If a=i+2j−3k,b=2i+j+k,c=i+3j−2k then (a×b)×(b×c)=

a×b=ijk12−3211=i(2+3)−j(1+6)+k(1−4)=5i−7j−3kb×c=ijk21113−2=i(−2−3)−j(−4−1)+k(6−1)=−5i+5j+5k(a×b)×(b×c)=ijk5−7−3−555=i(−35+15)−j(25−15)+k(25−35)

If a=i+2j−3k,b=2i+j+k,c=i+3j−2k then (a×b)×(b×c)=

a×b=ijk12−3211=i(2+3)−j(1+6)+k(1−4)=5i−7j−3kb×c=ijk21113−2=i(−2−3)−j(−4−1)+k(6−1)=−5i+5j+5k(a×b)×(b×c)=ijk5−7−3−555=i(−35+15)−j(25−15)+k(25−35)

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If $a = i + 2 j - 3 k, b = 2 i + j + k, c = i + 3 j - 2 k$ then $(a \times b) \times (b \times c) =$

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$5 (2 i + j + k)$

$- 5 (2 i + j + k)$

$10 (2 i + j + k)$

$- 10 (2 i + j + k)$

answer is D.

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Detailed Solution

$\begin{array}{l} a \times b = |\begin{array}{rrr} i & j & k \\ 1 & 2 & - 3 \\ 2 & 1 & 1 \end{array}| = i (2 + 3) - j (1 + 6) + k (1 - 4) = 5 i - 7 j - 3 k \\ b \times c = |\begin{array}{rrr} i & j & k \\ 2 & 1 & 1 \\ 1 & 3 & - 2 \end{array}| = i (- 2 - 3) - j (- 4 - 1) + k (6 - 1) = - 5 i + 5 j + 5 k \\ (a \times b) \times (b \times c) = |\begin{array}{rrr} i & j & k \\ 5 & - 7 & - 3 \\ - 5 & 5 & 5 \end{array}| = i (- 35 + 15) - j (25 - 15) + k (25 - 35) \end{array}$