Question

Let $a,b,c$ be the three vectors such that $a.(b+c)=b.(c+a)=0$ and $\left|a\right|=1,\left|b\right|=4,\left|c\right|=8$, then $|a+b+c|=$

• A. $13$
• B. $81$
• C. $9$
• D. $5$

Answer 1

The correct option is C $9$

Adding $a.(b+c)=b.(c+a)=c.(a+b)=0$

we get $2(a.b+b.c+c.a)=0$

$\Rightarrow {|a+b+c|}^2={\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2+2(a.b+b.c+c.a)=1+16+64=81$

Hence $|a+b+c|=9$