Let l1 be the line in xy-plane with x and y intercepts 1/8 and 1/(4√2) respectively, and l2 be the l

Q: Wait, why can't I just calculate the distance between the two points I chose on the lines?

That is a very common trap! The points $\vec{a}_1$ and $\vec{a}_2$ you identified are simply *any* arbitrary points on the lines. The shortest distance is only achieved at the specific points where the connecting line segment is perpendicular to both lines simultaneously. The formula accounts for this by projecting the vector connecting *any* two points onto the common normal.

Q: How do I find the direction vector $\vec{b}$ from the line equation $8x + 4\sqrt{2}y = 1$?

Think of the direction vector as the vector that satisfies the equation's change. If you have $Ax + By = C$, the slope is $-A/B$. This means for a change of $1$ in $x$, $y$ changes by $-A/B$. Thus, the direction vector is proportional to $(1, -A/B, 0)$, or simply $(B, -A, 0)$. It is all about maintaining the ratio of change!

Q: Why is the final answer $51$ and not $\frac{1}{51}$?

This is the classic 'JEE trap'—not reading the final requirement! The question asks for $d^{-2}$, which is $\frac{1}{d^2}$. Since we found $d = \frac{1}{\sqrt{51}}$, then $d^2 = \frac{1}{51}$. Therefore, $d^{-2} = 51$. Always double-check what the question is asking for before you box your answer!

Question

Let $l_1$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4\sqrt{2}}$ respectively, and $l_2$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6\sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_1$ and $l_2$, then $d^{-2}$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of skew lines involves two lines in 3D space that are neither parallel nor intersecting. To find the shortest distance between them, we must translate their geometric descriptions into vector equations.

For line $l_1$ in the $xy$-plane, the intercepts are $\frac{1}{8}$ and $\frac{1}{4\sqrt{2}}$. Using the intercept form $\frac{x}{a} + \frac{y}{b} = 1$, we derive the equation:

$$8x + 4\sqrt{2}y = 1, \quad z=0$$

For line $l_2$ in the $zx$-plane, the intercepts are $-\frac{1}{8}$ and $-\frac{1}{6\sqrt{3}}$. This leads to the equation:

$$-8x - 6\sqrt{3}z = 1, \quad y=0$$

### The Vector Toolkit

To apply the shortest distance formula, we identify a point $\vec{a}$ and a direction vector $\vec{b}$ for each line. For $l_1$, the $x$-intercept provides $\vec{a}_1 = (\frac{1}{8}, 0, 0)$.

From the equation $8x + 4\sqrt{2}y = 1$, the slope in the $xy$-plane is $-\frac{8}{4\sqrt{2}} = -\sqrt{2}$. Thus, the direction vector is $\vec{b}_1 = \hat{i} - \sqrt{2}\hat{j}$.

For $l_2$, we have $\vec{a}_2 = (-\frac{1}{8}, 0, 0)$. By analyzing the coefficients of $x$ and $z$ in $-8x - 6\sqrt{3}z = 1$, we determine the direction vector $\vec{b}_2 = 3\sqrt{3}\hat{i} - 4\hat{k}$.

### The Common Normal

The shortest distance lies along the common normal, which is the cross product $\vec{n} = \vec{b}_1 	imes \vec{b}_2$. We calculate this using the determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -\sqrt{2} & 0 \ 3\sqrt{3} & 0 & -4 \end{vmatrix}$$

Expanding the determinant, we obtain $\vec{n} = 4\sqrt{2}\hat{i} + 4\hat{j} + 3\sqrt{6}\hat{k}$. The magnitude of this normal vector is:

$$|\vec{n}| = \sqrt{(4\sqrt{2})^2 + 4^2 + (3\sqrt{6})^2} = \sqrt{32 + 16 + 54} = \sqrt{102}$$

### Final Calculation

We apply the shortest distance formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot \vec{n}|}{|\vec{n}|}$. The vector connecting the two points is $\vec{a}_2 - \vec{a}_1 = -\frac{1}{4}\hat{i}$.

The dot product is $|(-\frac{1}{4}\hat{i}) \cdot (4\sqrt{2}\hat{i} + 4\hat{j} + 3\sqrt{6}\hat{k})| = |-\sqrt{2}| = \sqrt{2}$. Substituting these values into the distance formula:

$$d = \frac{\sqrt{2}}{\sqrt{102}} = \sqrt{\frac{2}{102}} = \frac{1}{\sqrt{51}}$$

The problem asks for the value of $d^{-2}$. Therefore, $d^{-2} = (\sqrt{51})^2 = \mathbf{51}$.

Enter Numerical Value:

Visualized Solution

Visualizing the 3D Setup

Equation of $l_{1}$ in $x y$ -plane

Vector Form of $l_{1}$

Equation of $l_{2}$ in $z x$ -plane

Vector Form of $l_{2}$

The Shortest Distance Formula

Calculating $a_{2} - a_{1}$

Cross Product $b_{1} \times b_{2}$

Magnitude of Cross Product

Calculating the Numerator

Finding Distance $d$

Final Answer: $d^{- 2}$

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Vector Toolkit

The Common Normal

Final Calculation

Similar Questions

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ 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

Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

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 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 :

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ 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 :

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ and be two lines. Then which of the following points lies on the line of the shortest distance between L₁ and $L_{2}$ ?

Enter Numerical Value:

Visualized Solution

Visualizing the 3D Setup

Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​ in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xy-plane

Vector Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​

Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​ in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } zx-plane

Vector Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​

The Shortest Distance Formula

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

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