Question

Let $\vec{A},\vec{B},\vec{C}$ be 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 length of vector$\vec{A}+\vec{B}+\vec{C}$ is....

Answer 1

$| \overrightarrow{a} |=3, | \overrightarrow{d} |=4,| \overrightarrow{b} |=5$
Since, $\overrightarrow{A}.(\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{B}(\overrightarrow{C}+\overrightarrow{A})=\overrightarrow{C}.(\overrightarrow{A}+\overrightarrow{B})=0 ...(i)$
$\therefore | \overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} |^2=| \overrightarrow{A} |+| \overrightarrow{B} |^2+| \overrightarrow{C} |^2+(\overrightarrow{A}.\overrightarrow{B}+\overrightarrow{B}.\overrightarrow{C}+\overrightarrow{C}.\overrightarrow{A})$
$= 9 + 16 + 25 + 0$
[ from Eq.$(i)\overrightarrow{A}.\overrightarrow{B}+\overrightarrow{B}.\overrightarrow{C}+\overrightarrow{C}.\overrightarrow{A}=0]$
$\therefore | \overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} |^2=50$
$\Rightarrow | \overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C} |=5\sqrt{2}$