Question

Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear and find the ratio in which B divides AC.

Answer 1

The given points are A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7)
AB = PV of B - PV of A $=(5\hat{i}+0\hat{j}-2\hat{k})-(\hat{i}-2\hat{j}-8\hat{k})$
$=(5-1)\hat{i}+(0+2)\hat{j}+(-2+8)\hat{k}=4\hat{i}+2\hat{j}+6\hat{k}$
$|AB|=\sqrt{4^2+2^2+6^2}=\sqrt{16+4+36}=\sqrt{56}=2\sqrt{14}$
BC = PV of C - PV of B $=(11\hat{i}+3\hat{j}+7\hat{k})-(5\hat{i}+0\hat{j}-2\hat{k})$
$=(11-5)\hat{i}+(3-0)\hat{j}+(7+2)\hat{k}=6\hat{i}+3\hat{j}+9\hat{k}|BC|=\sqrt{6^2+3^2+9^2}=\sqrt{36+9+81}=\sqrt{126}=3\sqrt{14}$
AC=PV of C - PV of A $=(11\hat{i}+3\hat{j}+7\hat{k})-(\hat{i}-2\hat{j}-8\hat{k})$
$=(11-1)\hat{i}+(3+2)\hat{j}+(7+8)\hat{k}=10\hat{i}+5\hat{j}+15\hat{k}|AC|=\sqrt{{10}^2+5^2+{15}^2}=\sqrt{100+25+225}=\sqrt{350}=5\sqrt{14}\therefore |AC|=\sqrt{{10}^2+5^2+{15}^2}=\sqrt{100+25+225}=\sqrt{350}=5\sqrt{14}\therefore |AC|=|AB|+|BC|$
Thus, the given points A, B and C are collinear
Let P be the point (on the line AC) which divides |AC| in the ratio $\lambda :1$, then PV of the point P $=\frac{\lambda \times \text{PV of C+1}\times \text{PV of A}}{\lambda +1}$
$=\frac{1}{\lambda +1}\left[\lambda \right(11\hat{i}+3\hat{j}+7\hat{k})+1(\hat{i}-2\hat{j}-8\hat{k}\left)\right]=\left(\frac{11\lambda +1}{\lambda +1}\right)\hat{i}+\left(\frac{3\lambda -2}{\lambda +1}\right)\hat{j}+\left(\frac{7\lambda -8}{\lambda +1}\right)\hat{k}$
B lies on line AC, i.e., B is collinear with A and C, if P = B for a unique $\lambda$
$\Rightarrow \left(\frac{11\lambda +1}{\lambda +1}\right)\hat{i}+\left(\frac{3\lambda -2}{\lambda +1}\right)\hat{j}+\left(\frac{7\lambda -8}{\lambda +1}\right)\hat{k}=5\hat{i}+0\hat{j}-2\hat{k}\Rightarrow \frac{11\lambda +1}{\lambda +1}=5,\frac{3\lambda -2}{\lambda +1}=0\text{and}\frac{7\lambda -8}{\lambda +1}=-2\Rightarrow 11\lambda +1=5\lambda +5,3\lambda =2,7\lambda -8=-2\lambda -2\Rightarrow 6\lambda =4,\lambda =\frac{2}{3},9\lambda =6\Rightarrow \lambda =\frac{2}{3}$
Hence, A, B, C are collinear and B divides [AC] in the ratio $\frac{2}{3}:1$ i.e., 2 : 3