if `veca xx vecb = vecc ,vecb xx vecc = veca , " where " vecc ne vec0` then (a) `|veca|= |vecc|` (b) `|veca|= |vecb|` (c) `|vecb|=1` (d) `|veca|=|vecb|= |vecc|=1`

Options

A. `|veca|= |vecc|`
B. `|veca|= |vecb|`
C. `|vecb|=1`
D. `|veca|=vecb|= |vecc|=1`

Answer

a,c

Explanation

`veca xx vecb = vecc, vecb xx vecc = veca`
taking cross with `vecb` in the first equation, we get
`vecb xx (veca xx vecb) = vecb xx vecc = veca `
`or |vecb|^(2) veca - (veca .vecb) vecb =veca`
` Rightarrow |vecb|=1 and veca.vecb=0`
Also `|veca xx vecb| = |vecc| or |veca| |vecb| sin (pi/2) = |vecc|`
`or |veca| = |vecc|`