Question

Only answer please. This is about vectors

Answer 1

c x a = (3i-5j+7k)(2i-j+3k)
c x a = 6i^2-3ij+9ik-10ij+5j^2-15jk+14ik-7jk+21k^2
c x a = 6i^2-3ij-10ij+9ik+14ik+5j^2-15jk-7jk+21k^2
c x a = 6i^2-13ij+23ik+5j^2-22jk+21k^2

Answer 2

@ganeshie8

Answer 3

Thats not the correct answer it should be using this formula c × a = [c2a3 - a2c3, c3a1 - a3c1, c1a2 - a1c2] but i don't know how to use this equation when there is those letters in the question (i,j,k)

Answer 4

ijk are just there to indicate "like terms", like vector parts

Answer 5

a = 3i - 5j + 7k

a = (3,-5,7)

Answer 6

oh ok so the answer would then be as i previoulsy attempted as (8,-5,-7)

Answer 7

maybe, id have to run the calculation to be sure

Answer 8

-(-3z-15x+14y)
x y z x y
2 -1 3 2 -1
3 -5 7 3 -5
+(-7x+9y-10z)

(-7x + 9y -10z)
-(-15x+14y-3z)
----------------
8x -5y -7z

yes, 8,-5,-7

Answer 9

ok thanks allot .. even though i have no idea how you came up with that process hahah :)

Answer 10

over the years you gain alot of trivial knowledge lol

Answer 11

this is an answer my friend did so i guess its wrong and i should inform him?

Answer 12

can you spot the error in it?

Answer 13

no not really he used i process i didn't even understand ... looks more like yours in a way

Answer 14

c = 2i − j + 3k

he used a bad j element: c = 2i + j + 3k
^^

Answer 15

other than that, his subdeterminants look good, and he has remembered the +-+-+... adjustments

Answer 16

oh yea haha .. thanks again

Answer 17

just became a fan might help me again in near future :)

Answer 18

;) good luck

Answer 19

thanks

Answer 20

i^2=i , j^2=j , k^2=k ,

jk=i , ki=j , ij=k ,

kj=-i , ik=-j , ji=-k ,

Answer 21

\[\left|\begin{array}{ccc} a & b & c \\ d & e & f \\ g& h & i \end{array} \right|=a\left|\begin{array}{ccc} e & f \\ h & i \end{array} \right|-b\left|\begin{array}{ccc} d & f \\ g & i \end{array} \right|+c\left|\begin{array}{ccc} d & e \\ g & h\end{array} \right|\\\qquad\qquad \quad =a(ei-fh)-b(di-fg)+c(dh-eg)\]