If `vecu=veca-vecb,vecv=veca+vecb and |veca|=|vecb|=2,` then `|vecuxxvecv|` is equal to

Options

A. `2sqrt(16(a*b)^(2))`
B. `sqrt(16(a*b)^(2))`
C. `2sqrt(4-(a*b)^(2))`
D. `2sqrt(4+(a*b)^(2))`

Answer

A

Explanation

`|uxxv|=|(a-b)xx(a+b)|`
`=2|axxb|" "(becauseaxxa=bxxb=0)`
`and |axxb|^(2)+(a*b)^(2)=(ab sin theta)^(2)+(ab cos theta)^(2)=a^(2)b^(2)`
`implies|axxb|=sqrt(a^(2)b^(2)-(a*b)^(2))`
So, `|uxxv|=2|axxb|`
`=2sqrt(a^(2)b^(2)-(a*b)^(2))`
`=2sqrt(2^(2)2^(2)-(a*b)^(2))`
`=2sqrt(16-(a*b)^(2))because|a|=|b|=2`