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Question

If a,b,c are mutually perpendicular vectors of equal magnitude, then a+b+c is equally inclined to

A
a only
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B
a,c only
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C
All a,b,c
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D
none of these
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Solution

The correct option is D All a,b,c
Given

a.b=b.c=c.a=0
And
|a|=|b|=|c|=k (say)

a.(a+b+c)=|a||a+b+c|cosα
Where α is the angle between a and a+b+c

a.a+a.b+a.c=k|a+b+c|cosα

|a|2+a.b+a.c=k|a+b+c|cosα

k2=k|a+b+c|cosα

cosα=k|a+b+c| ----- (i)

b.(a+b+c)=|b||a+b+c|cosβ
Where β is the angle between b and (a+b+c)

b.a+b.b+b.c=k|a+b+c|cosβ

|b|2=k|a+b+c|cosβ

k2=k|a+b+c|cosβ

cosβ=k|a+b+c| ----- (ii)
Similarly,
c.(a+b+c)=|c||a+b+c|cosγ
Where γ is the angle between c and (a+b+c)

c.a+c.b+c.c=k|a+b+c|cosγ

|c|2=k|a+b+c|cosγ

k2=k|a+b+c|cosγ

cosγ=k|a+b+c| ----- (iii)
From (i),(ii), and (iii)

cosα=cosβ=cosγ

Hence a+b+c is equally inclined to all a,b and c

Hence, Option C is the correct answer.

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