# Thread: shortest diatance between plane and tower

1. ## shortest diatance between plane and tower

Here's a calc III problem(that was my approach) if anyone would like a shot. Not particularly difficult but kind of interesting.

"An airplane approaches the end of a runway from the east with an angle of 20 degrees. A tower 50 feet high is located 400 feet south and 800 feet east of the runway. How close does the plane come to the top of the tower?".

Here's a 'paint' drawing. It's a little tilted, but it should suffice.

2. No one wanted to tackle the problem, huh?.

Well, here's is the way I done it. I was wondering if someone else had a different method or verification of my solution.

We want the distance of the point $\displaystyle P_{1}(400,800,50)$ from the line of descent of the plane. The point $\displaystyle P_{2}(0,0,0)$ is on the line and the line has direction vector $\displaystyle V=cos(20)j+sin(20)k$

Then we have $\displaystyle P_{2}P_{1}=400i+800j+50k$

$\displaystyle P_{1}P_{2}\times{V}=\begin{vmatrix}i&j&k\\400&800& 50\\0&cos(20)&sin(20)\end{vmatrix}$

$\displaystyle =(800sin(20)-50cos(20))i-400sin(20)j+400cos(20)k$

The distance of P1 from the line using the distance from a point to a line formula.

$\displaystyle \frac{|P_{2}P_{1}\times{V}|}{|V|}$

$\displaystyle \frac{|(800sin(20)-50cos(20))|i-400sin(20)j+400cos(20)k|}{|cos(20)j+sin(20)k|}$

$\displaystyle =\frac{\sqrt{(800sin(20)-50cos(20))^{2}+(-400sin(20))^{2}+(400cos(20))^{2}}}{\sqrt{cos^{2}(2 0)+sin^{2}(20)}}$

$\displaystyle \approx\boxed{459.74} \;\ feet$