Question

Find a vector perpendicular to i +2j -k and 3i-j+k.
---> i - 4j -7k (I know how to get this)

Hence, find the cartesian equation of the plane parallel to both i +2j-k and 3i-j+k which passes through the point (2,0,-3)

Answer 1

@myko

Answer 2

(x,y,z)= (2,0,-2)+a(1,2-1)+b(3,-1,1)

Answer 3

x-2=a+b
y = 2a-b
z+2 = -a +b
eliminate a and b from here and you get the equation

Answer 4

ok?

Answer 5

How to eliminate a and b?

Answer 6

adding for example 1º and 3º equation you get:
x-2 +z+2 = 2b
so b = (x-2 +z +2 )/2
adding 1º and 2º:
x-2 +y = 3a
so a = (x-2 +y )/3
Now put this values in some equation and put it in the form Ax+By+Cz = D

Answer 7

ok?

Answer 8

i think if i didn't make any mistakes:
x-2y-3z =8
check it just in case...

Answer 9

ups, i found mistake:
x-2=a+b
y = 2a-b
z+2 = -a +b
should be:
x-2=a+3b !!!!!!!!
y = 2a-b
z+2 = -a +b
so in the rest fix this value

Answer 10

result is x-4y-7z =-16
this should be right

Answer 11

got it?

Answer 12

No.. Not really....
You should get at the end.

x-4y-7z=23

Answer 13

impossible, doesn't contain the point (2,0,-2)

Answer 14

That's what it says

Answer 15

try it your self. Put (2,0,-2) in the equation x-4y-7z=23 and it doesn't match it

Answer 16

Oh, I made a type. It's meant to be (2, 0, -3)

Answer 17

damm...:)

Answer 18

ya, than it gives
x-4y-7z=23

Answer 19

@order

Answer 20

How?

Answer 21

x-2=a+3b
y = 2a-b
z+3 = -a +b
from here: a =y+z+3, b = (x-2+z+3)/4
puting this values in the last equation:
z+3=-y-z-3+ (x-2+z+3)/4 and regrouping gives
x-4y-7z=23