Question

what is the equation of the plane that passes through the point (1, -2, 4) and is perpendicular to the vector -i-3k

Answer 1

you have vector <-1,0,3> as the normal line and a point (1,-2,4) on the plane

Answer 2

a place equation is given by
ax+by+cz=d
where a,b,c is from the normal line to this plane <a,b,c>

Answer 3

thus we have
-1x+0y+3z=d
now to find d, plug in your point and see the d that satisfied
(1,-2,4)
-1*1+0*-2+3*4=d
d=10

-1x+0y+3z=10
x+3z=10 is our plane

Answer 4

in the 2nd post that should say, a plane* equation

Answer 5

do you have any questions

Answer 6

how can you find the vector (-1, 0, 3)

Answer 7

they gave u that

Answer 8

perpendicular to the vector -i-3k

Answer 9

i dont understand how i can know from the vector they give -i-3k will be (-1, 0,3)

Answer 10

i j k are like x,y,z

Answer 11

i j k are unit vectors along each axis here

Answer 12

so they say u are going -1 on the i direction, and -3 on the k direction, by not specifying what j is in this case, its already taken to be 0

Answer 13

oh i see it know

Answer 14

thx so much

Answer 15

i can give u a more dot product way of solving for this, if you a bit unsettled by why the normal line is the coeffecients a,b,c

Answer 16

it is good to understand both

Answer 17

equation for a plane, normal n, and position vector r0
\[\large n*(r-r _{o)}=0\]

Answer 18

yes i was gonna go into that definition, if he wanted

Answer 19

just remember that, need a point on the plane and a normal direction
\[<a,b,c>*(<x,y,z>-<x _{o},y _{0},z _{o}>)=0\]

Answer 20

it will expand out to the scalar equation for the plane

Answer 21

are they the same with parallel
line?

Answer 22

if this plane passes through the origin then the vectors on this plane are of the form <x,y,z>

and we know dot product of 2 perpendicular vectors give u 0
so
<-1,0,-3> dot <x,y,z> = 0
-1x -3z=0
but remember this is going through the origin, so to simply displace this to the point we want
(1, -2, 4)
-1(x-1) -3(z-4) = 0

Answer 23

if you remember or understand the line equation form
y-b= m(x-a) is a line of the slope m at point (a,b)

Answer 24

this is exactly what we are doing to displace this origin passing plane to make the new 'origin' at the point we want

Answer 25

u can think about it as a new set of variables u,v,t, for the x,y,z... i dunno i wont go into all this lol, its something for you to make sense of, not my own way of making it comfortable for myself

Answer 26

thank you i get it now