Question

Let $\overline{\alpha },\overline{\beta },\overline{\gamma }$ be three vectors such that $\overline{\alpha }\cdot (\overline{\beta }+\overline{\gamma })+\overline{\beta }\cdot (\overline{\gamma }+\overline{\alpha })+\overline{\gamma }\cdot (\overline{\alpha }+\overline{\beta })=0$ and $\left|\overline{\alpha }\right|=\sqrt{3},\left|\overline{\beta }\right|=2$ and $\left|\overline{\gamma }\right|=3$ then $|\overline{\alpha }+\overline{\beta }+\overline{\gamma }|$ is

• A. $\sqrt{17}$
• B. $4$
• C. $5$
• D. $\sqrt{15}$

Answer 1

The correct option is B $4$

${|\overline{\alpha }+\overline{\beta +\overline{\gamma }}|}^2=(\overline{\alpha }+\overline{\beta }+\overline{\gamma })\cdot (\overline{\alpha }+\overline{\beta }+\overline{\gamma })$

${|\overline{\alpha }+\overline{\beta +\overline{\gamma }}|}^2={\left|\overline{\alpha }\right|}^2+{\left|\overline{\beta }\right|}^2+{\left|\overline{\gamma }\right|}^2+\sum \overline{\alpha }\cdot (\overline{\beta }+\overline{\gamma })$

${|\overline{\alpha }+\overline{\beta +\overline{\gamma }}|}^2=3+4+9+0=16$

$|\overline{\alpha }+\overline{\beta }+\overline{\gamma }|=4$