Question

$If\overrightarrow{a}=<3,-2,1>\overrightarrow{b}=<-1,1,1>$ then the unit vector parallel to the vector $\overrightarrow{a}+\overrightarrow{b}$ is

• A. $<\frac{2}{3},-\frac{1}{3},\frac{2}{3}>$
• B. $<\frac{2}{5},-\frac{1}{5},\frac{2}{5}>$
• C. $<\frac{2}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}>$
• D. $<-\frac{2}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{2}{\sqrt{3}}>$

Answer 1

The correct option is A $<\frac{2}{3},-\frac{1}{3},\frac{2}{3}>$

$\overrightarrow{a}+\overrightarrow{b}=<3,-2,1>+<-1,1,1>=<2,-1,2>\Rightarrow |\overrightarrow{a}+\overrightarrow{b}|=\sqrt{4+1+4}=\sqrt{9}=3$
Unit vector parallel to $\overrightarrow{a}+\overrightarrow{b}$ is $\pm \frac{a+b}{|\overrightarrow{a}+\overrightarrow{b}|}=\pm \frac{<2,-1,2>}{3}$