The line l1 passes through the point (2, 6, 2) and is perpendicular to the plane 2x + y - 2z = 10. T

Q: Wait, why can't we just find the intersection point of the two lines?

That is a fantastic question! In 3D space, two lines don't always intersect. These are called 'skew lines'—they are neither parallel nor intersecting. If you try to solve for an intersection point, you will find that no such point exists that satisfies both line equations simultaneously. That is exactly why we need the shortest distance formula!

Q: How do I know if the lines are parallel or skew?

Great intuition. Check the direction vectors $\vec{b}_1$ and $\vec{b}_2$. If one is a scalar multiple of the other, the lines are parallel. If they are not, they are either intersecting or skew. Since we are asked for the shortest distance, we assume they are skew and proceed with the formula.

Q: Does the order of subtraction in $\vec{a}_2 - \vec{a}_1$ matter?

Not at all! Because the formula uses the absolute value of the dot product, $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|$, the sign of the vector difference will be neutralized. Whether you calculate $\vec{a}_2 - \vec{a}_1$ or $\vec{a}_1 - \vec{a}_2$, the final distance will remain the same positive value.

Question

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$ is :

Options:

(A) 7

(B) $19/3$

(C) $19/3$

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have a plane, a flat, infinite sheet defined by the equation $2x + y - 2z = 10$.

Piercing through this plane at a perfect ninety-degree angle is a line, $l_1$. Somewhere else in this infinite space, another line, $l_2$, is floating, seemingly oblivious to the first.

Our mission is to find the shortest distance between these two lines. This is not just a calculation; it is a journey through the elegance of vector algebra.

### Decoding the Lines

First, we must understand our players. For line $l_1$, we are given a point $A(2, 6, 2)$.

The problem states $l_1$ is perpendicular to the plane $2x + y - 2z = 10$. In the language of vectors, the normal vector to the plane, $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$, is the key.

Since $l_1$ is perpendicular to the plane, its direction vector $\vec{b}_1$ must be parallel to this normal vector. Thus, $\vec{b}_1 = 2\hat{i} + \hat{j} - 2\hat{k}$.

With a point and a direction, we define $l_1$ as:

$$\frac{x-2}{2} = \frac{y-6}{1} = \frac{z-2}{-2}$$

Now, consider $l_2$, given by $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$. From this, we extract a point $\vec{a}_2 = -\hat{i} - 4\hat{j} + 0\hat{k}$ and a direction vector $\vec{b}_2 = 2\hat{i} - 3\hat{j} + 2\hat{k}$.

### The Shortest Distance Formula

Why do we use the formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$?

Think of it this way: the shortest distance between two skew lines is the length of the segment that is perpendicular to both lines. The vector $\vec{b}_1 	imes \vec{b}_2$ gives us the direction of this common perpendicular.

By projecting the vector connecting any two points on the lines, $\vec{a}_2 - \vec{a}_1$, onto this common perpendicular, we find the shortest distance.

### The Calculation

Let us execute this with precision. First, the difference vector:

$$\vec{a}_2 - \vec{a}_1 = (-1 - 2)\hat{i} + (-4 - 6)\hat{j} + (0 - 2)\hat{k} = -3\hat{i} - 10\hat{j} - 2\hat{k}$$

Next, we calculate the cross product $\vec{b}_1 	imes \vec{b}_2$:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 2 & -3 & 2 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(4 - (-4)) + \hat{k}(-6 - 2) = -4\hat{i} - 8\hat{j} - 8\hat{k}$$

The magnitude of this cross product is:

$$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$$

Finally, the dot product is:

$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (-3)(-4) + (-10)(-8) + (-2)(-8) = 12 + 80 + 16 = 108$$

### The Grand Finale

Substituting these values into our formula, we get:

$$d = \frac{|108|}{12} = 9$$

The shortest distance is exactly **9 units**. It is a beautiful result, isn't it?

Through the power of vectors, we have bridged the gap between two lines in 3D space. Keep practicing, and you will find that these problems are not obstacles, but stepping stones to mastering the language of the universe.

Animated Solution for Mathematics - Three Dimensional Geometry: The line $l_{1}$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2 x + y - 2 z = 10$ . Then the shortest distance between the line $l_{1}$ and the line $\frac{x + 1}{2} = \frac{y + 4}{- 3} = \frac{z}{2}$ is :

Select Answer:

Visualized Solution

Visualizing the 3D Geometry

Direction of Line $l_{1}$

Equation of Line $l_{1}$

Analyzing Line $l_{2}$

The Shortest Distance Formula

Calculating Vector $a_{2} - a_{1}$

Cross Product of Directions

Magnitude of Cross Product

Calculating the Dot Product

Final Calculation

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

Decoding the Lines

The Shortest Distance Formula

The Calculation

The Grand Finale

Similar Questions

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

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

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:

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 $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______.

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ 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 :

Select Answer:

Visualized Solution

Visualizing the 3D Geometry

Direction of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​

Analyzing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​

The Shortest Distance Formula

Calculating Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​−a1​

Cross Product of Directions

Magnitude of Cross Product

Calculating the Dot Product

Final Calculation

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

Decoding the Lines

The Shortest Distance Formula

The Calculation

The Grand Finale

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Direction of Line $l_{1}$

Equation of Line $l_{1}$

Analyzing Line $l_{2}$

Calculating Vector $a_{2} - a_{1}$

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:

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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:

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

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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:

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