Question

Using vectors show that the points A (−2, 3, 5), B (7, 0, −1) C (−3, −2, −5) and D (3, 4, 7) are such that AB and CD intersect at the point P (1, 2, 3).

Answer 1

We have,
$$\begin{gathered} \overrightarrow{\mathrm{AP}}=\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{P}-\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{A} \\ \Rightarrow \overrightarrow{\mathrm{AP}}=(\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j}+5\hat{k}) \\ =3\hat{i}-\hat{j}-2\hat{k} \\ \overrightarrow{\mathrm{PB}}=\mathrm{position}\mathrm{vector}\mathrm{of}B-\mathrm{position}\mathrm{vector}\mathrm{of}P \\ \Rightarrow \overrightarrow{\mathrm{PB}}=(7\hat{i}-0\hat{j}-\hat{k})-(\hat{i}+2\hat{j}+3\hat{k}) \\ =6\hat{i}-2\hat{j}-4\hat{k} \\ \mathrm{Since}\overrightarrow{\mathrm{PB}}=2\overrightarrow{\mathrm{AP}}.\mathrm{So},\mathrm{vectors}\overrightarrow{PB}and\overrightarrow{AP}\mathrm{are}\mathrm{collinear}.\mathrm{But}\mathrm{P}\mathrm{is}\mathrm{a}\mathrm{point} \\ \mathrm{common}\mathrm{to}\overrightarrow{\mathrm{PB}}\mathrm{and}\overrightarrow{\mathrm{AP}}. \end{gathered}$$

Hence, P, A, B are collinear points.

$$\begin{gathered} \mathrm{Now},\overrightarrow{\mathrm{CP}}=(-3\hat{i}-2\hat{j}-5\hat{k})-(\hat{i}+2\hat{j}+3\hat{k}) \\ =(-4\hat{i}-4\hat{j}-8\hat{k}) \\ \overrightarrow{\mathrm{PD}}=(\hat{i}+2\hat{j}+3\hat{k})-(3\hat{i}+4\hat{j}+7\hat{k}) \\ =(-2\hat{i}-2\hat{j}-4\hat{k}) \\ \mathrm{Thus},\overrightarrow{\mathrm{CP}}=2\overrightarrow{\mathrm{PD}}. \\ \mathrm{So}\mathrm{the}\mathrm{vectors}\overrightarrow{\mathrm{CP}}\mathrm{and}\overrightarrow{\mathrm{PD}}\mathrm{are}\mathrm{collinear}.\mathrm{But}\mathrm{P}\mathrm{is}\mathrm{a}\mathrm{common}\mathrm{point}\mathrm{to} \\ \overrightarrow{\mathrm{CP}}\mathrm{and}\overrightarrow{\mathrm{PD}} \end{gathered}$$
Hence, C,P,D are collinear points.
Thus A, B, C, D and P are points such that A,P,B and C,P,D are two sets of collinear points.
Hence, AB and CD intersect at point P.