The sum of all values of α, for which the shortest distance between the lines (x+1)/α = (y-2)/-1 = (

Q: Wait, why can't we just use the distance between the points $(-1, 2, 4)$ and $(0, 1, 1)$?

That is a classic trap! The distance between two specific points on the lines is not necessarily the shortest distance between the lines themselves. The shortest distance is the perpendicular distance, which requires the common perpendicular vector. Using the distance between arbitrary points will give you a value that depends on the parameters of the lines, not the constant shortest distance.

Q: Why can we cancel $3|\alpha|$? Is it possible $\alpha=0$?

Great observation! If $\alpha=0$, the original line equations would involve division by zero, which is undefined. Therefore, $\alpha$ cannot be zero. Since $\alpha \neq 0$, $|\alpha|$ is strictly positive, allowing us to safely cancel it from the numerator and denominator.

Q: Do I need to solve for $\alpha$ individually using the quadratic formula?

You absolutely could, but there is a more elegant way! The question asks for the sum of all values of $\alpha$. Once you have the quadratic equation $\alpha^2 + 6\alpha - 7 = 0$, you can use Vieta's formulas, which tell us the sum of the roots is $-b/a$. It saves time and reduces the chance of calculation errors.

Question

The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x+1}{\alpha} = \frac{y-2}{-1} = \frac{z-4}{-\alpha}$ and $\frac{x}{\alpha} = \frac{y-1}{2} = \frac{z-1}{2\alpha}$ is $\sqrt{2}$, is

Options:

(A) 6

(B) 8

(C) -8

(D) -6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a vast, three-dimensional space. You see two lines, $L_1$ and $L_2$, that refuse to meet and refuse to be parallel. They are skew lines, dancing around each other in the void.

Our mission is to find the parameter $\alpha$ that dictates their separation, specifically when the shortest distance between them is exactly $\sqrt{2}$. This is not just algebra; it is the study of how lines exist in space.

### Extracting the DNA

Every line has a signature. For $L_1$, the symmetric form

$$\frac{x+1}{\alpha} = \frac{y-2}{-1} = \frac{z-4}{-\alpha}$$

tells us it passes through $\vec{a}_1 = (-1, 2, 4)$ with direction $\vec{b}_1 = (\alpha, -1, -\alpha)$.

For $L_2$, the equation

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

reveals it passes through $\vec{a}_2 = (0, 1, 1)$ with direction $\vec{b}_2 = (\alpha, 2, 2\alpha)$. These vectors are the DNA of our lines.

### The Common Perpendicular

To find the shortest distance, we need a bridge—a vector perpendicular to both lines. We find this using the cross product: $\vec{b}_1 	imes \vec{b}_2$.

Calculating this determinant:

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

This vector, $(0, -3\alpha^2, 3\alpha)$, is the common perpendicular. Its magnitude is:

$$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{0^2 + (-3\alpha^2)^2 + (3\alpha)^2} = 3|\alpha|\sqrt{\alpha^2+1}$$

### The Scalar Triple Product

Now, we project the vector connecting the two points, $\vec{a}_2 - \vec{a}_1 = (1, -1, -3)$, onto this perpendicular. The numerator of our distance formula is the scalar triple product:

$$|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)| = |(1)(0) + (-1)(-3\alpha^2) + (-3)(3\alpha)| = |3\alpha^2 - 9\alpha|$$

### The Algebraic Dance

Putting it all together, the distance $d$ is:

$$\frac{|3\alpha^2 - 9\alpha|}{3|\alpha|\sqrt{\alpha^2+1}} = \sqrt{2}$$

We factor the numerator: $|3\alpha(\alpha-3)| = 3|\alpha||\alpha-3|$. Cancelling the $3|\alpha|$ term, we get:

$$\frac{|\alpha-3|}{\sqrt{\alpha^2+1}} = \sqrt{2}$$

Squaring both sides yields:

$$\frac{(\alpha-3)^2}{\alpha^2+1} = 2 \Rightarrow (\alpha-3)^2 = 2(\alpha^2+1)$$

Expanding this, we get $\alpha^2 - 6\alpha + 9 = 2\alpha^2 + 2$, leading to the quadratic:

$$\alpha^2 + 6\alpha - 7 = 0$$

### The Elegant Conclusion

We do not need to solve for $\alpha$ individually. By Vieta's formulas, the sum of the roots is $-b/a = -6/1 = -6$.

The beauty of this problem lies in how the complex geometry collapses into a simple, elegant quadratic equation. You have mastered the skew lines! The **sum of the possible values of $\alpha$ is $-6$**.

Animated Solution for Mathematics - Three Dimensional Geometry: The sum of all values of $α$ , for which the shortest distance between the lines $\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is

Select Answer:

Visualized Solution

Visualizing the Skew Lines

Extracting Parameters

The Shortest Distance Formula

Position Vector Difference

Computing the Cross Product

Magnitude of the Cross Product

Scalar Triple Product

Setting up the Distance Equation

Simplifying the Equation

Squaring Both Sides

Forming the Quadratic Equation

Finding the Sum of Values of $α$

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

Extracting the DNA

The Common Perpendicular

The Scalar Triple Product

The Algebraic Dance

The Elegant Conclusion

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