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

Answer: (B)

14

Answer 2

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 $14$.