Question

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 $L$?

• A. \(\left( \frac{1}{3}, 1, 1 \right)\)
• B. \(\left( \frac{1}{3}, 1, -1 \right)\)
• C. \(\left( \frac{1}{3}, -1, -1 \right)\)
• D. \(\left( \frac{1}{3}, -1, 1 \right)\)

Answer 1

The Correct Option is B

Given the two lines: \( L_1 : \frac{x - 2}{1} = \frac{y}{-1} = \frac{z - 1}{1} = \lambda \) and \( L_2 : \frac{x + 1}{2} = \frac{y - 1}{-1} = \frac{z + 1}{1} = \mu \).

The direction ratios (d.r.) of line \( MN \) are proportional to \( \langle 3 + \lambda - \frac{\mu}{2}, -1 - \lambda - \frac{\mu}{2}, 2 + \lambda - \mu \rangle \) and \( \langle 3, 1, 2 \rangle \).

Therefore:

\( \frac{3 + \lambda - \frac{\mu}{2}}{3} = \frac{-1 - \lambda - \frac{\mu}{2}}{1} = \frac{2 + \lambda - \mu}{2} \).

Equating the first two ratios yields \( 4\lambda + \mu = -6 \).

Equating the second and third ratios yields \( 4 + 3\lambda = 0 \).

Solving these equations gives \( \lambda = -\frac{4}{3} \) and \( \mu = -\frac{2}{3} \).

The coordinates of point \( M \) are \( M \left( \frac{4}{3}, \frac{4}{3}, -\frac{1}{3} \right) \).

The equation of the required line is \( \frac{x}{3} = \frac{y - 4/3}{1} = \frac{z + 1/3}{2} = k \).

Any point on this line can be expressed as \( \left( \frac{2}{3} + 3k, \frac{4}{3} + k, -\frac{1}{3} + 2k \right) \).

For \( k = -\frac{1}{3} \), the point on the line is \( \left( -\frac{1}{3}, 1, -1 \right) \).