Question

The distance of the point \( (7, -2, 11) \) from the line \( \frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3} \) along the line \( \frac{x - 5}{2} = \frac{y - 1}{-3} = \frac{z - 5}{6} \), is:

• A. 12
• B. 14
• C. 18
• D. 21

Answer 1

The Correct Option is B

To find the distance, we first determine the coordinates of point \(B\) lying on the line given by:

\(\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}\)

Assume:
\(B = (2\lambda + 7, -3\lambda - 2, 6\lambda + 11)\)

Point \(B\) also lies on the line:

\(\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}\)

Substitute the coordinates of \(B\):
\(2\lambda + 7 = 6 \quad \Rightarrow \quad -3\lambda - 2 = 4 \quad \Rightarrow \quad 6\lambda + 11 = 8 = 0\)
Solving:

\(-3\lambda - 6 = 0 \quad \Rightarrow \quad \lambda = -2\)

Substituting \(\lambda = -2\) gives:
\(B = (3, 4, -1)\)
To find the distance \(AB\) between points \(A = (7, -2, 11)\) and \(B = (3, 4, -1)\):
\(AB = \sqrt{(7 - 3)^2 + (-2 - 4)^2 + (11 - (-1))^2}\)

\(AB = \sqrt{(4)^2 + (-6)^2 + (12)^2}\)

\(AB = \sqrt{16 + 36 + 144}\)

\(AB = \sqrt{196} = 14\)
Thus, the distance of the point \((7, -2, 11)\) from the given line is \(14\).