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:
Points $P$ and $Q$ lie on two given lines. We express $P$ and $Q$ in terms of parameters $\lambda$ and $\mu$. The vector $\vec{PQ}$ must be parallel to the given direction ratios $(1, -1, 2)$.

Step 2: Key Formula or Approach:
1. $P = (2\lambda, 3\lambda, 3\lambda - 1)$. 2. $Q = (-\mu - 1, \mu + 2, 4\mu)$. 3. Direction ratios of $PQ \propto (1, -1, 2) \implies \frac{x_Q - x_P}{1} = \frac{y_Q - y_P}{-1} = \frac{z_Q - z_P}{2} = k$.

Step 3: Detailed Explanation:
1. Equations: - $-\mu - 1 - 2\lambda = k$ - $\mu + 2 - 3\lambda = -k$ - $4\mu - (3\lambda - 1) = 2k$ 2. Adding first two: $1 - 5\lambda = 0 \implies \lambda = 1/5$. 3. Substitute $\lambda$: $-\mu - 1 - 2/5 = k \implies k = -\mu - 7/5$. 4. Substitute into third: $4\mu - 3/5 + 1 = 2(-\mu - 7/5) \implies 4\mu + 2/5 = -2\mu - 14/5$. 5. $6\mu = -16/5 \implies \mu = -8/15$. 6. $k = 8/15 - 21/15 = -13/15$. 7. Length $PQ = \alpha = \sqrt{k^2 + (-k)^2 + (2k)^2} = \sqrt{6k^2} = |k|\sqrt{6}$. 8. $\alpha^2 = 6k^2 = 6(169/225)$. 9. $225\alpha^2 = 225 \times 6 \times \frac{169}{225} = 1014$. (Correction: Checking the specific ratio constant used in typical exams).

Step 4: Final Answer:
The value is 1014.