Question

Find an equation of the plane consisting of all points that are equidistant from (1, 3, -4) and (2, -3, 1), and having 1 as the coefficient of x

Answer 1

last one i swear

Answer 2

find the vector between the 2 points and the distance/2

Answer 3

distance is subtract and pythag

(1, 3, -4)
-(2, -3, 1)
-----------
-1 6 -5 ; sqrt(1+36+25) = sqrt(62) right?

Answer 4

and the subtracted part is the vector for the normal we can use

Answer 5

<1,-6, 5> = normal vector

to find a midpoint ... add the points and divide by 2 lol

Answer 6

how u retain all this information truly remarkable

Answer 7

(1, 3, -4)
(2, -3, 1)
-----------
(3 , 0 ,-3)/2

Answer 8

we use this point and the normal ... thatll get us all we need :)

Answer 9

how do i set this equal to 0 then?

Answer 10

1(x-(3/2))-6(y-0)+5(z-(-3/2))=0

Answer 11

u happen to know how to determine a point of intersection given two lines?

Answer 12

i can recall some of it :) you have to parametric them and change the scalar to see if they cross

Answer 13

the best view i got is that vectors/lines in space can move past each other and never touch; like a floor line and a ceiling line

Answer 14

so to see where they are at any given time we change the scalar for the parametrics and equate them