Let a line passing through the point (-1, 2, 3) intersect the lines L1 : (x-1)/3 = (y-2)/2 = (z+1)/-

Q: Wait, why do we need two different parameters, $\lambda$ and $\mu$? Can't we just use one?

That is a fantastic question! Because $M$ and $N$ lie on two distinct, independent lines in 3D space, their positions are not linked by the same parameter. If you used only one parameter, you would be forcing $M$ and $N$ to move in a synchronized way that doesn't exist in this problem. Using $\lambda$ for $L_1$ and $\mu$ for $L_2$ allows them to be anywhere on their respective lines, which is exactly what we need to find the specific points that satisfy the collinearity condition.

Q: Why can't we just use the distance formula to solve this?

The distance formula is a powerful tool, but it often introduces square roots that make algebraic manipulation messy and prone to error. In 3D geometry, especially with lines, the vector approach—using direction ratios and proportionality—is almost always more elegant and direct. It keeps the equations linear and much easier to solve.

Q: Is there a faster way to find the sum of the coordinates without finding $\alpha, \beta, \gamma$ and $a, b, c$ individually?

Yes, absolutely! Notice that the sum of the coordinates is a linear function of the parameter. By summing the parametric expressions for $x, y,$ and $z$ first, you get a single expression in terms of $\lambda$ (or $\mu$). This saves you from calculating each coordinate individually until the very last step, reducing the chance of arithmetic errors.

Question

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1 : \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2 : \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$. Then the value of $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty room. You see a point $P(-1, 2, 3)$ suspended in the air. Two lines, $L_1$ and $L_2$, stretch out across the room like beams of light.

A third line, invisible but real, shoots out from point $P$, piercing through $L_1$ at a point $M$ and $L_2$ at a point $N$. Because $P, M,$ and $N$ all lie on this single, straight line, they are collinear. This simple fact is our key to unlocking the entire puzzle.

### The Power of Parametrization

To conquer this, we must first tame the lines. We use parameters to describe the positions of $M$ and $N$.

For $L_1$, we set:
$$\frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2} = \lambda$$

This allows us to write the coordinates of $M(\alpha, \beta, \gamma)$ as:
$$M = (3\lambda+1, 2\lambda+2, -2\lambda-1)$$

Similarly, for $L_2$, we set:
$$\frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4} = \mu$$

This gives us the coordinates of $N(a, b, c)$ as:
$$N = (-3\mu-2, -2\mu+2, 4\mu+1)$$

### The Elegance of Summation

The problem asks for the ratio of the squares of the sums of the coordinates. Instead of finding each coordinate individually, let's be clever.

The sum of the coordinates of $M$ is:
$$\alpha + \beta + \gamma = (3\lambda+1) + (2\lambda+2) + (-2\lambda-1) = 3\lambda + 2$$

Similarly, for $N$, the sum is:
$$a + b + c = (-3\mu-2) + (-2\mu+2) + (4\mu+1) = -\mu + 1$$

By working with these sums directly, we have already simplified our target expression significantly.

### The Vector Bridge

Now, we invoke the collinearity condition. If $P, M,$ and $N$ are collinear, the vectors $\vec{PM}$ and $\vec{PN}$ must be parallel.

We calculate the vectors:
$$\vec{PM} = M - P = (3\lambda+2, 2\lambda, -2\lambda-4)$$
$$\vec{PN} = N - P = (-3\mu-1, -2\mu, 4\mu-2)$$

The condition for these vectors to be parallel is that the ratios of their components are equal:
$$\frac{3\lambda+2}{-3\mu-1} = \frac{2\lambda}{-2\mu} = \frac{-2\lambda-4}{4\mu-2}$$

### Solving the System

Look at the middle ratio: $\frac{2\lambda}{-2\mu} = -\frac{\lambda}{\mu}$. Equating the first ratio to this, we get:
$$\frac{3\lambda+2}{-3\mu-1} = -\frac{\lambda}{\mu}$$

Cross-multiplying yields $\mu(3\lambda+2) = -\lambda(-3\mu-1)$, which simplifies to $3\lambda\mu + 2\mu = 3\lambda\mu + \lambda$, or simply $\lambda = 2\mu$.

Now, substitute $\lambda = 2\mu$ into the ratio $-\frac{\lambda}{\mu} = \frac{-2\lambda-4}{4\mu-2}$:
$$-2 = \frac{-2(2\mu)-4}{4\mu-2}$$

This simplifies to $-2(4\mu-2) = -4\mu-4$. Solving this linear equation, we find $\mu = 2$, and consequently, $\lambda = 4$.

### Final Calculation

With $\lambda = 4$ and $\mu = 2$, we return to our sums. The sum for $M$ is $3(4) + 2 = 14$. The sum for $N$ is $-(2) + 1 = -1$.

The problem asks for the ratio of the squares of these sums:
$$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} = \frac{14^2}{(-1)^2} = \frac{196}{1} = 196$$

We have navigated the 3D space, used the power of parameters, and arrived at the solution with precision. The final answer is **196**.

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Parametrizing Point $M$ on $L_{1}$

Sum of Coordinates for $M$

Parametrizing Point $N$ on $L_{2}$

Sum of Coordinates for $N$

Vector $P M$ Calculation

Vector $P N$ Calculation

The Collinearity Condition

Simplifying the Ratios

Solving for $μ$ and $λ$

Calculating the Final Sums

The Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Power of Parametrization

The Elegance of Summation

The Vector Bridge

Solving the System

Final Calculation

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