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
- $7 \sqrt{3}$
- $6 \sqrt{3}$
- $4 \sqrt{3}$
- $5 \sqrt{3}$
The Correct Option is B
Solution and Explanation
To find the shortest distance between two skew lines in space, we can use the formula for the shortest distance between two lines \(L_1\) and \(L_2\) given by their vector equations:
- Line \(L_1\): \(\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}\)
- Line \(L_2\): \(\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}\)
Step 1: Find the direction vectors of the lines
The direction vector of line \(L_1\) is \(\mathbf{a_1} = \langle 1, 2, -3 \rangle\), and for line \(L_2\) the direction is \(\mathbf{a_2} = \langle 1, 4, -5 \rangle\).
Step 2: Find a vector between a point on each line
A point on line \(L_1\) can be taken as \(\mathbf{P_1} = (5, 2, 4)\) and on line \(L_2\) as \(\mathbf{P_2} = (-3, -5, 1)\). The vector between these two points is:
\(\mathbf{b} = \mathbf{P_2} - \mathbf{P_1} = \langle -3 - 5, -5 - 2, 1 - 4 \rangle = \langle -8, -7, -3 \rangle\)
Step 3: Calculate the cross product of the direction vectors
The cross product \(\mathbf{a_1} \times \mathbf{a_2}\) is:
\(\mathbf{n} = \mathbf{a_1} \times \mathbf{a_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix} = \langle (2)(-5) - (-3)(4), -((1)(-5) - (-3)(1)), (1)(4) - (1)(2) \rangle = \langle -10 + 12, -(-5 - 3), 4 - 2 \rangle = \langle 2, 8, 2 \rangle\)
Step 4: Compute the magnitude of the cross product
The magnitude of \(\mathbf{n}\) is:
\(|\mathbf{n}| = \sqrt{2^2 + 8^2 + 2^2} = \sqrt{4 + 64 + 4} = \sqrt{72} = 6\sqrt{2}\)
Step 5: Calculate the shortest distance
The shortest distance \(d\) between the lines is:
\(d = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{n}|}\)
Where \(\mathbf{b} \cdot \mathbf{n} = \langle -8, -7, -3 \rangle \cdot \langle 2, 8, 2 \rangle = (-8)(2) + (-7)(8) + (-3)(2) = -16 - 56 - 6 = -78\).
Thus, the shortest distance is:
\(d = \frac{|-78|}{6\sqrt{2}} = \frac{78}{6\sqrt{2}} = \frac{39}{3\sqrt{2}} = \frac{13\sqrt{2}}{2} \times \frac{\sqrt{2}}{\sqrt{2}} = 6\sqrt{3}\)
Hence, the shortest distance between the lines is \(6\sqrt{3}\).
Top Questions on Distance between Two Lines
- 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
- JEE Main - 2026
- Mathematics
- Distance between Two Lines
- 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:- JEE Main - 2024
- Mathematics
- Distance between Two Lines
- 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.- JEE Main - 2024
- Mathematics
- Distance between Two Lines
- The shortest distance (in units) between the lines \(\frac{1 - x}{1} = \frac{2y - 10}{2} = \frac{z + 1}{1}\) and \(\frac{x - 3}{-1} = \frac{y - 5}{1} = \frac{z - 0}{1}\) is:
- CUET (UG) - 2024
- Mathematics
- Distance between Two Lines
- Want to practice more? Try solving extra ecology questions todayView All Questions
