Question

If the shortest distance between the lines \(\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{-3} and \frac{x - \lambda}{2} = \frac{y + 1}{4} = \frac{z - 2}{-5}\) is \(\frac{6}{\sqrt{5}}\), then the sum of all possible values of \(\lambda\) is:

• A. \( 5 \)
• B. \( 8 \)
• C. \( 7 \)
• D. \( 10 \)

Answer 1

The Correct Option is B

Given:
\(\int_0^1 \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} \, dx = a + b \sqrt{2} + c \sqrt{3}\)

where \(a, b, c\) are rational numbers.

Step 1. Simplifying the Integral: Consider
\(\int_0^1 \frac{1}{\sqrt{3 + x} + \sqrt{1 + x}} \, dx\)

Rationalizing the denominator:
\(\int_0^1 \frac{\sqrt{3 + x} - \sqrt{1 + x}}{(3 + x) - (1 + x)} \, dx = \int_0^1 \frac{\sqrt{3 + x} - \sqrt{1 + x}}{2} \, dx\)

  Therefore:
 \(= \frac{1}{2} \int_0^1 \left( \sqrt{3 + x} - \sqrt{1 + x} \right) \, dx\)

Step 2. Separating the Integral:
 \(= \frac{1}{2} \left( \int_0^1 \sqrt{3 + x} \, dx - \int_0^1 \sqrt{1 + x} \, dx \right)\)
Step 3. Evaluating Each Integral:
- For   \(\int_0^1 \sqrt{3 + x} \, dx \:\)
  \(\int \sqrt{3 + x} \, dx = \frac{2}{3} (3 + x)^{3/2}\)
     Evaluating from 0 to 1:

    \(\frac{2}{3} \left( (3 + x)^{3/2} \right) \Big|_0^1 = \frac{2}{3} \left( (4)^{3/2} - (3)^{3/2} \right) = \frac{2}{3} (8 - 3\sqrt{3})\)
  

  For\(\int_0^1 \sqrt{1 + x} \, dx\):
   
    \(\int \sqrt{1 + x} \, dx = \frac{2}{3} (1 + x)^{3/2}\)
 

    Evaluating from 0 to 1:
\(\frac{2}{3} \left( (1 + x)^{3/2} \right) \Big|_0^1 = \frac{2}{3} \left( (2)^{3/2} - (1)^{3/2} \right) = \frac{2}{3} (2\sqrt{2} - 1)\)

Step 4. Combining the Results:

  \(\frac{1}{2} \left( \frac{2}{3} (8 - 3\sqrt{3}) - \frac{2}{3} (2\sqrt{2} - 1) \right)\)

  Simplifying:

 \(\frac{1}{3} (8 - 3\sqrt{3} - 2\sqrt{2} + 1) = \frac{1}{3} (9 - 3\sqrt{3} - 2\sqrt{2})\)

Thus:

\(a = 3, \quad b = -\frac{2}{3}, \quad c = -1\)
 

Step 5. Calculating \( 2a + 3b - 4c \):
\(2a + 3b - 4c = 2 \times 3 + 3 \times \left( -\frac{2}{3} \right) - 4 \times (-1)\)
\(= 6 − 2 + 4 = 8\)