Para encontrarmos as coordenadas do ponto B, realizaremos os cálculos abaixo:

\(\begin{align} & a=\frac{A}{|A|} \\ & a=\frac{1}{\sqrt{{{i}^{2}}+{{j}^{2}}+{{k}^{2}}}}A \\ & \\ & b=\frac{2B}{3}{{a}_{x}}-\frac{2B}{3}{{a}_{y}}+\frac{B}{3}{{a}_{z}} \\ & \left( 6-\frac{2B}{3} \right){{a}_{x}}+\left( -2+\frac{2B}{3} \right){{a}_{y}}+\left( -4-\frac{B}{3} \right){{a}_{z}}=10 \\ & {{\left( 6-\frac{2B}{3} \right)}^{2}}+{{\left( -2+\frac{2B}{3} \right)}^{2}}+{{\left( -4-\frac{B}{3} \right)}^{2}}={{10}^{2}} \\ & \left( {{6}^{2}}+12-\frac{2B}{3}+\frac{4{{B}^{2}}}{9} \right)+\left( 4+\left( 2-\frac{4B}{3} \right)+\frac{4{{B}^{2}}}{9} \right)+\left( 16-8-\frac{B}{3}+\frac{{{B}^{2}}}{3} \right)=100 \\ & 36-8B+\frac{4{{B}^{2}}}{9}+4-\frac{8B}{3}+\frac{4{{B}^{2}}}{9}+16+\frac{8B}{3}+\frac{{{B}^{2}}}{9}=100 \\ & 56+8B+{{B}^{2}}=100 \\ & {{B}^{2}}-8B-44=0 \\ & B=7,83{{a}_{x}}-7,83{{a}_{y}}+3,92{{a}_{z}} \\ \end{align}\ \)