Question

Find all vectors v such that (1,2,1) x v = (3,1,-5) (x=cross product)

Answer 1

teh cross product is used to find the normal of a plane; therefore all the vectors the reside in the plane will cross to get the desired results

Answer 2

the equation of the plane is the set of all vectors that can be crossed :)

Answer 3

if we cross these 2 that are given, we will find at least one of the vectors that reside in the plane

Answer 4

would the answer be the plane 3(x-1)+1(y-2)-5(z-1)?

Answer 5

id say yes; since any 2 vectors on the place cross to get the normal

Answer 6

wait, we havnt actually got any points to define a plane by tho; so this needs a little more thoughts

Answer 7

ok, well v has to be orthogonal to (3,1,-5)

Answer 8

(1,2,1)
(3,1,-5)
-------
x= -11
y = 8
z = -4

yes, i agree with that

Answer 9

you did the cross product of the two?

Answer 10

yeah, that gives me a vector that is at least in the same plane as the other :)

Answer 11

these (...) are vectors right? i mean im just assuming that they are. im used to a vector being denoted as <...>

Answer 12

could it be the only possible vector? yes they are vectors

Answer 13

no, it cant be, we can add this new one to the first and cross it to get the normal just as well

Answer 14

( 1, 2 ,1 )
(-11,8,-4)
----------
<-10,10,-3> = (1/10)<-1,1,-.3> would you agree?

Answer 15

hmm, i see <0, -7, 30>, cross product right?

Answer 16

oh sorry, I see, yes

Answer 17

why did u add them?

Answer 18

1 ,2 ,1
-10,10,-3
-------------
x = -16
y = -7
z = 30

<-16,-7, 30>

<16, 7, -30>
/6 /6 /6
-----------
hmmmm <3,1,-5>

am i dong the math right is what i gottta wonder now

Answer 19

i add them to find another vector that i know is in the plane to use as a test to cross with

Answer 20

the problem with vectors is that they can exist anyplace in the plane; they move about

Answer 21

if we suppose the first two vectors are in the xy plane, then the 2nd vector can be any non zero vector other than the 1st vector right?

Answer 22

sure, i dont see why not at the moment

Answer 23

I think this is easier than we think, perhaps you just do cross product of <1,2,1> and v and make it equal to <3,1,-5>

Answer 24

then do matrix

Answer 25

(1,2,1)
(3,1,-5)
-------
x= -10 - 1 = -11
y = 3 +5 = 8
z = 1- 6 = -5

that might be better

Answer 26

my z last time was 1 - (3+2) instead of 1 - 3(2) lol

Answer 27

so v can = t<11,-8,5>

Answer 28

ok cool thanks

Answer 29

but all the vectors? as far as i can see it, there are infinite number of them all in planes that are parallel to each other

Answer 30

ok thatt's a possible answer, the next part asks why there is no vector v such that =<3,1,5>

Answer 31

(1,2,1)
(a, b, c)
-------
x = 2c -b = 3
y = a - c = 1
z = b - 2a = -5


how about them apples?

Answer 32

niccceee

Answer 33

at the moment thats the best I got :)

Answer 34

thanks, Ive got to head to bed tho, have a good one

Answer 35

and i dont think 3,1,5 is possible since it aint gonna be orthogonal to the 3,1,-5