Let a line having direction ratios 1, -4, 2 intersect the lines (x-7)/3 = (y-1)/-1 = (z+2)/1 and x/2

Q: Wait, why can't we just use the shortest distance formula between two skew lines?

That is a fantastic question! The shortest distance formula is specifically for the *perpendicular* distance between skew lines. In this problem, the line segment $AB$ is not necessarily the shortest distance; it is a specific line segment with a *given* direction ratio $(1, -4, 2)$. Therefore, we must find the specific points $A$ and $B$ that satisfy this condition.

Q: Why do we use two different parameters, $\lambda$ and $\mu$?

Because points $A$ and $B$ are independent. Point $A$ is constrained to move only along line $L_1$, and point $B$ is constrained to move only along line $L_2$. Using the same parameter would incorrectly imply that $A$ and $B$ are linked in a way that doesn't exist in the problem.

Q: Why is the proportionality constant $k$ not necessarily 1?

Direction ratios define the *direction* of a line, not its magnitude. If a line has direction ratios $(a, b, c)$, any vector parallel to it will have direction ratios $(ka, kb, kc)$ for some non-zero constant $k$. We equate the ratios to a constant to account for this scaling factor.

Question

Let a line having direction ratios $1, -4, 2$ intersect the lines $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}$ and $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}$ at the point $A$ and $B$. Then $(AB)^2$ is equal to ____.

Sigma Percentile · Accepted Answer

### The Geometry of Skew Lines

Imagine you are standing in a vast, three-dimensional space. You see two laser beams, $L_1$ and $L_2$, crossing each other but never touching. These are skew lines.

We are tasked with finding the square of the distance between two points, $A$ on $L_1$ and $B$ on $L_2$, such that the line segment $AB$ has a specific orientation defined by the direction ratios $(1, -4, 2)$. This is not just a calculation; it is a journey of finding order in the chaos of 3D space.

### Phase 1: The Parametric GPS

To navigate this space, we need a way to pinpoint any location on these lines. We use the parametric form.

For line $L_1: \frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}$, we introduce a parameter $\lambda$. By setting this equal to $\lambda$, we can express any point $A$ as $(3\lambda + 7, -\lambda + 1, \lambda - 2)$.

Similarly, for line $L_2: \frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}$, we use a different parameter $\mu$ to define point $B$ as $(2\mu, 3\mu + 7, \mu)$. Think of $\lambda$ and $\mu$ as the GPS coordinates for points $A$ and $B$ on their respective lines.

### Phase 2: The Vector Connection

Now that we have the coordinates for $A$ and $B$, we need to define the line segment $AB$. The direction ratios of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are simply the differences in their coordinates: $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

Applying this to our points, the direction ratios of $AB$ are:
$$(3\lambda - 2\mu + 7, -\lambda - 3\mu - 6, \lambda - \mu - 2)$$
This vector represents the 'compass heading' of our segment $AB$.

### Phase 3: The Proportionality Trap

The problem gives us the direction ratios of $AB$ as $(1, -4, 2)$. A common mistake is to assume these are equal to our calculated ratios.

However, direction ratios are proportional. We must set up the proportionality:
$$\frac{3\lambda - 2\mu + 7}{1} = \frac{-\lambda - 3\mu - 6}{-4} = \frac{\lambda - \mu - 2}{2}$$
This is the heart of the problem. By equating these ratios, we create a system of linear equations that will lock our points $A$ and $B$ into their correct positions.

### Phase 4: Solving the System

Let's break down the algebra. Taking the first two parts of the ratio and cross-multiplying, we get:
$$-12\lambda + 8\mu - 28 = -\lambda - 3\mu - 6$$
This simplifies beautifully to $\lambda - \mu + 2 = 0$.

Now, taking the second and third parts, we get:
$$-2\lambda - 6\mu - 12 = -4\lambda + 4\mu + 8$$
This simplifies to $\lambda - 5\mu - 10 = 0$.

Solving this system is straightforward: subtracting the second from the first yields $4\mu + 12 = 0$, so $\mu = -3$. Substituting this back gives $\lambda = -5$. We have found the exact parameters that define our points.

### Phase 5: The Final Distance

With $\lambda = -5$ and $\mu = -3$, we find the coordinates:
$$A = (-8, 6, -7) \quad 	ext{and} \quad B = (-6, -2, -3)$$

The final step is to calculate the square of the distance $(AB)^2 = (x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2$. Plugging in our values:
$$(AB)^2 = (-6 - (-8))^2 + (-2 - 6)^2 + (-3 - (-7))^2$$
$$(AB)^2 = (2)^2 + (-8)^2 + (4)^2 = 4 + 64 + 16 = 84$$

The elegance of the final result is a testament to the power of parametric geometry. The square of the distance is **84**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let a line having direction ratios $1, - 4, 2$ intersect the lines $\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1}$ and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}$ at the point $A$ and $B$ . Then $(A B)^{2}$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Parametric Form of Point $A$

Parametric Form of Point $B$

Direction Ratios of $A B$

Using the Given Direction Ratios

Forming the First Equation

Forming the Second Equation

Solving for $μ$

Solving for $λ$

Coordinates of Point $A$

Coordinates of Point $B$

Distance Formula Setup

Final Calculation

The Sigma Insight: Equation of a Line in Space

The Geometry of Skew Lines

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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