Question

If "a" vector = 2 i+3j+k
a.b=1
axb=j-k
Find b vector..

Answer 1

@hartnn

Answer 2

pehchan kaun? :D

Answer 3

someone who knows hindi :P

Answer 4

okay ques btado

Answer 5

take b as xi+yj+zk

Answer 6

if we take cross with a?

Answer 7

if we assume xi+yj+zk then what?

Answer 8

did u take cross with a ?

Answer 9

then dot product 1 gives
2x+3y+z=1

Answer 10

(xi+yj+zk)(2i+3j+k)=1
2x+3y+z=1 okay then?

Answer 11

know how to find aXb ?

Answer 12

determinant?

i(3z-y)-j(2z-x)+k(2y-3x)=j-k

Answer 13

lamba nahi ho raha?
they have taken cross with a in the question
axb=j-k mein then we get (a.b)a thing and we are given its value

Answer 14

naah, thats what you need to do, solve the determinant and equate it to j-k

Answer 15

then how to solve further?

Answer 16

what you got for the determinant.?
then you compare the values with i,j,k on both sides ,

value with i on left = 0
value with j = 1
value with k =-1

Answer 17

i(3z-y)-j(2z-x)+k(2y-3x)=j-k
2z-x=1...(i)..........x=2z-1
3z-y=0.......y=3z...(ii)
2y-3x=-1...(iii)

6z-3x=-1
6z-3(2z-1)=-1
6z-6z+3=-1
z=0 :o

Answer 18

getting weird values ~_~

Answer 19

why can't 'z' be 0 ?
you have the answer already ?
i didn't check the calculation though...

Answer 20

i have the answer and z not equal to 0

Answer 21

@hartnn we will compare j to -1 i guess

Answer 22

b is a vector @Loser66

Answer 23

aXb = j-k
so, compare j to 1 only ....not -1

Answer 24

oh you're giving xyz?

Answer 25

\[\Large b=\frac{1}{14}(6i+j-k)\]
x should be 3/7
y should be 1/14
k should be -1/14 according to this

Answer 26

2z-x=1...(i)..........x=2z-1
3z-y=0.......y=3z...(ii)
2y-3x=-1...(iii)

are actually ONLY 2 independent equations
you cannot solve them unless you take 2x+3y+z=1
into consideration
then you have 3 independent equation, 3 unknowns

Answer 27

2x+3y+z=1
y=3z
x=2z-1

4z-2 +9z +z =1
14z = 3
z=3/14

samja ?

Answer 28

but z should be -1/14

Answer 29

abee yaar....
-j (2z-x) ha, right = j
so,
x-2z = 1 hoga na

Answer 30

2x+3y+z=1
y=3z
x=2z+1

4z+2 +9z +z =1
14z = -1
z=-1/14

Answer 31

y= 3z = -3/14

Answer 32

wahi to bola tha :/ -j se compare karenge

Answer 33

6z-3x=-1<<<<<<NO aisa kyun par

Answer 34

thats correct, i saw y=3z afterwards....

Answer 35

okay then evt is okay but if we take
axb=j-k
iska cross with a then evt reduces to simple smhw

Answer 36

a x a xb
(a.b)a-(a.a)b
a-14b
now just compute axb and equate to a-14b

Answer 37

(a.b)a = a..hmm

Answer 38

a.b to 1 hai

Answer 39

so, a=14b = aX (j-k)

Answer 40

a-14b**

Answer 41

but (a.b)a ko kaise khola

Answer 42

a.a bhi 14 hai, 4+9+1=14

Answer 43

yes a.a=14 and a.b=1

Answer 44

(a.b)a-(a.a)b
a - 14b

Answer 45

latter one is okay but former me confusion hai

Answer 46

whats the confusion xactly ?

Answer 47

how did you open (a.b) a like a.a+ab smt?

Answer 48

no!
a.b =1 is given :P
(a.b)a = 1a =a

Answer 49

omg :/

Answer 50

okay another thing when i say take cross with "a" aren't we doing in on the whole equation? like we divide the whole equation by a no. etc
we wont take (j-k)xa or smt?

Answer 51

you did aXaXb right ?
so you crossed with a in the beginning,
so it will be
aX (j-k) ............NOT (j-k)Xa

Answer 52

see if you get aX(j-k) as -4i +2j +2k
i did it on paper

Answer 53

yes wahi hai

Answer 54

then
a-14b = that
find a-14b
then compare i, j , k

Answer 55

ha hogaya thanks ^.^

Answer 56

welcome ^_^