Question #9079e

Question

278 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-05T17:03:42

A plane #Pi# can be defined by a normal vector and one point.

Given three points

#p_1,p_2,p_3# not aligned, then

#vec n = (p_2-p_1) xx (p_3-p_1)# is the normal vector, and we can choose any one of the three point also to make

#Pi-> << p-p_1, vec n >> = 0# where #p = (x,y,z)#

In our case

#vec n = (3,64,-74)# and then

#3x+64y-74z-35=0# or

#3x+64y-74z=35#

Answer 2 · 2018-02-05T17:14:13

# 3x+64y-74z=35#.

The eqn. of a plane through points

#P(x_1,y_1,z_1), Q(x_2,y_2,z_2) and R(x_3,y_3,z_3)# is given by,

#|(x-x_1,y-y_1,z-z_1), (x_2-x_1,y_2-y_1,z_2-z_1),(x_3-x_1,y_3-y_1,z_3-z_1)|=0#.

#:.|(x-5,y+2,z+2),(-5-5,-5+2,-5+2),(-3-5,3+2,2+2)|=0#.

#:. |(x-5,y+2,z+2),(-10,-3,-3),(-8,5,4)|=0#.

#:.(x-5)(-12+15)-(y+2)(-40-24)+(z+2)(-50-24)=0#.

#:. 3(x-5)+64(y+2)-74(z+2)=0#.

#rArr 3x+64y-74z=35#, is the desired eqn. of the plane.