Question:medium
If the lines $\dfrac{x-4}{m} = \dfrac{y-3}{2} = \dfrac{z+2}{1}$ and $\dfrac{x-3}{1} = \dfrac{y-4}{1} = \dfrac{z+3}{m}$ are coplanar, then the values of $m$ are
If the lines $\dfrac{x-4}{m} = \dfrac{y-3}{2} = \dfrac{z+2}{1}$ and $\dfrac{x-3}{1} = \dfrac{y-4}{1} = \dfrac{z+3}{m}$ are coplanar, then the values of $m$ are
Show Hint
Two lines in 3D are coplanar iff $(\overrightarrow{AB})\cdot(\vec{d_1}\times\vec{d_2})=0$. Set up the $3\times3$ determinant with $\overrightarrow{AB}$, $\vec{d_1}$, $\vec{d_2}$ as rows and equate to zero.
Updated On: Apr 25, 2026
- $4,\,1$
- $1,\,-4$
- $3,\,1$
- $5,\,-1$
- $3,\,-1$
Show Solution
The Correct Option is B
Solution and Explanation
Was this answer helpful?
0
Top Questions on Coplanarity of Two Lines
- Let the lines l1: \(\frac{x+5}{3} = \frac{y+4}{1}=\frac{z−α}{−2}\) and l2: 3x + 2y + z – 2 = 0 = x – 3y + 2z - 13 be coplanar. If the point P(a, b, c) on l1 is nearest to the point Q(- 4, -3, 2), then |a| + |b|+|c| is equal to
- JEE Main - 2023
- Mathematics
- Coplanarity of Two Lines
Two lines:
L₁: \(x = 5, \; \frac{y}{3 - \alpha} = \frac{z}{-2}\)
L₂: \(x = \alpha, \; \frac{y}{-1} = \frac{z}{2 - \alpha}\)
are coplanar. Then \(\alpha\) can take value(s):- BITSAT - 2019
- Mathematics
- Coplanarity of Two Lines
- Two lines \(L_1:\;x=5,\; \dfrac{y}{3-\alpha}=\dfrac{z}{-2}\) \(L_2:\;x=\alpha,\; \dfrac{y}{1}=\dfrac{z}{2-\alpha}\) are coplanar. Then \(\alpha\) can take value(s)
- BITSAT - 2015
- Mathematics
- Coplanarity of Two Lines
- If vectors \( 2i + j + k \), \( 2j - 3k \), and \( 3i + j + 5k \) are coplanar, then the value of \( a \) is
- BITSAT - 2013
- Mathematics
- Coplanarity of Two Lines
- Want to practice more? Try solving extra ecology questions todayView All Questions
