Let →a=^i+^j+^k,→b=2^i+^k,→c=^i+2^j+3^k, then the ascending order the following isA.[→a×→b →b×→c →c×→a]B.[→a+→b →b+→c →c+→a]C.[→a→b→c][→a1 →b1 →c1]D.[→a−→b →b−→c →c−→a]

Let →a=^i+^j+^k,→b=2^i+^k,→c=^i+2^j+3^k, then the ascending orde

1.	Let →a=^i+^j+^k,→b=2^i+^k,→c=^i+2^j+3^k, then the ascending order the following isA.[→a×→b →b×→c →c×→a]B.[→a+→b →b+→c →c+→a]C.[→a→b→c][→a1 →b1 →c1]D.[→a−→b →b−→c →c−→a]
Answer» Let $\to a =^i +^j +^k, \to b = 2^i +^k, \to c =^i + 2^j + 3^k$ , then the ascending order the following is $A . [\to a \times \to b \to b \times \to c \to c \times \to a]$ $B . [\to a + \to b \to b + \to c \to c + \to a]$ $C . [\to a \to b \to c] [{\to a}^{1} {\to b}^{1} {\to c}^{1}]$ $D . [\to a - \to b \to b - \to c \to c - \to a]$

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Let →a=^i+^j+^k,→b=2^i+^k,→c=^i+2^j+3^k, then the ascending order the following isA.[→a×→b →b×→c →c×→a]B.[→a+→b →b+→c →c+→a]C.[→a→b→c][→a1 →b1 →c1]D.[→a−→b →b−→c →c−→a]

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