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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).
Solution
We have,
\[\overrightarrow{AP} =\text{ position vector of P - position vector of A}\]
\[ \Rightarrow \overrightarrow{AP} = ( \hat{i} + 2 \hat{j} + 3 \hat{k} ) - ( - 2 \hat{i} + 3 \hat{j} + 5 \hat{k} )\]
\[ = 3 \hat{i} - \hat{j} - 2 \hat{k} \]
\[ \overrightarrow{PB} =\text{ position vector of B - position vector of P}\]
\[ \Rightarrow \overrightarrow{PB} = (7 \hat{i} - 0 \hat{j} - \hat{k} ) - ( \hat{i} + 2 \hat{j} + 3 \hat{k} ) \]
\[ = 6 \hat{i} - 2 \hat{j} - 4 \hat{k} \]
\[\text{ Since }\overrightarrow{PB} = 2 \overrightarrow{AP} .\text{ So, vectors }\overrightarrow{PB} \text{ and }\overrightarrow{AP}\text{ are collinear . But P is a point}\]
\[\text{ common to }\overrightarrow{PB} \text{ and }\overrightarrow{AP} .\]
Hence, P, A, B are collinear points.
\[\text{ Now, }\overrightarrow{CP} = ( - 3 \hat{i} - 2 \hat{j} - 5 \hat{k} ) - ( \hat{i} + 2 \hat{j} + 3 \hat{k} )\]
\[ = ( - 4 \hat{i} - 4 \hat{j} - 8 \hat{k} )\]
\[ \overrightarrow{PD} = ( \hat{i} + 2 \hat{j} + 3 \hat{k} ) - (3 \hat{i} + 4 \hat{j} + 7 \hat{k} )\]
\[ = ( - 2 \hat{i} - 2 \hat{j} - 4 \hat{k} )\]
\[\text{ Thus, } \overrightarrow{CP} = 2 \overrightarrow{PD} . \]
\[\text{ So the vectors }\overrightarrow{CP}\text{ and }\overrightarrow{PD}\text{ are collinear . But P is a common point to }\]
\[ \overrightarrow{CP}\text{ and }\overrightarrow{PD} \]
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.
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