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Question

Let a, b and c be 3 mutually perpendicular unit vectors. If d is a unit vector which is equally inclined to a, b and c, then the value of |a+b+c+d|2 is p+q then p+q is

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Solution

Since, the vector d is a unit vector which is equally inclined to a, b and c
d=a+b+c3

d.a=d.b=d.c=13

|a+b+c+d|2=|a+b+c|2+|d|2+2d.(a+b+c)

=3+1+2×(33)

=4+23

=4+12

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