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Question

If a,b,c are three mutually perpendicular vectors equally inclined to a+b+c at angle θ then find the value of 36cos22θ.


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Solution

Finding the value of 36cos22θ:

Since the vectors are mutually perpendicular,

a.b=b.c=c.a=0

(a+b+c).(a+b+c)=a+b+c2a2+|b|2+c2=a+b+c2(1)a.(a+b+c)=|a||a+b+c|cosθa2=|a||a+b+c|cosθ|a|=|a+b+c|cosθ

In the same manner,

b=a+b+ccosθ|c|=a+b+ccosθa=b=c

From equation 1, we get

3|a|2=|a+b+c|23a+b+c2cos2θ=|a+b+c|2cos2θ=131+cos2θ=2cos2θ36cos22θ=36(2×1/3-1)2=369=4

Hence, the required answer is 4.


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