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)]
$=200$
$\Rightarrow |a+b+c|=10\sqrt{2}$