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