Question

Find the equation of the plane in R3 that contains the point A(-6, -2, -2), and is perpendicular to the line through to points (0, -2, -6) and (2, -4, -2

Answer 1

To find the equation of plane , you need a point and normal vector

point
A(-6, -2, -2)


normal vector

from (0, -2, -6) to (2, -4, -2)

<2,-2,4>

Answer 2

the line given is the normal right?

Answer 3

a(x+6)+b(y+2)+c(z+2)

Answer 4

yes

Answer 5

<2,-2,4>
could be scaled down to

<1,-1,2>

Answer 6

(0, -2, -6) and (2, -4, -2)
+2 +6 +2 +6
------------------------
( 0 , 0 , 0) (2,-2,4) .... yep good eye

Answer 7

how you calculate the normal..?

Answer 8

I believe we just did

1(x+6)-1(y+2)+2(z+2)

Answer 9

the normal is perpendicular to the plane; so you just need to know the vector that points from the 2 other given points since that IS perpendicular to the plane

Answer 10

and it's equal to zero..?

Answer 11

a little more specific please

Answer 12

confused.. may I noe the formula for the normal?

Answer 13

1(x+6)-1(y+2)+2(z+2)=0

x-y+2 z=-8

Answer 14

subtract one point from the other that constitutes the perpendicular line to the plane as stated in the problem

Answer 15

point1 = (0, -2, -6)
-point2 = -(2, -4, -2)
---------------------
normal = <-2, 2,-4>, or its negation

Answer 16

this line is parallel to the normal of your plane

Answer 17

how you decide if the line is parallel or perpendicular?

Answer 18

..... i used the information given inthe stated problem.

Answer 19

ok. let me think about it for awhile.. :)

Answer 20

"...and is perpendicular to the line through to points (0, -2, -6) and (2, -4, -2)" means:

whose normal vector is parallel to the line through to points (0, -2, -6) and (2, -4, -2)

Answer 21

:) ah. okay! get it now.. :))

Answer 22

good :)

Answer 23

once you know the normal; its just the vector that directs you from one point to the next, you can apply it to the equation of a plane with the given point

Answer 24

right?

Answer 25

yup.

Answer 26

i get the answer now..

Answer 27

Find the equation of the plane in R3 that contains the point A(2, 5, 5), and is perpendicular to the line
x = 6 - 3 t
y = -4 - 3 t
z = -3 t

Answer 28

do u noe how to do this..? im stuck for looking the value for the normal..

Answer 29

the line they give you is in parametric form; which means that it has a point and a vector hidden within the information itself. And its the vector thats hidden in there that we want to use.

Answer 30

remember that a line can be form by starting at a given point and scaling a vector to any length. In other words, the constant parts are the point components, and the coefficients of the variables are actually are vector components.

the point given in the line is: (6,-4,-3)

the vector attached to that point is: <-3t,-3t,-3t>, or simply -3t<1,1,1>

our normal to the plane for the question is then <1,1,1>