Question

how to do this?
Find the scalar equation of the plane that is perpendicular to the plane x + 2y + 4 = 0, contains the origin, and whose normal makes an angle of 30 with the z-axis.

Answer 1

i know that the normal of the required plane is perpendicular to the normal of the given plane, so their dot product must be zero
i let <a,b,c> be the normal of the required plane
so i got a+2b=0

Answer 2

-(2/√5) x + (1/√5) y + √3 z =0

this is the answer

Answer 3

I'm just here for a medal *_* sorry @Data_LG2 but your question is Greek to meh!

Answer 4

Hmm I'm having a problem drawing. When I click 'draw', it posts whatever I have typed in. But if you sketch the line, you can imagine how the plane looks like. So I guess you can find the line that's perpendicular to this line and goes through (0,0), and somehow just add a z component? That z component must make an angle to z-axis

Answer 5

I would, for the z part, imagine/sketch it on a z-y or z-x plane. I'm sure I'm making no sense.

Answer 6

i did this: http://sketchtoy.com/58404478
but it's wrong :/

Answer 7

this is the plane right
x + 2y + 4 = 0
so its normal is
<1,2,0>

Answer 8

yes

Answer 9

lets assume the normal of the plane u wanna is
<a,b,c>

Answer 10

okay, i did that....
then?

Answer 11

sry i dint know u do it
then
<a,b,c>.<1,2,0>=0
so
a+2b+0c=0
right ??
this is equation 1

Answer 12

that's fine :)
yes..i posted my solution before, but i still don't get the right answer
here: http://sketchtoy.com/58404478
and.....?

Answer 13

ohk :)
u have the angle of 30 with the z-axis.
which is DIRECTION ANGLE whith z axis

that mean
cos 30 = c /||n||

Answer 14

mhmm

Answer 15

||n||=c/cos 30
||n|| =root a^2+b^2+c^2
c/cos 30=root a^2+b^2+c^2 equation 2
got it ??

Answer 16

which part confusing u ?
ill explain

Answer 17

i understand it,
but i did not get the right answer, that's why i'm asking how to do it
maybe my method is wrong

so what will be your next step?

Answer 18

so u have 2 equation
a+2b=0
c^2=3a^2+3b^2
right ?

Answer 19

aasume c humm any num u wanna
then solve for a and b

Answer 20

wait! how did you come up with c^2=3a^2+3b^2 from c/cos 30=root a^2+b^2+c^2??

Answer 21

aahh got it..

Answer 22

cos 30 = root 3 /2

c/ cos 30 = 2c/root 3=root a^2+b^2+c^2
take ^2 to both side
4 c^2 /3 =a^2+b^2+c^2
4 c^2= 3a^2+3b^2+3c^2
c^2= 3a^2+3b^2

Answer 23

so any number for "c"??

Answer 24

any number of c take 2 /root 3 for example
it will give u unit normal
or take any integer u eant it dosnt mater

Answer 25

ok, give me a sec, i'll try to solve it

Answer 26

becouse of this
<a,b,c>.<1,2,0>=0
a+2b+0c=0
so any num of c wont be wrong

Answer 27

so i assume c=4
i solved for b, and i got 4/√15
and for a i got -8/√15

which means the equation will be: -8/√15 x + 4/√15 y + 4z =0?
if i simplified it , it will be -2/√15 x + 1/√15 y + z =0

ALMOST same with the right answer but instead of √15 it should be √5 and the z should have a coefficient of √3 ..
the right answer is : -(2/√5) x + (1/√5) y + √3 z =0
..mm.. should i multiply the equation with something to get it √5 ?

Answer 28

humm wait ill check

Answer 29

assume c= root 3
it will give u the same answer

Answer 30

c=√3
3a^2+3b^2=3
a^2+b^2=1
a=-2b
(-2b)^2+b^2=1
b= (1/√5)
a=-(2/√5)

Answer 31

but yess if u simplify u answer ull get the same equation

Answer 32

ooh..
i think it's the same thing!!
for my answer before, i just need to multiply 1/√5 to get the same equation
and for the second one that i did, i need to multiply it by √3!
whoa i was right all this time, smh


thank you very much!

Answer 33

good :)
ur wlc

Answer 34

it was nice qs