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Question

If a×b=2i+2jk, a+b=i3j4k then

A
a=14i+14j+k;b=34i134j5k
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B
a=j+2k;b=i4j6k
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C
a=14i+74j+2k;b=54i194j7k
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D
a=12i12j;b=34i134j4k
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Solution

The correct options are
A a=14i+14j+k;b=34i134j5k
B a=j+2k;b=i4j6k
D a=14i+74j+2k;b=54i194j7k
Checking option (A)
a=14^i+14^j+^k
b=34^i134^j5^k
a+b=14^i+14^j+^k+34^i134^j5^k
^i3^j4^k=^i3^j4^k
valid option

Now for option (B)
a=^j+2^k
b=^i4^j6^k
a+b=^j+2^k+^i4^j6^k
^i3^j4^k=^i3^j4^k
valid option

Now for option (c) is wrong correct b=54^i194^j6^k
a=14^i+74^j+2^k
b=54^i194^j6^k
a+b=14^i+74^j+2^k+54^i194^j6^k
^i3^j4^k=^i3^j4^k
valid option

Now for option (D)
a=12^i+12^j
b=34^i134^j4^k
a+b=12^i+12^j+34^i134^j4^k
54^i114^j4^k^i3^j4^k
not valid option



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