Question

Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.

Answer 1

The given points are A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1).
$\therefore$ AB = PV of B - PV of A $=(2\hat{i}+6\hat{j}+3\hat{k})-(1\hat{i}+2\hat{j}+7\hat{k})$
$=(2-1)\hat{i}+(6-2)\hat{j}+(3-7)\hat{k}=\hat{i}+4\hat{j}-4\hat{k}$
Magnitude of AB, $|AB|=\sqrt{(1{)}^2+(4{)}^2+(-4{)}^2}=\sqrt{1+16+16}=\sqrt{33}$
BC = PV of C - PV of B $=(3\hat{i}+10\hat{j}-1\hat{k})-(2\hat{i}+6\hat{j}+3\hat{k})$
$=(3-2)\hat{i}+(10-6)\hat{j}+(-1-3)\hat{k}=\hat{i}+4\hat{j}-4\hat{k}$
Magnitude of BC, $|BC|=\sqrt{(1{)}^2+(4{)}^2+(-4{)}^2}=\sqrt{1+16+16}=\sqrt{33}$
AC = PV of C - PV of A $=(3\hat{i}+10\hat{j}-1\hat{k})-(1\hat{i}+2\hat{j}+7\hat{k})$
$=(3-1)\hat{i}+(10-2)\hat{j}+(-1-7)\hat{k}=2\hat{i}+8\hat{j}-8\hat{k}$
Magnitude of AC, $|AC|=\sqrt{2^2+8^2+(-8{)}^2}=\sqrt{4+64+64}=\sqrt{132}=2\sqrt{33}$
$=\sqrt{33}+\sqrt{33}$
$\therefore |AC|=|AB|+|BC|$. Hence, the given points A,B and C are collinear.