Seemingly easy question... too easy.

• Aug 23rd 2009, 09:38 AM
Ares_D1
Seemingly easy question... too easy.
Suppose the Earth is a plane described by the equation $x + y + z = 18$. Now suppose that at some arbitrary point in time there will be an explosion at the point $P = (1,1,1)$ in space. "City A" will be the first place on Earth that is destroyed by the explosion. With this information, calculate the coordinates of "City A".

It would seem that the first place on Earth destroyed would be directly underneath the explosion, in which case you have (1,1,0). But this is far too easy. Anyone have any other ideas?
• Aug 23rd 2009, 09:58 AM
earboth
Quote:

Originally Posted by Ares_D1
Suppose the Earth is a plane described by the equation $x + y + z = 18$. Now suppose that at some arbitrary point in time there will be an explosion at the point $P = (1,1,1)$ in space. "City A" will be the first place on Earth that is destroyed by the explosion. With this information, calculate the coordinates of "City A".

It would seem that the first place on Earth destroyed would be directly underneath the explosion, in which case you have (1,1,0) <<<<< Why? . But this is far too easy. Anyone have any other ideas?

1. The "Earth" is not parallel or equal to the x-y-plane!

2. The nearest point on the "Earth" is placed on a line perpendicular to the "Earth" passing through (1,1,1).

3. $l: (x,y,z)=(1,1,1)+t \cdot (1,1,1)$

4. to calculate the point of intersection between the "Earth" and the line l plug in 1+t for each variable:

$1+t+1+t+1+t = 18~\implies~t=5$

5. Therefore the coordinates of the nearest city are (6,6,6)
• Aug 23rd 2009, 10:05 AM
Ares_D1
Quote:

Originally Posted by earboth
1. The "Earth" is not parallel or equal to the x-y-plane!

2. The nearest point on the "Earth" is placed on a line perpendicular to the "Earth" passing through (1,1,1).

3. $l: (x,y,z)=(1,1,1)+t \cdot (1,1,1)$

4. to calculate the point of intersection between the "Earth" and the line l plug in 1+t for each variable:

$1+t+1+t+1+t = 18~\implies~t=5$

5. Therefore the coordinates of the nearest city are (6,6,6)

Thanks for that. But where did you get the equation from in 3?
• Aug 23rd 2009, 09:36 PM
earboth
Quote:

Originally Posted by Ares_D1
Thanks for that. But where did you get the equation from in 3?

From the equation of the plane:

$x+y+z = 18~\implies~(1,1,1) \cdot (x,y,z)=18$

you know that the normal vector to the plane is $(1,1,1)$.

I used this normal vector to get the equation of the line perpendicular to the plane and passing through the point P(1,1,1).

Thus:

$l: (x,y,z)=\underbrace{(1,1,1)}_{coords\ of\ point}+t \cdot \underbrace{(1,1,1)}_{normal\ vector\ to\ plane}$