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Q.

Given  a→=xi^+yj^+2k^,b→=i^−j^+k^,c→=i^+2j^;a→⊥b→,a→⋅c→=4 Then

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a

[a→b→c→]2=|a→|

b

[a→b→c→]=|a→|

c

[a→b→c→]=0

d

[a→b→c→]=|a→|2

answer is D.

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Detailed Solution

a→⊥b→⇒x−y+2=0a→⋅c→=4⇒x+2y=4Solving we get x=0; y=2 ⇒ a→=2j^+2k^⇒ [a→b→c→]=0221−11120=8=|a→|2
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