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Question

If a,b and c be three vectors such that |a|=3,|b|=4,|c|=5 and each one is perpendicular to the sum of the other two vectors, then find |a+b+c|2.

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Solution

Given:

|a|=3,b=4,|c|=5

Also,

Each one of them being perpendicular to the sum of the other two vectors

Therefore,

a is perpendicular to b+c

a(b+c)=0

ab+ac=0 …… (1)

b is perpendicular to a+c

b(a+c)=0

ba+bc=0 …… (2)

c is perpendicular to a+b

c(a+b)=0

ca+cb=0 …… (3)

Now,

a+b+c2=(a+b+c)(a+b+c)

a+b+c2=aa+ab+ac+ba+bb+bc+ca+cb+cc

a+b+c2=aa+(ab+ac)+bb+(ba+bc)+cc+(ca+cb)

From equations (1), (2) and (3), we have

a+b+c2=aa+(0)+bb+(0)+cc+(0)

a+b+c2=aa+bb+cc

a+b+c2=|a|2+b2+|c|2

a+b+c2=32+42+52

a+b+c2=50

So,

a+b+c=50

a+b+c=52

Hence, this is the required result.

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