Distance from point to plane

• Sep 4th 2009, 04:47 AM
purplerain
Distance from point to plane
Find the distance from the point [11, -10, 17] to the plane
-4478 - 283 x + 347 y + 416 z = 0

what i did was i multipled the variables in the plane by their corresponding variables in the point, and then divided that answer by the distance or magnitude of the plane. I got 611.1906 but the computer keep saying that this is not the correct value. Anyone has an alternative solution? Thanks.
• Sep 4th 2009, 05:31 AM
earboth
Quote:

Originally Posted by purplerain
Find the distance from the point [11, -10, 17] to the plane
-4478 - 283 x + 347 y + 416 z = 0

what i did was i multipled the variables in the plane by their corresponding variables in the point, and then divided that answer by the distance or magnitude of the plane. <<< what exactly is this? I got 611.1906 but the computer keep saying that this is not the correct value. Anyone has an alternative solution? Thanks.

I assume that you calculated:
$\displaystyle d = \dfrac{-4478 - 283 \cdot 11 + 347 \cdot (-10) + 416 \cdot 17}{\sqrt{283^2+347^2+416^2}} = \dfrac{-3989 \cdot \sqrt{41506}}{124516} \approx -6.52661$

The negantive sign of d indicates that the given point and the origin are on different sides of the plane.
• Sep 4th 2009, 05:34 AM
Calculus26
If the equation of a plane is ax + by + cz + d = 0

Then the distance of a point(x0,y0,z0) to the plane is

D = |ax0 + by0 + cz0 + d|/sqrt(a^2+b^2+c^2)

I get 6.527
• Sep 4th 2009, 07:32 PM
purplerain
Oops, yall are right, i recalulated it myself and got approximately 6.480549334. I think the sign should be positive.