Question

Consider the lines $L_1$ and $L_2$ given by 

$L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2} $ 

$ L_2: \frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}$

A line $L_3$ having direction ratios $1,-1,-2$, intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively Then the length of line segment $P Q$ is

• A. $3 \sqrt{2}$
• B. 4
• C. $2 \sqrt{6}$
• D. $4 \sqrt{3}$

Answer 1

The Correct Option is C

\(Let \,P=(2λ+1,λ+3,2λ+2) \)

\(Let\, Q=(μ+2,2μ+2,3μ+3) \)

\(⇒\frac{12λ−μ−1​}1=\frac{−1λ−2μ+1}{-1}​=\frac{2λ−3μ−1​}{-2} \)
\(⇒λ=μ=3⇒P(7,6,8) \,\,and ,\,Q(5,8,12) \)

\(PQ=2\sqrt6\)

Answer 2

1. The equation of line \(L_3\) is given by: \[ \frac{x-\lambda}{1} = \frac{y-\mu}{-1} = \frac{z-\nu}{-2}. \] 2. \(L_3\) intersects \(L_1\) at point \(P\). Equate the parametric equations of \(L_1\) and \(L_3\): \[ \frac{\lambda - 1}{2} = \frac{\mu - 3}{2} = \frac{\nu - 2}{2}. \] Let \(\frac{\lambda - 1}{2} = t\). Solve for \(\lambda, \mu, \nu\): \[ \lambda = 1 + 2t, \quad \mu = 3 + 2t, \quad \nu = 2 + 2t. \] 3. For \(L_3\), the parametric equations are: \[ x = \lambda, \quad y = \mu, \quad z = \nu. \] Substitute into \(L_3\): \[ \frac{1 + 2t - \lambda}{1} = \frac{3 + 2t - \mu}{-1} = \frac{2 + 2t - \nu}{-2}. \] Solve for \(t\) to find the coordinates of \(P(7, 6, 8)\). 4. Similarly, \(L_3\) intersects \(L_2\) at point \(Q\). Repeat the above process for \(L_2\) to find: \[ Q(5, 8, 12). \] 5. Find the length of segment \(PQ\): \[ PQ = \sqrt{(7-5)^2 + (6-8)^2 + (8-12)^2}. \] Simplify: \[ PQ = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}. \] The problem involves finding the intersection points of two lines with a third line and calculating the distance between them using the distance formula.