List of top Mathematics Questions on Distance between Two Lines asked in JEE Main

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

The shortest distance between the lines
\[\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}\]and
\[\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}\] is:

If the shortest distance between the lines.
L1: $\vec{r} = (2 + \lambda)\hat{i} + (1 - 3\lambda)\hat{j} + (3 + 4\lambda)\hat{k}$, $\lambda \in \mathbb{R}$.
L2: $\vec{r} = 2(1 + \mu)\hat{i} + 3(1 + \mu)\hat{j} + (5 + \mu)\hat{k}$, $\mu \in \mathbb{R}$ is $\frac{m}{\sqrt{n}}$, where gcd(m, n) = 1, then the value of m + n equals.

If the shortest distance between the lines \[ \frac{x - \lambda}{3} = \frac{y - 2}{-1} = \frac{z - 1}{1} \] and \[ \frac{x + 2}{-3} = \frac{y + 5}{2} = \frac{z - 4}{4} \] is \[ \frac{44}{\sqrt{30}}, \] then the largest possible value of $|\lambda|$ is equal to ________.

The shortest distance between the line:
\[ \frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5} \] and \[ \frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1} \] is:

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

Let the line of the shortest distance between the lines \(L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})\)and  \(L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})\) intersect \(L_1\) and \(L_2\) at \(P\) and \(Q\), respectively.  If \((\alpha, \beta, \gamma)\) is the midpoint of the line segment \(PQ\), then \(2(\alpha + \beta + \gamma)\) is equal to \(\_\_\_\_\).

If the shortest distance between the lines \(\frac{x - \lambda}{-2} = \frac{y - 2}{1} = \frac{z - 1}{1}\) and \(\frac{x - \sqrt{3}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1}\) is 1, then the sum of all possible values of \( \lambda \) is:

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

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 \(\alpha\), then \(28 \alpha^2\) is equal to

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 $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $\frac{7-x}{2}=y-2=z-6$ is

Let the line \(\frac{x}{1}=\frac{6−y}{2}=\frac{z+8}{5}\)  intersect the lines  \(\frac{x-5}{1}=\frac{y-7}{3}=\frac{z+2}{1}\) and \(\frac{x+3}{6}=\frac{3−y}{3}=\frac{z-6}{1}\) at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2x – 2y + z = 14 is 

Let the plane P : 4x – y + z = 10 be rotated by an angle \(\frac{π}{2}\) about its line of intersection with the plane x + y – z = 4. If α is the distance of the point (2, 3, -4) from the new position of the plane P, then 35α is

The distance of the point ( -1,2,3) from the plane \(\overrightarrow{r}( \^i−2 \^j+3 \^k )=10\) parallel to the line of the shortest distance between the lines \(\overrightarrow{r}=(\^i=\^j)+λ(2\^i+\^k)\) and  \(\overrightarrow{r}=(2\^i=\^j)+μ(\^i+\^j+\^k)\)  is 

Let the image of the point \(\left(\frac{5}{3},\frac{5}{3},\frac{8}{3}\right)\) in the plane x - 2y + z - 2 = 0 be P. If the distance of the point Q(6,-2,α), α > 0, from P is 13, then α is equal to___.

Let the equations of two adjacent sides of a parallelogram ABCD be 2x – 3y = –23 and 5x + 4y = 23. If the equation of its one diagonal AC is 3x + 7y = 23 and the distance of A from the other diagonal is d, then 50 d2 is equal to____.

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

Two lines passing through (2, 3) parallel to coordinate axes. A circle of unit radius touches both the lines and lies on the origin side. Then the shortest distance of point (5,5) from the circle is: