Let the line $L$ intersect the lines
$x - 2 = -y = z - 1$, $\quad 2(x + 1) = 2(y - 1) = z + 1$
and be parallel to the line
$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$.
Then which of the following points lies on $L$?
Let the line $L$ intersect the lines
$x - 2 = -y = z - 1$, $\quad 2(x + 1) = 2(y - 1) = z + 1$
and be parallel to the line
$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$.
Then which of the following points lies on $L$?
- \(\left( \frac{1}{3}, 1, 1 \right)\)
- \(\left( \frac{1}{3}, 1, -1 \right)\)
- \(\left( \frac{1}{3}, -1, -1 \right)\)
- \(\left( \frac{1}{3}, -1, 1 \right)\)
The Correct Option is B
Approach Solution - 1
Given the two lines:
\( L_1 : \frac{x - 2}{1} = \frac{y}{-1} = \frac{z - 1}{1} = \lambda \)
\( L_2 : \frac{x + 1}{2} = \frac{y - 1}{-1} = \frac{z + 1}{1} = \mu \)
The direction ratios (d.r.) of the line \( MN \) will be:
\( \langle 3 + \lambda - \frac{\mu}{2}, -1 - \lambda - \frac{\mu}{2}, 2 + \lambda - \mu \rangle \)
and it will be proportional to \( \langle 3, 1, 2 \rangle \).
Hence,
\( \frac{3 + \lambda - \frac{\mu}{2}}{3} = \frac{-1 - \lambda - \frac{\mu}{2}}{1} = \frac{2 + \lambda - \mu}{2} \)
From the first two ratios:
\( 4\lambda + \mu = -6 \)
From the second and third ratios:
\( 4 + 3\lambda = 0 \)
Solving these equations:
\( \lambda = -\frac{4}{3}, \quad \mu = -\frac{2}{3} \)
Therefore, the coordinates of point \( M \) are:
\( M \left( \frac{4}{3}, \frac{4}{3}, -\frac{1}{3} \right) \)
and the equation of the required line is:
\( \frac{x}{3} = \frac{y - 4/3}{1} = \frac{z + 1/3}{2} = k \)
So, any point on this line will be:
\( \left( \frac{2}{3} + 3k, \frac{4}{3} + k, -\frac{1}{3} + 2k \right) \)
Substituting \( k = -\frac{1}{3} \):
\( \text{Point on the line} = \left( -\frac{1}{3}, 1, -1 \right) \)
Approach Solution -2
The given line \( L \) is parallel to the line:
\[ \frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}, \]
which implies the direction ratios of \( L \) are \( 3 : 1 : 2 \).
Assume that \( L \) intersects the first line \( x - 2 = -y = z - 1 \) at some point \( P \). From the equation of the line:
\[ x - 2 = -y = z - 1 = k. \]
This gives:
\[ x = 2 + 3k, \quad y = -k, \quad z = 1 + k. \]
Next, assume \( L \) also intersects the second line \( 2(x + 1) = 2(y - 1) = z + 1 \) at some point \( Q \). From the equation of the line:
\[ 2(x + 1) = 2(y - 1) = z + 1 = m. \]
This gives:
\[ x = m - 1, \quad y = \frac{m}{2} + 1, \quad z = m - 1. \]
Now, the line \( L \) passes through both points \( P \) and \( Q \), and we know it is parallel to the direction ratios \( 3 : 1 : 2 \). Substituting the parametric forms into the equations of the line \( L \) and solving for \( k \) and \( m \), we determine the valid points that lie on \( L \).
After solving, it is found that the point:
\[ \left( -\frac{1}{3}, 1, -1 \right) \]
satisfies the equation of \( L \).
Thus, the correct answer is option (2).
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