Question

Let $\vec{a},\vec{b},\vec{c}$ be three vectors of magnitudes $3,4$ and $5$ respectively. If each one is perpendicular to the sum of the other two vectors, then $|\vec{a}+\vec{b}+\vec{c}|=$

• A. $5$
• B. $3\sqrt{2}$
• C. $5\sqrt{2}$
• D. $12$

Answer 1

Answer: (C)

We have $|\vec{a}|=3,|\vec{b}|=4$ and $|\vec{c}|=5$. It is given that
$\vec{a}\perp (\vec{b}+\vec{c}),\vec{b}\perp (\vec{c}+\vec{a})$ and $\vec{c}\perp (\vec{a}+\vec{b})$
$\Rightarrow \vec{a}\cdot (\vec{b}+\vec{c})=0,\vec{b}\cdot (\vec{c}+\vec{a})=0,\vec{c}\cdot (\vec{a}+\vec{b})=0$
$\Rightarrow \vec{a}\cdot \vec{b}+\vec{a}\cdot \vec{c}=0,\vec{b}\cdot \vec{c}+\vec{b}\cdot \vec{a}=0,\vec{c}\cdot \vec{a}+\vec{c}\cdot \vec{b}=0$
Adding all these, we get, $2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$
$\Rightarrow \vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=0$
Now, $|\vec{a}+\vec{b}+\vec{c}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2$
$+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=3^2+4^2+5^2+0=50$
$\Rightarrow |\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=5\sqrt{2}$