Question

A line with direction ratios 1, -1, 2 intersects the lines \(\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}\) and \(\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}\) at the points P and Q, respectively. If the length of the line segment PQ is \(\alpha\), then \(225\alpha^2\) is equal to:

• A. 1024
• B. 1014
• C. 1104
• D. 1204

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a line with direction ratios \((1, -1, 2)\) that intersects two skew lines. We can find the general coordinates of points \(P\) and \(Q\) on the respective lines and use the direction ratios to find their exact positions.
Step 2: Key Formula or Approach:
Let the first line be \(L_1\): \(\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3} = \lambda\).
Any point \(P\) on \(L_1\) is \((2\lambda, 3\lambda, 3\lambda - 1)\).
Let the second line be \(L_2\): \(\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4} = \mu\).
Any point \(Q\) on \(L_2\) is \((-\mu - 1, \mu + 2, 4\mu)\).
The direction ratios of the vector \(\vec{PQ}\) are proportional to \((1, -1, 2)\).
Step 3: Detailed Explanation:
The direction ratios of \(\vec{PQ}\) are:
\[ \vec{PQ} = (-\mu - 1 - 2\lambda, \mu + 2 - 3\lambda, 4\mu - (3\lambda - 1)) = (-2\lambda - \mu - 1, -3\lambda + \mu + 2, -3\lambda + 4\mu + 1) \] Since \(\vec{PQ}\) is parallel to \((1, -1, 2)\), their ratios are equal:
\[ \frac{-2\lambda - \mu - 1}{1} = \frac{-3\lambda + \mu + 2}{-1} = \frac{-3\lambda + 4\mu + 1}{2} = k \] From the first two terms:
\[ -2\lambda - \mu - 1 = 3\lambda - \mu - 2 \implies -5\lambda = -1 \implies \lambda = \frac{1}{5} \] From the first and third terms:
\[ 2(-2\lambda - \mu - 1) = -3\lambda + 4\mu + 1 \implies -4\lambda - 2\mu - 2 = -3\lambda + 4\mu + 1 \] \[ -6\mu = \lambda + 3 \] Substitute \(\lambda = \frac{1}{5}\):
\[ -6\mu = \frac{1}{5} + 3 = \frac{16}{5} \implies \mu = -\frac{16}{30} = -\frac{8}{15} \] Now, compute coordinates of \(P\) and \(Q\):
\(P = \left(2(\frac{1}{5}), 3(\frac{1}{5}), 3(\frac{1}{5}) - 1\right) = \left(\frac{2}{5}, \frac{3}{5}, -\frac{2}{5}\right)\)
\(Q = \left(-(-\frac{8}{15}) - 1, -\frac{8}{15} + 2, 4(-\frac{8}{15})\right) = \left(-\frac{7}{15}, \frac{22}{15}, -\frac{32}{15}\right)\)
Calculate the vector \(\vec{PQ}\):
\[ \vec{PQ} = \left(-\frac{7}{15} - \frac{6}{15}, \frac{22}{15} - \frac{9}{15}, -\frac{32}{15} - (-\frac{6}{15})\right) = \left(-\frac{13}{15}, \frac{13}{15}, -\frac{26}{15}\right) \] The length of \(PQ\) is \(\alpha\):
\[ \alpha^2 = \left(-\frac{13}{15}\right)^2 + \left(\frac{13}{15}\right)^2 + \left(-\frac{26}{15}\right)^2 = \frac{169 + 169 + 676}{225} = \frac{1014}{225} \] Step 4: Final Answer:
We are asked to find \(225\alpha^2\):
\[ 225\alpha^2 = 225 \times \frac{1014}{225} = 1014 \]