Question

can anyone help? Find the distance the point P(1, -3, -5), is to the plane through the three points
Q(4, -4, 0), R(8, 1, 3), and S(3, 0, 2).

Answer 1

find 2 vectors from the given points to cross and find a normal to the plane

Answer 2

then you can define a line using the given point and the unit normal vector

Answer 3

that line will pierce the plane at the shortest distance ... since it creates a perp line; and the distance between the given point and the piercing point is the solution

Answer 4

so i find |PR| and |PS|?

Answer 5

im sure theres a formula for this, but as i said prior, I got no idea what it would be :)

Answer 6

ok thanks :) so i find the plane and then use the normal vector to find the distance of the point?

Answer 7

Q(4, -4, 0) R(8, 1, 3)
-S( 3, 0, 2) -S(3, 0, 2)
-------------------------
<1,-4-2> <5,1,1>

yes :)

Answer 8

howd you get -S?

Answer 9

i get where ya got it now :) but how would i find is the point is on it?

Answer 10

-S is just subtracting the point S from the other 2 points to define the 2 vectors to cross

Answer 11

normal I find is: <2,11,-21>

we can either make that a unit vector or just use it as is to find the piercing point

P(1, -3, -5)

x = 1 + 2t
y = -3+11t
z = -5 - 21t

lets use any of the 3 given points in the plane to define the plane equation with; ill use S(3,0,2) i spose

1(x-3)+11(y-0)-21(z-2)=0

and using the xyz values in terms of t

1(1 + 2t-3)+11(-3+11t-0)-21(-5 - 21t-2)=0

and the calculations give me t = -28/121

use that to define the piercing point if you want and distance it from the given P

Answer 12

or we can unit the normal vector and just solve for t :) divide normal by sqrt(566)

x = 1 + 2t/sqrt(566)
y = -3+11t/sqrt(566)
z = -5 - 21t/sqrt(566)

1(1 + 2t/sqrt(566)-3)+11(-3+11t/sqrt(566)-0)-21(-5 - 21t/sqrt(566)-2)=0


do it your way and lets see if you get something close to 5 :)

Answer 13

why sqrt9566)

Answer 14

the normal is <2,11,-21> which has a length of sqrt(566)

so to make a unit normal we have to divide all the parts by its length

Answer 15

then its just a matter of finding out how many t units there are

Answer 16

i still dont understand how to find the distance of P to the plane?

Answer 17

my idea, and granted its not like the textbook since I cant recall the textbook formula :)

is to define the normal of the plane you want to measure from.

Create a line equation using the P point so that you can determine the point on the plane that this line will pierce thru at.

The distance between point P and the piercing point is the distance P is from the plane.