Question:
If for some \(m\in R\) the lines \(L_1:\frac{x+1}{m} = \frac{y-m}{-1} = \frac{z-1}{1}\) and \(L_2:\frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}\) are coplanar, then line \(L_1\) passes through the point
If for some \(m\in R\) the lines \(L_1:\frac{x+1}{m} = \frac{y-m}{-1} = \frac{z-1}{1}\) and \(L_2:\frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}\) are coplanar, then line \(L_1\) passes through the point
Show Hint
Use the coplanarity determinant to find \(m\), then test which point lies on \(L_1\).
Updated On: Oct 1, 2026
- \((-7,2,-5)\)
- \((7,-2,5)\)
- \((7,2,5)\)
- \((7,-2,-5)\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Two lines are coplanar when the scalar triple product of the joining vector and the two directions is zero.
Step 2: Key Formula or Approach
\(P_1(-1,m,1)\) with direction \((m,-1,1)\); \(P_2(-2,-1,-1)\) with direction \((-4,9,1)\). Joining vector \(\overrightarrow{P_1P_2}=(-1,-1-m,-2)\).
Step 3: Detailed Explanation
\[ \begin{vmatrix}-1&-1-m&-2\\m&-1&1\\-4&9&1\end{vmatrix}=m^2-13m+22=0 \]
So \(m=2\) or \(m=11\).
Take \(m=2\): \(L_1:\dfrac{x+1}{2}=\dfrac{y-2}{-1}=\dfrac{z-1}{1}\). At the point \((7,-2,5)\): \(\dfrac82=4\), \(\dfrac{-4}{-1}=4\), \(\dfrac41=4\). All equal, so \((7,-2,5)\) lies on \(L_1\).
Final Answer:
The line \(L_1\) passes through \((7,-2,5)\), option (B). \[ \boxed{(7,-2,5)\ \text{(B)}} \]
Two lines are coplanar when the scalar triple product of the joining vector and the two directions is zero.
Step 2: Key Formula or Approach
\(P_1(-1,m,1)\) with direction \((m,-1,1)\); \(P_2(-2,-1,-1)\) with direction \((-4,9,1)\). Joining vector \(\overrightarrow{P_1P_2}=(-1,-1-m,-2)\).
Step 3: Detailed Explanation
\[ \begin{vmatrix}-1&-1-m&-2\\m&-1&1\\-4&9&1\end{vmatrix}=m^2-13m+22=0 \]
So \(m=2\) or \(m=11\).
Take \(m=2\): \(L_1:\dfrac{x+1}{2}=\dfrac{y-2}{-1}=\dfrac{z-1}{1}\). At the point \((7,-2,5)\): \(\dfrac82=4\), \(\dfrac{-4}{-1}=4\), \(\dfrac41=4\). All equal, so \((7,-2,5)\) lies on \(L_1\).
Final Answer:
The line \(L_1\) passes through \((7,-2,5)\), option (B). \[ \boxed{(7,-2,5)\ \text{(B)}} \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Coplanarity of Two Lines Questions
- If the vectors \(m\hat{i} + m\hat{j} + n\hat{k}, \hat{i} + \hat{k}, n\hat{i} + n\hat{j} + p\hat{k}\) lie in a plane then...
- The lines $\frac{x-3}{1} = \frac{y-2}{1} = \frac{z-5}{-k}$ and $\frac{x-4}{k} = \frac{y-3}{1} = \frac{z-3}{2}$ are coplanar, hence $k =$}
- The lines $\frac{x-3}{1} = \frac{y-2}{1} = \frac{z-5}{-k}$ and $\frac{x-4}{k} = \frac{y-3}{1} = \frac{z-3}{2}$ are coplanar, hence $k =$}
- If the vectors $2\hat{i} - \hat{j} - \hat{k}$, $\hat{i} + 2\hat{j} - 3\hat{k}$ and $3\hat{i} + \lambda\hat{j} + 5\hat{k}$ are coplanar, then the value of $\lambda$ is
- If the lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$ ($ab \neq 0$) are coplanar, then \dots
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions