Let $\vec{A}, \vec{B}, \vec{C}$ be three vectors of length $3,4,5$ respectively. Let $\vec{A}$ be pe

Question

Let $\vec{A}, \vec{B}, \vec{C}$ be three vectors of length $3,4,5$ respectively. Let $\vec{A}$ be perpendicular to $\vec{B}+\vec{C}, \vec{B}$ to $\vec{C}+\vec{A}$ and $\vec{C}$ to $\vec{A}+\vec{B}$. Then the square of  length of the vector $\vec{A}+\vec{B}+\vec{C}=$ $\_\_\_\_$\

Professor PC Thomas Classes · Answer

Given that $|\vec{A}|=3,|\vec{B}|=4,|\vec{C}|=5$. We have, $(\vec{A}+\vec{B}+\vec{C})^{2}=\vec{A}^{2}+\vec{B}^{2}+\vec{C}^{2}$ $$ \begin{gathered} +2(\vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{C}+\vec{C} \cdot \vec{A}) \ \Rightarrow|\vec{A}+\vec{B}+\vec{C}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+|\vec{C}|^{2}+\vec{A} \cdot(\vec{B}+\vec{C}) \ +\vec{B} \cdot(\vec{C}+\vec{A})+\vec{C} \cdot(\vec{A}+\vec{B}) \end{gathered} ~ \begin{array}{r} \ \Rightarrow|\vec{A}+\vec{B}+\vec{C}|^{2}=(3)^{2}+(4)^{2}+(5)^{2}+0+0+0 \ (\because \vec{A} 	ext { is perpendicular to } \vec{B}+\vec{C} 	ext { etc }) \end{array} $$