Question

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

Answer 1

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