

Setting up the Determinant

Calculating the $\hat{i}$ component

Calculating the $\hat{j}$ component

Calculating the $\hat{k}$ component

Magnitude of the Normal Vector

The Shortest Distance Formula

Calculating the Dot Product

Final Result

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of the Infinite

Phase 1

Phase 2

Phase 3

Similar Questions

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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 4x−3​=−11y+7​=5z−1​ and 3x−5​=−6y−9​=1z+2​ is:

Select Answer:

Visualized Solution

Visualizing the Skew Lines

Extracting Parameters for Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1

Extracting Parameters for Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2