Question
Find the coordinate of the points which trisect the line segment joining the points$ \text{A}(2, 1, -3)$ and $\text{B}(5, -8, 3).$
✓
Answer
Let $\text{P}(x_1, y_1, z_1) $and $\text{Q}(x_2, y_2, z_2)$ trisect line segment $\text{AB}.$
Since point $\text{P}$ divides $\text{AB}$ in the ratio $1 : 2$ internally, we have
$\text{P}(\text{x}_1,\text{y}_1,\text{z}_1)\equiv\text{P}\Big(\frac{1(5)+2(2)}{2+1},\frac{1(-8)+2(1)}{2+1},\frac{1(3)+2(-3)}{2+1}\Big)$
$\equiv\text{P}(3,-2,-1)$
Since point $\text{Q}$ divides $\text{AB}$ in the ratio $2 : 1$ internally, we have
$\text{P}(\text{x}_2,\text{y}_2,\text{z}_2)\equiv\text{Q}\Big(\frac{2(5)+1(2)}{2+1},\frac{2(-8)+1(1)}{2+1},\frac{2(3)+(-3)}{2+1}\Big)$
$\equiv\text{Q}(4,-5,1)$
Since point $\text{P}$ divides $\text{AB}$ in the ratio $1 : 2$ internally, we have
$\text{P}(\text{x}_1,\text{y}_1,\text{z}_1)\equiv\text{P}\Big(\frac{1(5)+2(2)}{2+1},\frac{1(-8)+2(1)}{2+1},\frac{1(3)+2(-3)}{2+1}\Big)$
$\equiv\text{P}(3,-2,-1)$
Since point $\text{Q}$ divides $\text{AB}$ in the ratio $2 : 1$ internally, we have
$\text{P}(\text{x}_2,\text{y}_2,\text{z}_2)\equiv\text{Q}\Big(\frac{2(5)+1(2)}{2+1},\frac{2(-8)+1(1)}{2+1},\frac{2(3)+(-3)}{2+1}\Big)$
$\equiv\text{Q}(4,-5,1)$
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