Fundamentals of Physics Extended (10th Edition)

$5\left( \sqrt {3}-\sqrt {2}\right) \approx 1.59m$
If $\overrightarrow {r}=\overrightarrow {a}+\overrightarrow {b}$ $\Rightarrow r=(a_x\widehat {i}+a_y\widehat {j}+a_{z}\widehat {k})+(b_x\widehat {i}+b_y\widehat {j}+b_{z}\widehat {k})=(a_x +b_x)\widehat {i} +(a_y+b_y)\widehat {j}+(a_{z}+b_{z})\widehat {k}$ $r_{x}=\left( a_{x}+b_x\right) \widehat {i}$ Lets calculate $a_{y}\widehat {j}$ and $b_{y}\widehat {j}$, $a_x=a\times \cos \theta _{1}=10.0m\times \cos 30= 5 \sqrt 3 m;$ $b_x=10.0m\times \cos \left( \theta _{2}+\theta _{1}\right) =10.0m\times \cos \left( 105+30\right) =10.0m\times \cos 135=-5\sqrt {2}m$ $\Rightarrow r _{x}=a_{x}+bx=5\left( \sqrt {3}-\sqrt {2}\right) \approx 1.59m$