Question

consider the plane -x+4y+2z+6=0
a) determine a normal vector,n to the plane
b) determine the coordinates of two points ,s and t, on the plane
c)determine st
d)show that st is perpendicular to n

Answer 1

please help

Answer 2

take the gradient of the equation for the plane
do you know how to do that?

Answer 3

no

Answer 4

do you know how to do cross products?

Answer 5

no

Answer 6

...that will make this difficult
do you know how to take a partial derivative?

Answer 7

dervative i think so

Answer 8

do you know how to find the determinant of a matrix?

Answer 9

no havent studied that

Answer 10

its calculus 12

Answer 11

ok, I'm gonna try to do this the only way I know how
what is the partial derivative of the function for the plane with respect to x ?

Answer 12

ok

Answer 13

is it -1+4+2=0

Answer 14

?

Answer 15

in vector form, yes
-i+4j+2k

Answer 16

that is the gradient, so you apparently do know how to take it after all

Answer 17

you realise that the partial with respect to x is just -1
you did \(f_x\hat i+f_y\hat j+f_z\hat k\)

Answer 18

the gradient of a function is\[\nabla f=f_x\hat i+f_y\hat j+f_z\hat k\]

Answer 19

that is a vector that is perpendicular to the surface
so the vector you want is\[\nabla f(x,y,z)=\vec n=\langle-1,4,2\rangle\]

Answer 20

aha

Answer 21

so how do i determine the coordinates of s and t on plane

Answer 22

well, unless you did not provide some info, you just need to pick any two points in the plane
you can do that by just plugging in numbers
for example, let x=0, y=0, then you can figure out what z is
do that and get two points

Answer 23

aha but the quwstion doesnt have any graphs

Answer 24

so just pick two random points in the plane

Answer 25

aha

Answer 26

so for t i do the same?

Answer 27

yes, we need to get the coordinates of any two points so we can get a vector in the plane
we are then going to prove that that vector is perpendicular to \(\vec n\) for part c)

Answer 28

aha

Answer 29

so point s is (1,2,-6.5)

Answer 30

t is (2,3,-8)

Answer 31

if x=1 and y=2 I get z=-0.5
try to pick whole numbers though if you can
x=0, y=0 are obvious choices; or letting any two variables be zero, then finding the third

Answer 32

oh k thx

Answer 33

the (2,3,-8) works just fine though, don't get me wrong

Answer 34

yea nvm

Answer 35

ah you were right about the other point
so you want to use
s is (1,2,-6.5)
and
t is (2,3,-8)
???
can you form a vector between them?

Answer 36

yea st= t-s i guess

Answer 37

yep

Answer 38

i got (1,1,-14,5)

Answer 39

-14.5

Answer 40

-8-(-6.5)=-1.5

Answer 41

so how do we find st perpendicular to n

Answer 42

oh my bad

Answer 43

what do you know about the dot product of two perpendicular vectors?

Answer 44

d.m=0 so d is perpendicular to n

Answer 45

yes

Answer 46

so we use st to find d?

Answer 47

the last part just asks us to confirm that the vector st it perpendicular to \(\vec n\)
so dot them, and if it comes out to be zero we have proved that they are perpendicular

Answer 48

so in this case st.n=0

Answer 49

it should, let's check

Answer 50

\[\vec{st}\cdot\vec n=\langle1,1,-1.5\rangle\cdot\langle-1,4,2\rangle\]

Answer 51

i got [-1,4,-3]

Answer 52

careful, dot products are scalar\[\vec{st}\cdot\vec n=\langle1,1,-1.5\rangle\cdot\langle-1,4,2\rangle=-1+4-3=0\]

Answer 53

aha

Answer 54

it matchs

Answer 55

cool how even those random points you chose prove our theorem
it always matches, or else math fails :)

Answer 56

yea i think so thx for your help very much

Answer 57

welcome :)