BK=1/4AB , KD*KB-? .............

Решите задачу:

$A(4,0,0)\; ,\; \; B(0,4,0)\; ,\; \; D(0,0,3)\\\\AK=\frac{3}{4}\cdot AB\; \to \; \; AK:KB=3:1\; \; \to \; \; \lambda =\frac{3}{1}=3\\\\x_{K}=\frac{x_{A}+\lambda \cdot x_{B}}{1+\lambda }=\frac{4+3\cdot0}{1+3}=\frac{4}{4}=1\\\\y_{K}=\frac{y_{A}+\lambda \cdot y_{B}}{1+\lambda }=\frac{0+3\cdot 4}{1+3}=\frac{12}{4}=3\\\\z_{K}=0\; \; \; \to \; \; \; K(1,3,0)\\\\\overline {KD}=(0-1,0-3,3-0)=(-1,-3,3)\\\\\overline {KB}=(0-1,4+3,0-0)=(-1,7,0)\\\\\overline {KD}\cdot \overline {KB}=-1\cdot (-1)+7\cdot (-3)+0\cdot 3=1-21=-20$