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) (a) Find the foot of the perpendicular drawn from \((2, 4, -1)\) to the path represented by student A.

Hint:

Take Q(lambda - 5, 4 lambda - 3, 6 - 9 lambda) on A with p = 3 and make PQ perpendicular to the direction 1, 4, -9.

Answer 1

Step 1: Pick a point and the direction on the line:
From part (i), $p=3$, so A is the line through $A_0(-5,-3,6)$ with direction $\vec{d}=(1,4,-9)$.
The given point is $P(2,4,-1)$.

Step 2: Use the projection idea:
The foot is the point of the line closest to $P$.
It lies at $A_0+t\vec{d}$, where $t=\frac{\overrightarrow{A_0P}\cdot\vec{d}}{|\vec{d}|^2}$.

Step 3: Compute the numbers:
$\overrightarrow{A_0P}=(2+5,\ 4+3,\ -1-6)=(7,7,-7)$.
$\overrightarrow{A_0P}\cdot\vec{d}=7+28+63=98$.
$|\vec{d}|^2=1+16+81=98$.
So $t=1$.

Step 4: Locate the foot:
Foot $=(-5,-3,6)+1\cdot(1,4,-9)=(-4,1,-3)$.
The distance $PF$ is $\sqrt{36+9+4}=7$ units, and $\overrightarrow{PF}\cdot\vec{d}=0$, which confirms the foot.

Final Answer:
Projecting P on the line of A gives the foot $(-4,1,-3)$. \[ \boxed{(-4,\ 1,\ -3)} \]