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Three vectors $$\vec { A } =a\vec { i } +\vec { j } +\vec { k } ,\vec { B } =\vec { i } +b\vec { j } +\vec { k } ,\vec { C } =\vec { i } +\vec { j } +c\vec { k } $$ are mutually perpendicular ($$\vec { i } ,\vec { j } ,\vec { k } $$ are unit vectors along $$X,Y,Z$$ axis respectively). The respective values of $$a,b$$ and $$c$$ are


A
0,0,0
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B
12,12,12
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C
1,1,1
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D
12,12,12
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Solution

The correct option is B $$-\cfrac { 1 }{ 2 } , -\cfrac { 1 }{ 2 } , -\cfrac { 1 }{ 2 } $$
$$a+b+1=0$$  ------- (1)
$$1+b+c=0$$  --------(2)
$$a+1+c=0$$  ---------(3)
--------------------
$$2(a+b+c)+3=0$$
$$a+b+c=-\cfrac{3}{2}$$  ---- (4)
$$\Rightarrow$$ $$-1+c=\cfrac{-3}{2}$$ using (1)
$$c=-\dfrac{1}{2}$$    $$b= - \dfrac{1}{2}$$       $$a= -\dfrac{1}{2}$$

Physics

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