Question

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are perpendicular to $\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}$ and $\overrightarrow{a}+\overrightarrow{b}$ respectively and if $|\overrightarrow{a}+\overrightarrow{b}|=6,|\overrightarrow{b}+\overrightarrow{c}|=8$ and $|\overrightarrow{c}+\overrightarrow{a}|=10$, the $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|$ is equal to

• A. $5\sqrt{2}$
• B. $50$
• C. $10\sqrt{2}$
• D. $10$

Answer 1

Answer: (D)

$10$

Given, $|\overrightarrow{a}+\overrightarrow{b}|=6$
$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}=36$ ...... (i)
Similarly, $|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2\overrightarrow{b}\cdot \overrightarrow{c}=64$ ..... (ii)
and $|\overrightarrow{c}{|}^2+|\overrightarrow{a}{|}^2+2\overrightarrow{c}\cdot \overrightarrow{a}=100$ ..... (iii)
On adding Eqs. (i), (ii) and (iii), we get
$|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})=100$
$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2=100$ ..... (iv)
$(\because \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}=0)$
Now, $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})$
$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=100$ [from Eq. (iv)]
$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=10$