If `veca,vecb,vec(c)` are mutually perpendicular vectors having magnitudes 1,2,3 respectivley, then `[veca+vecb+vec(c)" "vecb-veca" "vec(c)]`=

Options

A. 0
B. 6
C. 12
D. 18

Answer

C

Explanation

We have,
`vec(a),vec(b), and vec(c)` are mutaully perpendicular vectors.
`therefore vec(a)*vec(b)=vec(b)*vec(c)=vec(a)*vec(c)=0`
and `|veca|=1,|vecb|=2,|vec(c)|=3`
Now, `[vec(a)+vec(b)+vec(c)" "vec(b)-vec(a)" "vec(c)]`
`=(vec(a)+vecb+vec(c))*((vecb-veca)xxvec(c))`
`=(vec(a)+vecb+vec(c))*[(vecbxxvec(c))-(vecaxxvec(c))]`
`=veca*(vecbxxvec(c))-veca*(vecaxxvec(c))+vecb*(vecbxxvec(c))-vecb*(vecaxxvec(c))`
`+vec(c)*(vecbxxvec(c))-vec(c)*(vecaxxvec(c))`
`=[veca vecb vec (c)]-vecavecavec(c)]=vec(b)vecbvec(c)]-[vecbvecavec(c)]+[vec(c)vecb vec(c)]-[vec(c)vecavec(c)]`
`=[vecavecbvec(c)]`
`=2veca*(vecbxxvec(c))=2veca*(|vecb||vec(c)|"sin"(pi)/(2)hatn)`
`=2veca*(2xx3xx1xxhatn)=12veca*hatn`
`=12|veca||hatn|cos0^(@)=12xx1xx1xx1=12`
Hence, option (c) is correct.