Question

Let $d$ be the distance of the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ from the point $(7, 8, 9)$. Then $d^2 + 6$ is equal to:

• A. 72
• B. 69
• C. 75
• D. 78

Answer 1

Answer: (C)

75

Answer 2

Let the equations of the lines be:
For Line 1:
$\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1} = \lambda$ Then $x = 3\lambda - 6$, $y = 2\lambda$, $z = \lambda - 1$.
For Line 2:
$\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2} = \mu$ Then $x = 4\mu + 7$, $y = 3\mu + 9$, $z = 2\mu + 4$.
By equating the coordinates, we get the system of equations:
$\begin{aligned} 3\lambda - 6 &= 4\mu + 7 \quad (1) \\\ 2\lambda &= 3\mu + 9 \quad (2) \\\ \lambda - 1 &= 2\mu + 4 \quad (3) \end{aligned}$ Solving these equations, we find the values of $\lambda$ and $\mu$ at the point of intersection as $\lambda = 3$ and $\mu = -1$. Thus, the intersection point is $(3, 6, 2)$.
The distance $d$ from the point $(7, 8, 9)$ to $(3, 6, 2)$ is:
$d = \sqrt{(7 - 3)^2 + (8 - 6)^2 + (9 - 2)^2} = \sqrt{16 + 4 + 49} = \sqrt{69}$ Therefore,
$d^2 + 6 = 69 + 6 = 75$