Question

If either `veca = vec0`  or `vecb = vec0`, then `veca xxvecb = vec0`. Is the converse true? Justify your answer with an example.

Answer 1

When `veca = vec0,` then `|veca| = 0.`

Let 'θ' be the angle between `veca  "and" vecb`

∴ `veca xx vecb = |veca|  |vecb| sin theta = vec0`

`= (0) |vecb| sin theta = vec0`

Similarly when `vecb = vec0, "then"  veca xx vecb = vec0`

Conversely: Let `veca = a_1 hati + a_2 hatj + a_3hatk`

and `vecb = lambda a_1 hati + lambda a_2 hatj + lambda a_3 hatk `

Clearly `vec a, vecb` are parallel

⇒ θ  = 0

When `|veca| ne 0` and `|vecb| ne  0`

But `veca xx vecb = vec0` even if sinθ = 0

Hence `veca xx vecb = vec0` even `veca ne vec0`  and  `vecb ne vec0`

Let `veca = 2 hati - hatj + hatk` and `hatb = 4hati - 2hatj + 2hatk`

∴ `veca xx vecb = abs((hati,hatj, hatk), (2, -1, 1), (4, -2, 2)) = 0`

⇒ `veca xx vecb = 0`

But `veca ne vec0`  and `vecb ne  0`