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, $|\vec{ a }+\vec{ b }+\vec{ c }|^{2}=|\vec{ a }|^{2}+|\vec{ b }|^{2}$
$+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=3^{2}+4^{2}+5^{2}+0=50$
$\Rightarrow |\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=5 \sqrt{2}$