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Question
If →a=→i+→j+→k,→a=2→i+→k,→c=2→i+→j+→k;→d=→i+→j+2→k, then verify that (→a×→b)×(→c×→d)=[→a→b→d]→c−[→a→b→c]→d
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Solution
→a×→b=∣∣
∣
∣∣→i→j→k111201∣∣
∣
∣∣=→i(I)−→j(1−2)+→k(0−2)=→i+→j−2→k
→c×→d=∣∣
∣
∣∣→i→j→k211112∣∣
∣
∣∣=→i(2−1)−→j(4−1)+→k(2−1)=→i−3→j+→k
(→a×→b)×(→c×→d)=∣∣
∣
∣∣→i→j→k11−21−31∣∣
∣
∣∣
=→i(1−6)−→j(1+2)+→k(−3−1)
=−5→i−3→j−4→k .....(1)
[→a→b→c=∣∣
∣∣111201112∣∣
∣∣=1(0−1)−1(4−1)+1(2−0)
=−1−3+2=−2
[→a→b→c]=∣∣
∣∣111201211∣∣
∣∣=1(0−1)−1(2−2)+1(2−0)=−1−0+2=1
R.H.S. =[→a→b→d]→c−[→a→b→c]→d
=(−2)(2→i+→j+→k)−(1)(→i+→j−2→k)
=−4→i−2→j−2→k−→i−→j+2→k
=−5→i−3→j−4→k ....(2)
From (1) and (2), we get
(→a×→b)×(→c×→d)=[→a→b→d]→c−[→a→b→c]→d.
→c×→d=∣∣ ∣ ∣∣→i→j→k211112∣∣ ∣ ∣∣=→i(2−1)−→j(4−1)+→k(2−1)=→i−3→j+→k
(→a×→b)×(→c×→d)=∣∣ ∣ ∣∣→i→j→k11−21−31∣∣ ∣ ∣∣
=→i(1−6)−→j(1+2)+→k(−3−1)
=−5→i−3→j−4→k .....(1)
[→a→b→c=∣∣ ∣∣111201112∣∣ ∣∣=1(0−1)−1(4−1)+1(2−0)
=−1−3+2=−2
[→a→b→c]=∣∣ ∣∣111201211∣∣ ∣∣=1(0−1)−1(2−2)+1(2−0)=−1−0+2=1
R.H.S. =[→a→b→d]→c−[→a→b→c]→d
=(−2)(2→i+→j+→k)−(1)(→i+→j−2→k)
=−4→i−2→j−2→k−→i−→j+2→k
=−5→i−3→j−4→k ....(2)
From (1) and (2), we get
(→a×→b)×(→c×→d)=[→a→b→d]→c−[→a→b→c]→d.
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