If the shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and x/1 = y/α = (z-5)/1 is 5/

Q: Why do we use the cross product to find the common normal?

The shortest distance vector must be perpendicular to both lines simultaneously. In 3D space, the cross product of two vectors is the only way to generate a third vector that is perpendicular to both of them. It is the geometric key to unlocking the direction of the shortest path.

Q: Can we use the distance formula for parallel lines here?

No, absolutely not! Parallel lines have proportional direction vectors. If you check the direction vectors $\vec{d}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{d}_2 = 1\hat{i} + \alpha\hat{j} + 1\hat{k}$, you will see they are not proportional for any value of $\alpha$. These are skew lines, which means they are neither parallel nor intersecting.

Q: Why did we square both sides of the equation?

We squared both sides to eliminate the square root in the denominator and the absolute value in the numerator. This transformed a complex radical equation into a standard quadratic equation, which is much easier to solve for $\alpha$.

Question

If the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x}{1} = \frac{y}{\alpha} = \frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

Options:

(A) $3/2$

(B) $-3/2$

(C) $3$

(D) $-3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You see two lines, $L_1$ and $L_2$, drifting through space like contrails of two jets. They do not intersect, and they are not parallel; these are skew lines.

Our mission is to find the shortest distance between them, a task that is a beautiful exercise in vector geometry.

### Identifying the Anchors

Every line in 3D space is defined by a point and a direction. For $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$, we can immediately see it passes through point $A(1, 2, 3)$.

For $L_2: \frac{x}{1} = \frac{y}{\alpha} = \frac{z-5}{1}$, it passes through point $B(0, 0, 5)$. These points are our anchors.

To bridge the gap between these lines, we define a vector $\vec{AB}$:

$$\vec{AB} = (0-1)\hat{i} + (0-2)\hat{j} + (5-3)\hat{k} = -\hat{i} - 2\hat{j} + 2\hat{k}$$

### The Common Normal

The shortest distance between these lines is the path that is perpendicular to both lines. We need a vector $\vec{n}$ that is perpendicular to both direction vectors $\vec{d}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{d}_2 = 1\hat{i} + \alpha\hat{j} + 1\hat{k}$.

We find this using the cross product $\vec{n} = \vec{d}_1 	imes \vec{d}_2$:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 1 & \alpha & 1 \end{vmatrix} = (3-4\alpha)\hat{i} + 2\hat{j} + (2\alpha-3)\hat{k}$$

This vector $\vec{n}$ represents the common normal to both lines.

### The Projection

The shortest distance $SD$ is the projection of our bridge vector $\vec{AB}$ onto the common normal $\vec{n}$. Mathematically, this is expressed as:

$$SD = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|}$$

Plugging in our values, the dot product $\vec{AB} \cdot \vec{n}$ becomes:

$$\vec{AB} \cdot \vec{n} = (-1)(3-4\alpha) + (-2)(2) + (2)(2\alpha-3) = 8\alpha - 13$$

The magnitude squared of the normal is:

$$|\vec{n}|^2 = (3-4\alpha)^2 + 2^2 + (2\alpha-3)^2 = 20\alpha^2 - 36\alpha + 22$$

### The Algebraic Finale

We are given $SD = \frac{5}{\sqrt{6}}$. Squaring both sides, we obtain:

$$\frac{(8\alpha-13)^2}{20\alpha^2 - 36\alpha + 22} = \frac{25}{6}$$

Cross-multiplying leads us to the following expansion:

$$6(64\alpha^2 - 208\alpha + 169) = 25(20\alpha^2 - 36\alpha + 22)$$

Simplifying this expression, we arrive at:

$$116\alpha^2 + 348\alpha - 464 = 0$$

Dividing by $116$, we get the elegant quadratic equation:

$$\alpha^2 + 3\alpha - 4 = 0$$

The sum of the possible values of $\alpha$ is given by $-b/a$, which results in **$-3$**. We have successfully conquered the geometry and the algebra.

Animated Solution for Mathematics - Three Dimensional Geometry: If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

Select Answer:

Visualized Solution

Identify Lines and Points

Direction Vectors

Connecting Vector $A B$

The Common Normal $n$

Calculating $n$

Shortest Distance Formula

Numerator: Dot Product $A B \cdot n$

Denominator: Magnitude Squared $∣ n ∣^{2}$

Equating and Squaring

Cross-Multiplication

Forming the Quadratic Equation

Sum of Possible Values

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

Identifying the Anchors

The Common Normal

The Projection

The Algebraic Finale

Similar Questions

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{3}$ , then the sum of all possible values of $λ$ is :

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If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{30}$ , then the largest possible value of $∣ λ ∣$ is equal to _________

Select Answer:

Visualized Solution

Identify Lines and Points

Direction Vectors

Connecting Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

The Common Normal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n