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

Q: Why do we use the cross product $\vec{b}_1 \times \vec{b}_2$ in the shortest distance formula?

The cross product $\vec{b}_1 \times \vec{b}_2$ gives us a vector that is perpendicular to both lines. Since the shortest distance must be measured along a line perpendicular to both, projecting the vector connecting any two points on the lines onto this common perpendicular direction gives us exactly that distance.

Q: How do I know where to split the integral of $[x^2]$?

The greatest integer function $[f(x)]$ changes its value whenever $f(x)$ hits an integer. Since we are integrating $[x^2]$ from $0$ to $1.5$, we look for values of $x$ where $x^2 = 1, 2, 3...$ within that range. Here, $x^2=1$ at $x=1$ and $x^2=2$ at $x=\sqrt{2}$.

Q: Can I just pull the constant out of the greatest integer function?

Absolutely not! The greatest integer function is not linear. You must treat it as a piecewise constant function. Always evaluate the value of the expression inside the brackets for each sub-interval first, then integrate that constant.

Q: Is it always safe to assume $\alpha = 2$ by comparing $\alpha - \sqrt{\alpha} = 2 - \sqrt{2}$?

In this specific context, yes. Since $\alpha - \sqrt{\alpha}$ is a strictly increasing function for $\alpha \ge 1$, the solution is unique. Comparing the rational and irrational parts is a standard and valid technique here.

Q: What if the lines were parallel? Would the formula still work?

Great catch! If the lines were parallel, $\vec{b}_1 \times \vec{b}_2$ would be the zero vector, making the formula undefined. For parallel lines, you would instead use the distance from a point on one line to the other line using the cross product of the direction vector and the position vector difference.

Question

If the shortest distance between the lines $\frac{x+2}{2} = \frac{y+3}{3} = \frac{z-5}{4}$ and $\frac{x-3}{1} = \frac{y-2}{-3} = \frac{z+4}{2}$ is $\frac{38}{3\sqrt{5}}k$, and $\int_{0}^{k} [x^2] dx = \alpha - \sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^3$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You see two lines, 
$$\frac{x+2}{2} = \frac{y+3}{3} = \frac{z-5}{4} \quad 	ext{and} \quad \frac{x-3}{1} = \frac{y-2}{-3} = \frac{z+4}{2}$$

They do not intersect and are not parallel. These are **skew lines**, and our mission is to find the shortest bridge between them.

### The Common Perpendicular

To find the distance between these lines, we need a vector perpendicular to both. We extract the direction vectors $\vec{b}_1 = (2, 3, 4)$ and $\vec{b}_2 = (1, -3, 2)$.

By calculating their cross product, $\vec{b}_1 	imes \vec{b}_2$, we find the direction of the bridge:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 1 & -3 & 2 \end{vmatrix} = 18\hat{i} - 9\hat{k} = 9(2\hat{i} - \hat{k})$$

The magnitude is $|\vec{b}_1 	imes \vec{b}_2| = 9\sqrt{5}$. We define the vector connecting a point on the first line, $\vec{a}_1 = (-2, -3, 5)$, to a point on the second, $\vec{a}_2 = (3, 2, -4)$, which is $\vec{a}_2 - \vec{a}_1 = (5, 5, -9)$.

The shortest distance $d$ is the projection of this vector onto our common perpendicular:

$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|} = \frac{|(5, 5, -9) \cdot (18, 0, -9)|}{9\sqrt{5}} = \frac{171}{9\sqrt{5}} = \frac{19}{\sqrt{5}}$$

Equating this to the given $\frac{38}{3\sqrt{5}}k$, we find **$k = \frac{3}{2}$**.

### Taming the Greatest Integer Integral

Now, we evaluate $I = \int_{0}^{3/2} [x^2] dx$. The greatest integer function $[x^2]$ acts like a staircase, jumping at specific values.

Since $x$ ranges from $0$ to $1.5$, $x^2$ ranges from $0$ to $2.25$. The jumps occur at $x^2 = 1$ (so $x=1$) and $x^2 = 2$ (so $x=\sqrt{2}$).

We split the integral into three distinct regions:

1. For $x \in [0, 1)$, $[x^2] = 0$.
2. For $x \in [1, \sqrt{2})$, $[x^2] = 1$.
3. For $x \in [\sqrt{2}, 1.5]$, $[x^2] = 2$.

This transforms our integral into a sum of areas:

$$I = \int_{0}^{1} 0 dx + \int_{1}^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{3/2} 2 dx$$

Evaluating these, we get $0 + (\sqrt{2} - 1) + 2(\frac{3}{2} - \sqrt{2}) = 2 - \sqrt{2}$.

### The Final Synthesis

We are given $\alpha - \sqrt{\alpha} = 2 - \sqrt{2}$. By inspection, it is clear that **$\alpha = 2$**.

The final step is a simple calculation:

$$6\alpha^3 = 6(2^3) = 6 	imes 8 = \mathbf{48}$$

Enter Numerical Value:

Visualized Solution

Identify Line Parameters

Shortest Distance Formula

Calculate Vector Difference $a_{2} - a_{1}$

Find Common Perpendicular $b_{1} \times b_{2}$

Magnitude of Cross Product

Calculate Shortest Distance $d$

Solve for Constant $k$

Set up the Integral

Identify Breakpoints for $[x^{2}]$

Split the Integral

Evaluate Piecewise Integral

Solve for $α$ and Final Answer

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Common Perpendicular

Taming the Greatest Integer Integral

The Final Synthesis

Similar Questions

If the shortest distance between the straight lines $3 (x - 1) = 6 (y - 2) = 2 (z - 1)$ and $4 (x - 2) = 2 (y - λ) = (z - 3), λ \in R$ is $\frac{1}{38}$ , then the integral value of $λ$ is equal to :

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

If the shortest distance between the line $r = (- \hat{i} + 3 \hat{k}) + λ (\hat{i} - a \hat{j})$ and $r = (- \hat{j} + 2 \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$ is $\frac{2}{3}$ , then the integral value of $a$ is equal to

Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $330$ . Then the positive value of $5 α + β$ is

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

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 :

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

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 _________

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______.

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

Enter Numerical Value:

Visualized Solution

Identify Line Parameters

Shortest Distance Formula

Calculate Vector Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​−a1​

Find Common Perpendicular .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​×b2​

Magnitude of Cross Product

Calculate Shortest Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Solve for Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Set up the Integral

Identify Breakpoints for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x2]

Split the Integral

Evaluate Piecewise Integral

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and Final Answer