Let the values of p, for which the shortest distance between the lines (x+1)/3 = y/4 = z/5 and r = (

Q: Wait, why do we use the cross product here? Can't we just find the distance between two arbitrary points?

Great question! The distance between two arbitrary points changes as you move along the lines. We need the *shortest* distance, which only occurs when the connecting segment is perpendicular to both lines. That's why we need the common normal vector, derived from the cross product, to define the direction of that unique shortest path.

Q: Why does the latus rectum formula change? I thought it was always $\frac{2b^2}{a}$.

Ah, the classic trap! The formula $\frac{2b^2}{a}$ assumes the major axis is along the x-axis. When $b > a$, the ellipse is stretched vertically along the y-axis. In that case, the formula becomes $\frac{2a^2}{b}$. Always check which axis is the major one before applying the formula!

Q: Is there a risk of mixing up the direction vectors in the cross product?

Absolutely. The cross product is anti-commutative, meaning $\vec{b_1} \times \vec{b_2} = -(\vec{b_2} \times \vec{b_1})$. However, because we take the absolute value in the shortest distance formula, the sign of the cross product won't affect the final distance. Just be consistent in your calculation!

Question

Let the values of $p$, for which the shortest distance between the lines $\frac{x+1}{3} = \frac{y}{4} = \frac{z}{5}$ and $\vec{r} = (p\hat{i} + 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ is $\frac{1}{\sqrt{6}}$, be $a, b (a < b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is :-

Options:

(A) $9$

(B) $3/2$

(C) $2/3$

(D) $18$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of skew lines represents the elegant, three-dimensional architecture of space. These lines are like the contrails of two airplanes that never intersect and are not parallel.

To find the shortest distance between them, we must identify the length of the unique segment that acts as a bridge, perpendicular to both lines simultaneously.

### Extracting the DNA of the Lines

Every line in 3D space is defined by a point and a direction. For our first line, $L_1$, given by:

$$\frac{x+1}{3} = \frac{y}{4} = \frac{z}{5}$$

We extract the point $\vec{a_1} = (-1, 0, 0)$ and the direction vector $\vec{b_1} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

For the second line, $L_2$, given by $\vec{r} = (p\hat{i} + 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$, we identify the point $\vec{a_2} = (p, 2, 1)$ and the direction vector $\vec{b_2} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

The vector connecting these two lines is:

$$\vec{a_2} - \vec{a_1} = (p+1)\hat{i} + 2\hat{j} + \hat{k}$$

### The Common Normal

The shortest distance is measured along the common normal. To find the direction of this normal, we calculate the cross product $\vec{n} = \vec{b_1} 	imes \vec{b_2}$.

Setting up the determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8)$$

This simplifies to $\vec{n} = \hat{i} - 2\hat{j} + \hat{k}$. The magnitude of this normal vector is $|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.

### The Shortest Distance

The formula for the shortest distance is the projection of the connecting vector onto the common normal:

$$SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot \vec{n}|}{|\vec{n}|}$$

Substituting our values, we get:

$$\frac{|(p+1)(1) + (2)(-2) + (1)(1)|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$$

The $\sqrt{6}$ terms cancel out, leaving us with $|p + 1 - 4 + 1| = 1$, or $|p - 2| = 1$. This yields two possibilities: $p = 3$ or $p = 1$. Since $a < b$, we assign $a = 1$ and $b = 3$.

### The Ellipse

We are given the ellipse equation:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Substituting our values, we get $\frac{x^2}{1^2} + \frac{y^2}{3^2} = 1$. Because the denominator under $y^2$ is larger ($b > a$), this is a vertical ellipse.

The length of the latus rectum for a vertical ellipse is given by $\frac{2a^2}{b}$. Plugging in our values, we obtain:

$$\frac{2(1)^2}{3} = \frac{2}{3}$$

The final result is **$\frac{2}{3}$**.

Select Answer:

Visualized Solution

Introduction to Skew Lines

Extracting Line Parameters

Vector Joining the Points

Finding the Common Normal

Magnitude of the Normal Vector

The Shortest Distance Formula

Solving for p

Analyzing the Ellipse

Calculating Latus Rectum

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

Extracting the DNA of the Lines

The Common Normal

The Shortest Distance

The Ellipse

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