Question

find the vector and parametric equations of the plane that is parallel to the yz plane and contains the point A (-1,2,1)

Answer 1

the solution given has to use cross product, which i dont understand why
because if it says "parallel", shouldnt u just use the vector in yz plane, such as [0,1,1] as the direction vector

Answer 2

yz plane equation is x=0
to make it contain point A, do x-1=0, x=1

Answer 3

but the solution says to do cross product

Answer 4

take vectors y [0,1,0] and z [0,0,1] and do their cross product to find perpendicular vector of the plane, which will be [1,0,0]

Answer 5

answer given is [-1,2,1] + , [ 0,0,1 ] + n [ 0,1,0]

Answer 6

and plain equation will be [1,0,0][x+1,y-2,z-1] =0
which is equal to: x+1 + 0(y-2) + 0(z-1) =0
or simpifying: x+1=0, x=-1
sry, i think i made sign mistake befor

Answer 7

dont really understand..ur way looks so complicated.. o_o
cuz what i did is cross -1,1,0 and -1,0,0 and get 0,0,1 as the vector..
but i dont know why i did that

Answer 8

shouldn't it be
[-1,2,1] + m[ 0,0,1 ] + n [ 0,1,0]
?

Answer 9

[0,0,1] is z. [0,1,0] is y

Answer 10

oh...so is the direction vector just the two vectors from y and z..? dont need to do cross product at all?

Answer 11

i think so, :)

Answer 12

ohh ok thanks. do u know what to do if it asks for perpendicular to yz plane?

Answer 13

then you make cross product of two non parallel vectors of yz plane

Answer 14

cross product gives you a vector which is perpendicular to bouth vectors that you crossing

Answer 15

then wouldnt u just get 1 vector if u cross the vector of y and z

Answer 16

yes

Answer 17

cuz equation of a plane needs 2 direction vector isnt it?

Answer 18

no

Answer 19

there is equation that use onle one perpendicular vector

Answer 20

ohh..my teacher only tell us plane -> 2 direction vector , line -> 1 direction vector ..
maybe urs is the higher class ones..

Answer 21

for example: ax+by+cz=0 is equation of the plane which is perpendicular to vector (a,b,c)

Answer 22

your teacher is probably speaking about parametric equations.

Answer 23

or dimensions

Answer 24

yeaa, we have to find parametric equations..
so then i couldnt find it with just 1 direction vector

Answer 25

like for this question, it asks for parametric equation
so if i only have 1 direction vector, should i just get 2 points and make a vector and cross product it again..

Answer 26

it is not colled direction vector in this case, becouse it's perpendicular, so not in the direction of the plane

Answer 27

no

Answer 28

OHH yeaa i get it .
that normal vector is for cartesian equation ,