Show that the points $A (-2 \hat{i}+3 \hat{j}+5 \hat{k})$,$B (\hat{i}+2 \hat{j}+3 \hat{k})$ and $C (7 \hat{i}-\hat{k})$ are collinear.
(N/A) We have the position vectors of points $A$,$B$,and $C$ as $\vec{a} = -2\hat{i} + 3\hat{j} + 5\hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$,and $\vec{c} = 7\hat{i} - \hat{k}$.
First,we calculate the vectors $\vec{AB}$ and $\vec{BC}$:
$\vec{AB} = \vec{b} - \vec{a} = (1 - (-2))\hat{i} + (2 - 3)\hat{j} + (3 - 5)\hat{k} = 3\hat{i} - \hat{j} - 2\hat{k}$
$\vec{BC} = \vec{c} - \vec{b} = (7 - 1)\hat{i} + (0 - 2)\hat{j} + (-1 - 3)\hat{k} = 6\hat{i} - 2\hat{j} - 4\hat{k}$
Observe that $\vec{BC} = 2(3\hat{i} - \hat{j} - 2\hat{k}) = 2\vec{AB}$.
Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ and they share a common point $B$,the vectors are parallel and the points $A$,$B$,and $C$ must be collinear.
First,we calculate the vectors $\vec{AB}$ and $\vec{BC}$:
$\vec{AB} = \vec{b} - \vec{a} = (1 - (-2))\hat{i} + (2 - 3)\hat{j} + (3 - 5)\hat{k} = 3\hat{i} - \hat{j} - 2\hat{k}$
$\vec{BC} = \vec{c} - \vec{b} = (7 - 1)\hat{i} + (0 - 2)\hat{j} + (-1 - 3)\hat{k} = 6\hat{i} - 2\hat{j} - 4\hat{k}$
Observe that $\vec{BC} = 2(3\hat{i} - \hat{j} - 2\hat{k}) = 2\vec{AB}$.
Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ and they share a common point $B$,the vectors are parallel and the points $A$,$B$,and $C$ must be collinear.
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