Question

9. A portion of the graph of a linear function is shown.
(a) Find an equation for the linear function.
(b) Find a vector perpendicular to the plane.
(c) Find the area of the shaded triangular region.

Answer 1

its a plane , so
to find the equation u need
normal vector + a point on the plane

Answer 2

to find normal , define any 2 vectors on the plane lets say v1,v2
normal=v1×v2

Answer 3

so u have those points in the plane
p1 (0,0,12)
p2 (0,3,0)
p3 (-5,0,0)
right ?

Answer 4

yes

Answer 5

nw find v1,v2 using the point
v1=<0--5,0-0,12-0>=<5,0,12?
v2=<0--5,3-0,0-0>=<5,3,0>
then find
normal=v1×v2
do u know how to do it ?

Answer 6

I know how to do cross but how did you get V1 and V2?

Answer 7

what are v1 and v2?

Answer 8

simple
"define any 2 vectors on the plane lets say v1,v2"
v1=p3p1
v2=p3p2

Answer 9

??

Answer 10

ok so how do I get the equation of the linear function?

Answer 11

and how do I find the vector perpendicular to the plane?

Answer 12

now the equation is of the formulla
a(x-x0)+b(y-y0)+c(z-z0)=0
s,t
normal=<a,b,c>
any point on the plane (p0)
P0=(x0,y0,z0)

Answer 13

normal u shud find it using cross product v1×v2
p0 any point from p1,p2,p3 ok ??

Answer 14

(b) Find a vector perpendicular to the plane.
the normal vector is also perpendiculer
so any perpendiculer wud be parallet to the normal
which mean
any perpendiculer vector = k<a,b,c> s,t k any number ^^

Answer 15

I'm sorry I still don't get how to find the equation for the linear function. what is "a" "x" and "x0"

Answer 16

I found the are of the shaded triangular region :)

Answer 17

lol !
ok did u find v1×v2 ?

Answer 18

and divided by 2

Answer 19

I got 35.78

Answer 20

normal=v1×v2=<a,b,c>

Answer 21

no no no
find v1×v2 not ||v1×v2||
it shud produce vector not scallor

Answer 22

so how do i find the equation of the linear function?

Answer 23

I'm so confuse

Answer 24

for P2P1= <0,-3,12> =V1
P2P3= <-5,-3,0>=V2
V1XV2= <36, -60, -15>

Answer 25

||V1XV2||= squr(5121)