Q100 / 180 · KCET 2016 · Mathematics
If \( |\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5 \) each one of \( \vec{a}, \vec{b} \& \vec{c} \) is perpendicular to the sum of the
remaining then \( |\vec{a}+\vec{b}+\vec{c}| \) is equal to
remaining then \( |\vec{a}+\vec{b}+\vec{c}| \) is equal to
- A\( \frac{5}{\sqrt{2}} \)
- B\( \frac{2}{\sqrt{5}} \)
- Correct answerC\( 5 \sqrt{2} \)
- D\( \sqrt{5} \)
Solution
Given that, each one of $\vec{a}, \vec{b}$ and $\vec{c}$ is perpendicular to the sum of the remaining. So,
$\vec{a} \cdot(\vec{b}+\vec{c})=0 \rightarrow(1)$
$\vec{b} \cdot|\vec{c}+\vec{a}|=0 \rightarrow(2)$
$\vec{c} \cdot \mid \vec{a}+\vec{b})=0 \rightarrow(3)$
Now, adding Eqs. (1), (2) and (3), we get
$2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
Now,
$|a+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow|a+\vec{b}+\vec{c}|^{2}=9+16+25$
$\Rightarrow|a+\vec{b}+\vec{c}|^{2}=50 \Rightarrow|a+\vec{b}+\vec{c}|^{2}=5 \sqrt{2}$
Hence, the correct answer is option: 'C'.