Given the vectors →a = a→i + 2→j − 3→k, →b = →i + 2a→j − 2→k, and →c = 2→i − a→j + →k, if (→a × →b) × (→b × →c) × (→c × →a) equals →0, what is the value of a?

a equals 2/3
If a equals 0, the resulting vector product is −60(2→i + →k)
(→a × →b) ⋅ (→c × →a) = 0 has no real solution for α
None of the above

Correct answer: a equals 2/3

Solution

The condition (→a × →b) × (→b × →c) × (→c × →a) = →0 implies that the scalar triple product of the vectors is zero. Solving this for the given vectors yields a = 2/3.

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