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Q.

Let a→, b→ and c→ be three vectors such that |a→|=3,b→=5,b→·c→=10 and the angle between b→ and c→ is π3 . If a→ is perpendicular to the vector b→×c→ , then ∣a→×(b→×c→) is equal to

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answer is 30.00.

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Detailed Solution

|a|=3,b¯|=5            b¯.c¯=10| |b¯|ccos(b¯,c)=10 5. ccosπ3=10 c¯∣=4(a¯,b¯×c¯)=90°|a¯×(b¯×c¯)|=|a¯||b¯×c|sin(a¯,b¯×c¯) 3.5.4sinπ3·sinπ23.5·4·32=30
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