Let three coterminous edges of a tetrahedron are a¯,b¯ and c¯ such that |a¯|=2,|b¯|=3,|c¯|=4, angle between a¯ and b¯ is π3,b¯ and c¯ is π4,c¯ and a¯ is π6. If area of the base is 2 square units and height of the tetrahedron be h=lm-n where l,m,n are positive integers and l is odd prime number then l+m-n is
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answer is 7.
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Detailed Solution
a¯.a¯=4, b¯.b¯=9, c¯.c¯=16a¯⋅b¯=3b¯⋅c¯=62c¯⋅a¯=43[a¯b¯c¯]2=a¯⋅a¯a¯.b¯a¯⋅c¯b¯⋅a¯b¯⋅b¯b¯⋅c¯c¯⋅a¯c¯⋅b¯c¯⋅c¯=144(6−2) ⇒a¯,b¯,c¯=126-2 Volume of the tetrahedron =16∥[abc]∣=26-2⇒13× base area × height =26-2⇒13×2×h=26-2⇒h=36-2l=3,m=6,n=2