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Question
If →a=3^i+2^j+2^k,→b=−^i+3^j−^k,→c=^i+^j+^k, then find the value of →a×(→b×→c) and (→a×→b)×→c
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Solution
→a=3^i+2^j+2^k,→b=−^i+3^j−^k,→c=^i+^j+^k,
(→a×→b)×→c=?
→a×(→b×→c)=?
→b×→c=∣∣
∣
∣∣^i^j^k−13−1111∣∣
∣
∣∣
=^i∣∣∣3−111∣∣∣−^j∣∣∣−1−111∣∣∣+^k∣∣∣−1311∣∣∣
=^i(3+1)−^j(−1+1)+^k(−1−3)=4^i−4^k
→a×(→b×→c)=∣∣
∣
∣∣^i^j^k32240−4∣∣
∣
∣∣
=^i∣∣∣220−4∣∣∣−^j∣∣∣324−4∣∣∣+^k∣∣∣3240∣∣∣=^i(−8−0)−^j(−12−8)+^k(0−8)
=−8^i+20^j−8^k
→a×(→b×→c)=−8^i+20^j−8^k
→a×→b=∣∣
∣
∣∣^i^j^k322−13−1∣∣
∣
∣∣
=^i∣∣∣223−1∣∣∣−^j∣∣∣32−1−1∣∣∣+^k∣∣∣32−13∣∣∣
=^i(−2−6)−^j(−3+2)+^k(9+2)
=−8^i+^j+11^k
(→a×→b)×→c=∣∣
∣
∣∣^i^j^k−8111111∣∣
∣
∣∣
=^i∣∣∣11111∣∣∣−^j∣∣∣−81111∣∣∣+^k∣∣∣−8111∣∣∣
=^i(1−11)−^j(−8−11)+^k(−8−1)
=−10^i+19^j−9^k
(→a×→b)×→c=?
→a×(→b×→c)=?
→b×→c=∣∣ ∣ ∣∣^i^j^k−13−1111∣∣ ∣ ∣∣
=^i∣∣∣3−111∣∣∣−^j∣∣∣−1−111∣∣∣+^k∣∣∣−1311∣∣∣
=^i(3+1)−^j(−1+1)+^k(−1−3)=4^i−4^k
→a×(→b×→c)=∣∣ ∣ ∣∣^i^j^k32240−4∣∣ ∣ ∣∣
=^i∣∣∣220−4∣∣∣−^j∣∣∣324−4∣∣∣+^k∣∣∣3240∣∣∣=^i(−8−0)−^j(−12−8)+^k(0−8)
=−8^i+20^j−8^k
→a×(→b×→c)=−8^i+20^j−8^k
→a×→b=∣∣ ∣ ∣∣^i^j^k322−13−1∣∣ ∣ ∣∣
=^i∣∣∣223−1∣∣∣−^j∣∣∣32−1−1∣∣∣+^k∣∣∣32−13∣∣∣
=^i(−2−6)−^j(−3+2)+^k(9+2)
=−8^i+^j+11^k
(→a×→b)×→c=∣∣ ∣ ∣∣^i^j^k−8111111∣∣ ∣ ∣∣
=^i∣∣∣11111∣∣∣−^j∣∣∣−81111∣∣∣+^k∣∣∣−8111∣∣∣
=^i(1−11)−^j(−8−11)+^k(−8−1)
=−10^i+19^j−9^k
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