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  4. 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|=

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Q. 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 $|a+b+c|=$

Vector Algebra

Solution:

$\because a \perp(b+c)$
$\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, $|a+b+c|=5 \sqrt{2}$