Question

In a school, the swimming pool is shaped as a cuboid, and the breadth of the pool is represented by line \(\frac{-x+2}{-5} = y = z-3\). Two students A and B are swimming along the length of the pool.

Based on the above information, answer the following questions :

(iii) (b) If a bird flies across the top of the swimming pool with its path represented by \(\frac{x+4}{2}=\frac{y+3}{8}=\frac{z-5}{-18}\), then find the shortest distance between its path and the path of student A.

Hint:

The paths are parallel (directions in ratio 2:1), so use d = |(a2 - a1) x b| / |b| with b = (1, 4, -9).

Answer 1

Step 1: Compare the directions:
The bird path has direction ratios $2,8,-18$, which is $2(1,4,-9)$.
Student A has direction ratios $1,4,-9$, so the two paths are parallel.
The shortest distance is then the perpendicular distance from any point of one path to the other path.

Step 2: Take a point on the bird path:
Take $B(-4,-3,5)$ on the bird path.
A point on the path of A is $Q(\lambda-5,\ 4\lambda-3,\ 6-9\lambda)$.
Then $\overrightarrow{BQ}=(\lambda-1,\ 4\lambda,\ 1-9\lambda)$.

Step 3: Make BQ perpendicular to the path of A:
Set $\overrightarrow{BQ}\cdot(1,4,-9)=0$.
\[ (\lambda-1)+16\lambda-9(1-9\lambda)=98\lambda-10=0 \]
So $\lambda=\frac{5}{49}$.

Step 4: Find the length BQ:
$\overrightarrow{BQ}=\left(-\frac{44}{49},\ \frac{20}{49},\ \frac{4}{49}\right)$.
\[ |BQ|=\frac{\sqrt{1936+400+16}}{49}=\frac{\sqrt{2352}}{49}=\frac{28\sqrt{3}}{49}=\frac{4\sqrt{3}}{7} \]

Final Answer:
The two paths are parallel and the distance between them is $\frac{4\sqrt{3}}{7}$ units. \[ \boxed{\frac{4\sqrt{3}}{7}\text{ units}} \]