Find the value of (4i^ )× ( -6k^ )
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15
As this question is about vector cross product
according to some general rules
a)i^×j^=k^
b)j^×k^=i^
c)k^×i^=j^
again if the multiplication is in opposite of this rule then a minus sign will be given to the resultant
so in your question 4i^×(-6k^)=24j^
as vector multiplication is in opposite so additional minus sign is introduced
---------____________
hope it will help you
according to some general rules
a)i^×j^=k^
b)j^×k^=i^
c)k^×i^=j^
again if the multiplication is in opposite of this rule then a minus sign will be given to the resultant
so in your question 4i^×(-6k^)=24j^
as vector multiplication is in opposite so additional minus sign is introduced
---------____________
hope it will help you
DebashishJoshi:
yes vector has only two priducts dot and cross and according to your question its a cross product
Answered by
6
!! Hey Mate !!
your answer is --
Let, A = 4i^ and B = (-6k^)
we can write it as
A = 4i^ + 0j^ + 0k^ and
B = 0i^ +0j^ - 6k^
such that , a1 = 4 , a2=0, a3=0 and
b1 = 0 , b2 = 0 , b3= -6
we know that ,
A × B = i^(a2b3-b2a3) - j^(a1b3-b1a3) +k^(a1b2-b1a2)
=> A×B = i^(0×-6-0×4)-j^(4×-6-0×0)+k^(4×0-0×0)
=> A×B = i^(0) - j^(-24) +k^(0)
=> A×B = 24j^
------------------------
Alternate method
we know that
k^ + i^ = j^
=> -6k^ × 4i^ = 24j^
【 Hope it helps you 】
your answer is --
Let, A = 4i^ and B = (-6k^)
we can write it as
A = 4i^ + 0j^ + 0k^ and
B = 0i^ +0j^ - 6k^
such that , a1 = 4 , a2=0, a3=0 and
b1 = 0 , b2 = 0 , b3= -6
we know that ,
A × B = i^(a2b3-b2a3) - j^(a1b3-b1a3) +k^(a1b2-b1a2)
=> A×B = i^(0×-6-0×4)-j^(4×-6-0×0)+k^(4×0-0×0)
=> A×B = i^(0) - j^(-24) +k^(0)
=> A×B = 24j^
------------------------
Alternate method
we know that
k^ + i^ = j^
=> -6k^ × 4i^ = 24j^
【 Hope it helps you 】
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