Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$. The distance of the point $$ \left( \frac{11}{3}, \frac{11}{3}, \frac{19}{3} \right) $$ from the line $L$ along the line $$ \frac{3x - 11}{2} = \frac{3y - 11}{1} = \frac{3z - 19}{2} $$ is equal to:
- 3
- 5
- 4
- 6
The Correct Option is A
Approach Solution - 1
We start with the given line equation:
\[ \frac{x - 1}{2 - 1} = \frac{y - 2}{3 - 2} = \frac{z - 3}{5 - 3} \]
This can be simplified as:
\[ \frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 3}{2} = \lambda \]
Hence, the direction ratios of the line are:
\[ \langle 1, 1, 2 \rangle \]
Let point \( B \) be represented as:
\[ B(1 + \lambda, 2 + \lambda, 3 + 2\lambda) \]
and point \( A \) be given by:
\[ A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right) \]
Now, the direction ratios of line \( AB \) are:
\[ \left\langle \frac{3\lambda - 8}{3}, \frac{3\lambda - 5}{3}, \frac{6\lambda - 10}{3} \right\rangle \]
Since line \( AB \) is parallel to the given line, we have:
\[ \frac{3\lambda - 8}{3\lambda - 5} = \frac{2}{1} \]
Solving this gives:
\[ 3\lambda - 8 = 6\lambda - 10 \Rightarrow 3\lambda = 2 \Rightarrow \lambda = \frac{2}{3} \]
Now, substituting the value of \( \lambda \) to find \( AB \):
\[ AB = \sqrt{\left(\frac{6 - 3}{3}\right)^2 + \left(\frac{9 - 6}{3}\right)^2 + \left(\frac{12 - 9}{3}\right)^2} \]
\[ AB = \frac{\sqrt{36 + 9 + 36}}{3} = \frac{9}{3} = 3 \]
Hence, the length of \( AB \) is:
\[ \boxed{3} \]
Approach Solution -2
Step 1: Direction ratios of the given line
The direction ratios of the line:
\[ \frac{3x - 11}{2} = \frac{3y - 11}{1} = \frac{3z - 19}{2} \]
are:
\[ \vec{d} = \langle 2, 1, 2 \rangle \]
Step 2: Representing point \( B \)
Let point \( B \) on the given line be:
\[ B = (1 + \lambda, 2 + \lambda, 3 + 2\lambda), \]
where \(\lambda\) is the parameter.
Step 3: Direction ratios of line \( AB \)
Point \( A = \left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right) \). The direction ratios of \( AB \) are:
\[ \text{D.R. of } AB = \left\langle \frac{3\lambda - 8}{3}, \frac{3\lambda - 5}{3}, \frac{6\lambda - 10}{3} \right\rangle \]
Step 4: Parallel condition
Since \( AB \) lies along the direction vector \(\vec{d} = \langle 2, 1, 2 \rangle\), we have:
\[ \frac{\frac{3\lambda - 8}{3}}{2} = \frac{\frac{3\lambda - 5}{3}}{1} = \frac{\frac{6\lambda - 10}{3}}{2} \]
Simplify the first ratio:
\[ \frac{3\lambda - 8}{3 \cdot 2} = \frac{3\lambda - 5}{3} \]
Cross-multiply:
\[ 3\lambda - 8 = 6\lambda - 10 \]
Solve for \(\lambda\):
\[ 3\lambda = 2 \implies \lambda = \frac{2}{3} \]
Step 5: Find \( B \)
Substitute \(\lambda = \frac{2}{3}\) into \( B = (1 + \lambda, 2 + \lambda, 3 + 2\lambda) \):
\[ B = \left(1 + \frac{2}{3}, 2 + \frac{2}{3}, 3 + 2 \cdot \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \]
Step 6: Find distance \( AB \)
The distance \( AB \) is given by:
\[ AB = \sqrt{\left(\frac{11}{3} - \frac{5}{3}\right)^2 + \left(\frac{11}{3} - \frac{8}{3}\right)^2 + \left(\frac{19}{3} - \frac{13}{3}\right)^2} \]
Simplify each term:
\[ AB = \sqrt{\left(\frac{6}{3}\right)^2 + \left(\frac{3}{3}\right)^2 + \left(\frac{6}{3}\right)^2} \]
\[ AB = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} \]
Thus:
\[ AB = 3 \]
Final Answer:
\(Option (1) : \; 3\)
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