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Q.

a¯=i¯j¯+k¯, b¯=i¯2j¯+k¯, c¯=Pi¯+2j¯+Qk¯ and d¯=Pi¯+Qj¯+2k¯ are given vectors.  If the projection of c¯ on a¯ is 53 units and if a¯,b¯,c¯ form a parallelepiped of volume 5 cubic units then Tan1b¯.d¯=

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a

π4

b

π3

c

π6

d

π2

answer is A.

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Detailed Solution

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c¯.a¯a¯=53

P+Q=171

v=a¯    b¯   c¯=5

PQ=52

From (1) (2)   P = 11, Q = 6

Tan1b¯.d¯=Tan11=π4

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