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

Answer: (B)

$8$

Answer 2

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