The 3-D Coordinate System — Question 3

Question 3 – Solution

Strategy Moving one-third of the way from $A$ to $B$ produces the required $1:2$ division.

Section formula Since $B-A=\langle 9,-6,-6\rangle$, $$P=A+\frac 13(B-A)=(-5,2,7)+(3,-2,-2)=\boxed{(-2,0,5)}.$$

Distance check We have $P-A=\langle 3,-2,-2\rangle$ and $B-P=\langle 6,-4,-4\rangle=2(P-A)$. Therefore $AP=\sqrt{17}$ and $PB=2\sqrt{17}$, giving $AP:PB=1:2$.

Key idea An internal ratio determines an affine weighted average; the distance check also confirms that $P$ lies between the endpoints.