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Mathematics Question on Product of Two Vectors

If a, ba , \,b and cc are three non-zero vectors such that each one of them being perpendicular to the sum of the other two vectors, then the value of ∣a+b+c∣2| a + b + c |^{2} is

A

∣a∣2+∣b∣2+∣c∣2\left|{a}\right|^{2} +\left|{b}\right|^{2} +\left|{c}\right|^{2}

B

∣a∣+∣b∣+∣c∣\left|{a}\right| +\left|{b}\right| +\left|{c}\right|

C

2(∣a∣2+∣b∣2+∣c∣2)2\left(\left|{a}\right|^{2}+\left|{b}\right|^{2}+\left|{c}\right|^{2}\right)

D

12(∣a∣2+∣b∣2+∣c∣2)\frac{1}{2}\left(\left|{a}\right|^{2}+\left|{b}\right|^{2}+\left|{c}\right|^{2}\right)

Answer

∣a∣2+∣b∣2+∣c∣2\left|{a}\right|^{2} +\left|{b}\right|^{2} +\left|{c}\right|^{2}

Explanation

Solution

According to the given condition, each vector is perpendicular to the sum of two vectors.
∴a⋅(b+c)=0,\therefore a \cdot( b + c )=0,
b⋅(a+c)=0b \cdot( a + c )=0
and c⋅(a+b)=0 c \cdot( a + b )=0,
⇒a⋅b+a⋅c=0,b⋅a+b⋅c=0\Rightarrow a \cdot b + a \cdot c =0, b \cdot a + b \cdot c =0
and c⋅a+c⋅b=0 c \cdot a + c \cdot b =0
⇒2(a⋅b+b⋅c+c⋅a)=0   ...(i)\Rightarrow 2( a \cdot b + b \cdot c + c \cdot a )=0\,\,\,...(i)
Now,∣a+b+c∣2=∣a∣2+∣b∣2+∣c∣2+2(a⋅b+b⋅c+c⋅a)Now ,| a + b + c |^{2}=| a |^{2}+| b |^{2}+| c |^{2}+2( a \cdot b + b \cdot c + c \cdot a )
=∣a∣2+∣b∣2+∣c∣2+2(0)   [=| a |^{2}+| b |^{2}+| c |^{2}+2(0) \,\,\,[ From E (i)]
=∣a∣2+∣b∣2+∣c∣2=| a |^{2}+| b |^{2}+| c |^{2}