Consider the lines L₁ : (x+1)/3 = (y+2)/1 = (z+1)/2 and L₂ : (x-2)/1 = (y+2)/2 = (z-3)/3.

1. The un

Q: Why do we use the cross product here? Can't we just use the dot product?

The dot product tells us about angles and projections, but the cross product is the geometric tool for finding a vector that is mutually perpendicular to two others. Since we need a direction orthogonal to both lines, the cross product is our only path.

Q: Why is the shortest distance the projection of the connecting vector onto the normal?

Imagine the normal vector as a bridge between the two lines. The shortest path between two skew lines must be perpendicular to both. Therefore, the length of this 'bridge' is simply the component of the vector connecting the lines that lies along the direction of the common normal.

Q: How do I know which point to use for the plane equation?

The plane passes through a point on one of the lines. You can pick any point on $L_1$ or $L_2$ that lies on the plane. Since the problem specifies the plane passes through the point $(-1, -2, -1)$ from $L_1$, we use that as our anchor point $(x_1, y_1, z_1)$.

Question

Passage:
Consider the lines $L_1 : \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2}$ and $L_2 : \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3}$.

Question 1:
The unit vector perpendicular to both $L_1$ and $L_2$ is
(A) $\frac{-\hat{i}+7\hat{j}+7\hat{k}}{\sqrt{99}}$
(B) $\frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$
(C) $\frac{-\hat{i}+7\hat{j}+5\hat{k}}{5\sqrt{3}}$
(D) $\frac{7\hat{i}-7\hat{j}-\hat{k}}{\sqrt{99}}$

Question 2:
The shortest distance between $L_1$ and $L_2$ is
(A) 0
(B) $\frac{17}{\sqrt{3}}$
(C) $\frac{41}{5\sqrt{3}}$
(D) $\frac{17}{5\sqrt{3}}$

Question 3:
The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_1$ and $L_2$ is
(A) $\frac{2}{\sqrt{75}}$
(B) $\frac{7}{\sqrt{75}}$
(C) $\frac{13}{\sqrt{75}}$
(D) $\frac{23}{\sqrt{75}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty room. Two long, thin rods are suspended in the air, not touching, not parallel, just drifting in their own orientations. These are skew lines.

In the world of JEE Advanced, these lines are defined by their direction vectors. Look at the equations:

$$L_1 : \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2}$$

$$L_2 : \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3}$$

From the denominators, we extract the direction vectors: $\vec{b_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b_2} = \hat{i} + 2\hat{j} + 3\hat{k}$. These vectors are the DNA of the lines; they tell us exactly where they are heading.

### The Common Perpendicular

Now, we seek a vector $\vec{n}$ that is perpendicular to both. This is the 'common perpendicular.' The cross product is our magic wand here.

We set up the determinant:

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

Expanding this carefully, we get $\vec{n} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1)$, which simplifies to $\vec{n} = -\hat{i} - 7\hat{j} + 5\hat{k}$. This vector is the backbone of our geometry.

To find the unit vector, we simply divide by the magnitude:

$$|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{75} = 5\sqrt{3}$$

Thus, the unit vector is $\frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$.

### The Shortest Bridge

How do we measure the gap between these two rods? We take a point $A_1(-1, -2, -1)$ on $L_1$ and $A_2(2, -2, 3)$ on $L_2$.

The vector connecting them is $\vec{a_2} - \vec{a_1} = 3\hat{i} + 0\hat{j} + 4\hat{k}$. The shortest distance is the projection of this connecting vector onto our normal vector $\vec{n}$.

Using the formula $d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot \vec{n}|}{|\vec{n}|}$, we calculate the dot product:

$$|(3\hat{i} + 4\hat{k}) \cdot (-\hat{i} - 7\hat{j} + 5\hat{k})| = |-3 + 0 + 20| = 17$$

Dividing by the magnitude $5\sqrt{3}$, we find the distance is **$\frac{17}{5\sqrt{3}}$**.

### The Plane of Existence

Finally, we construct a plane containing $L_1$ and parallel to $L_2$. The plane passes through $A_1(-1, -2, -1)$ and has the normal vector $\vec{n} = -\hat{i} - 7\hat{j} + 5\hat{k}$.

Using the equation $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$, we get:

$$-1(x+1) - 7(y+2) + 5(z+1) = 0$$

This simplifies to the plane equation $x + 7y - 5z + 10 = 0$.

To find the distance of point $P(1, 1, 1)$ from this plane, we use the standard formula:

$$D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2+b^2+c^2}}$$

Substituting the values, we get:

$$D = \frac{|1(1) + 7(1) - 5(1) + 10|}{\sqrt{1^2 + 7^2 + (-5)^2}} = \frac{13}{\sqrt{75}} = \frac{13}{5\sqrt{3}}$$

You have just navigated the complex terrain of 3D geometry. Keep this intuition alive—it is the key to mastering JEE Advanced.

Animated Solution for Mathematics - Three Dimensional Geometry: Comprehension Passage

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

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The shortest distance between $L_{1}$ and $L_{2}$ is

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The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

Select Answer:

Visualized Solution

Identify Direction Vectors $b_{1}$ and $b_{2}$

Cross Product Setup for Normal Vector

Evaluate the Cross Product

Calculate the Unit Vector $\overset{n}{^}$

Identify Points and Connecting Vector

Shortest Distance Formula

Calculate Shortest Distance

Setup Equation of the Plane

Derive Cartesian Equation of Plane

Distance of Point from Plane Setup

Final Distance Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Common Perpendicular

The Shortest Bridge

The Plane of Existence

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

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 $L_{2}$ ?

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 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 :

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

Animated Solution for Mathematics - Three Dimensional Geometry: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between L1​ and L2​ is

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines L1​ and L2​ is

Select Answer:

Visualized Solution

Identify Direction Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2​​

Cross Product Setup for Normal Vector

Evaluate the Cross Product

Calculate the Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n^

Identify Points and Connecting Vector

Shortest Distance Formula

Calculate Shortest Distance

Setup Equation of the Plane

Derive Cartesian Equation of Plane

Distance of Point from Plane Setup

Final Distance Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Common Perpendicular

The Shortest Bridge

The Plane of Existence

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 0x−1​=−1y+1​=1z​ and x+y+z+1=0,2x−y+z+3=0 is:

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The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

Identify Direction Vectors $b_{1}$ and $b_{2}$

Calculate the Unit Vector $\overset{n}{^}$

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 $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}$ and $x + y + z + 1 = 0, 2 x - y + z + 3 = 0$ is:

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