List of top Mathematics Questions on Distance between Two Lines

In a school, the swimming pool is shaped as a cuboid, and the breadth of the pool is represented by line $\frac{-x+2}{-5} = y = z-3$. Two students A and B are swimming along the length of the pool.

Based on the above information, answer the following questions :

(iii) (b) If a bird flies across the top of the swimming pool with its path represented by $\frac{x+4}{2}=\frac{y+3}{8}=\frac{z-5}{-18}$, then find the shortest distance between its path and the path of student A.

Let $ P(\alpha, \beta, \gamma) $ be the point on the line \[ \frac{x-1}{2} = \frac{y+1}{-3} = z \] at a distance $ 4\sqrt{14} $ from the point $ (1,-1,0) $ and nearer to the origin. Then the shortest distance between the lines \[ \frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3} \quad \text{and} \quad \frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1} \] is equal to

Check whether the lines given by $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$ and $\dfrac{x-4}{5} = \dfrac{y-1}{2} = z$ are parallel or not. If parallel, find the distance between them, otherwise find their point of intersection, if the lines are intersecting.

The shortest distance between the lines \[ x = y + 2 = 6z - 6 \] and \[ x + 1 = 2y = -12z \] is:

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

The shortest distance between the lines $\overset{⃗}{r} = (4\hat{i}-\hat{j})+λ(\hat{i}+2\hat{j}-3\hat{k})$ and $\overset{⃗}{r} = (\hat{i}-\hat{j}+2\hat{k})+μ(\hat{i}+4\hat{j}-5\hat{k})$ is...

If $p$ is the shortest distance between the lines $\frac{x+1}{7} = \frac{y+1}{-6} = z+1$ and $\overset{⃗}{r} = (3\hat{i}+5\hat{j}+7\hat{k})+μ(\hat{i}-2\hat{j}+\hat{k})$ then $[p]$ is... ,(where $[\,.\,]$ denotes the greatest integer function.)

The values of p and q so that the line joining the points (7, p, 2) and (q, -2, 5) may be parallel to the line joining the points (2, -3, 5) and (-6, -15, 11) are

The lines $\frac{x-1}{-1} = \frac{y+2}{1} = \frac{z-3}{-2}$ and $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z+1}{-2}$ are

The shortest distance between the lines, where the first line passes through $(0,0,0)$ and $(2,0,3)$ and the second line passes through $(2,5,0)$ and $(0,4,0)$ is

Find the distance between the planes $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$.

Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

The shortest distance between the straight line \[ \vec{r} = 3\hat{i}+12\hat{j}+5\hat{k} +\lambda(2\hat{i}), \qquad \lambda\in\mathbb{R}, \] and the $x$-axis is:

The shortest distance between the lines $\vec{r}=\hat{i}+\hat{j}+3\hat{k}+\lambda(2\hat{i}+2\hat{j}+\hat{k})$ and $\vec{r}=(2\mu+1)\hat{i}+(2\mu-1)\hat{j}+(\mu+1)\hat{k}$ where $\lambda$ and $\mu$ are parameters, is

The shortest distance between the lines $\vec{r} = -\hat{i} + t\hat{k}, \; t \in \mathbb{R}$ and $\vec{r} = -\hat{j} + s\hat{i}, \; s \in \mathbb{R}$ is:

Determine the shortest distance between the lines $ \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4} $ and $ \dfrac{x-2}{3} = \dfrac{y-4}{4} = \dfrac{z-5}{5} $.

If the shortest distance between the lines $\frac{x - k}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$, then $k =$}

The shortest distance between the lines $\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})$ and $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})$ is

The lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$ are \dots}

If the shortest distance between the lines $\bar{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k}), \lambda \in \mathbb{R}, \alpha > 0$ and $\bar{r}_2 = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k}), \mu \in R$, is $9$, then the value of $\alpha$ is

The shortest distance between the point $ (2,3,4) $ and the line $ \dfrac{x-4}{-2}=\dfrac{y-4}{2}=\dfrac{z-6}{1} $ is

If the shortest distance between the lines $\bar{r}_1 = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k}), \lambda \in \mathbb{R}, \alpha > 0$ and $\bar{r}_2 = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k}), \mu \in R$, is $9$, then the value of $\alpha$ is

Shortest distance between the lines $ \vec{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k} $ and $ \vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k}) $ is

A line $L_1$ passing through the point $A$ with position vector $ \vec{a} = 4\hat{i} + 2\hat{j} + 2\hat{k} $ is parallel to the vector $ \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} $. The length of the perpendicular drawn from a point $P$ with position vector $ \vec{p} = \hat{i} + 2\hat{j} + 3\hat{k} $ to $L_1$ is

The shortest distance between the lines $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$ is :