Question

If $a,b,c$ are three vectors such that $|a|=3,|b|=4$, $|c|=5$ and $a,b,c$ are perpendicular to $b+c,c+a,a$ $+b$ respectively, then $|a+b+c|=$

• A. $6\sqrt{2}$
• B. $4\sqrt{2}$
• C. $3\sqrt{2}$
• D. $5\sqrt{2}$

Answer 1

Answer: (D)

$\because a\perp (b+c)$
$\therefore a\cdot (b+c)=0$
$\Rightarrow a\cdot b+a\cdot c=0(1)$
Similarly $b\perp (c+a)$
$\Rightarrow b\cdot c+b\cdot a=0(2)$
and $c\perp (a+b)=0$
$\Rightarrow c\cdot a+c\cdot b=0(3)$
Adding (1), (2), (3), we get
$2(a\cdot b+b\cdot c+c\cdot a)=0$
Now, $(a+b+c{)}^2$
$=a^2+b^2+c^2+2(a\cdot b+b\cdot c+c\cdot a)$
$=|a{|}^2+|b{|}^2+\mid c{/}^2+0$
$=(3{)}^2+(4{)}^2+(c{)}^2=50$
Hence, $|a+b+c|=5\sqrt{2}$