A) \[\sqrt{12}\]
B) \[2\sqrt{12}\]
C) \[3\sqrt{14}\]
D) \[2\sqrt{14}\]
Correct Answer: D
Solution :
We have, \[a=i+2j+3k\] \[\therefore \]\[b=i\times (a\times i)+j\,\times \,(a\times j)+k\times (a\times k)\] ?(i) Now, \[i\times (a\times i)=(i.i)a-(i.a)i\] \[=1(i+2j+3k)-(1)i\] \[=2j+3k\] Similarly, \[j\times (a\times j)=i+3k\] and \[k\times (a\times k)=i+2j\] \[\therefore \]From Eq. (i), \[b=2j+3k+i+3k+i+2j\] \[=2i+4j+6k\] \[\Rightarrow \] \[|b|=\sqrt{4+16+36}=2\sqrt{14}\]You need to login to perform this action.
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