Question

It is given that $a ,\, b ,\, c$ are vectors of lengths $6,\,8 ,\, 10$ respectively. If $a$ is perpendicular to $( b + c ), b$ is perpendicular $( c + a ) ;$ and $c$ is perpendicular to $( a + b )$, then the length of the vector of $a + b + c$ is

• A. $6 \sqrt{2}$
• B. $12 \sqrt{2}$
• C. $5 \sqrt{2}$
• D. $10 \sqrt{2}$

Answer 1

Answer: (D)

We have,
$| a |=6,| b |=8, \,| c |=10, a \perp( b + c )$
b $1(c +a)$ and $c \perp(a +b)$
From the given condition, we have
$a \cdot( b + c )=0$
$\Rightarrow a \cdot b + a \cdot c =0$ ...(i)
$\Rightarrow b \cdot( c + a )=0$
$\Rightarrow b \cdot c + b \cdot a =0$ ...(ii)
and $c \cdot( a + b )=0$
$c \cdot a + c \cdot b =0$ ...(iii)
On adding Eqs. (i), (ii) and (iii), we get
$2( a \cdot b + b \cdot c + c \cdot a )=0$
$[\because$ dot product is commutative $]$
$\Rightarrow a \cdot b + b \cdot c + c \cdot a =0$ ...(iv)
Now, consider.
$| a + b + c |^{2}=( a + b + c ) \cdot( a + b + c )$
$=| a |^{2}+| b |^{2}+| c |^{2}+2 \cdot( a \cdot b + b \cdot c \cdot a )$
$=36+64+100+2 \times 0$ [using Eq .(iv)]
$=200$
$\Rightarrow |a + b + c |=10 \sqrt{2}$