मराठी

Show that the Points A, B, C with Position Vectors → a − 2 → B + 3 → C , 2 → a + 3 → B − 4 → C and − 7 → B + 10 → C Are Collinear.

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प्रश्न

Show that the points A, B, C with position vectors \[\vec{a} - 2 \vec{b} + 3 \vec{c} , 2 \vec{a} + 3 \vec{b} - 4 \vec{c}\] and \[- 7 \vec{b} + 10 \vec{c}\] are collinear.

बेरीज
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उत्तर

We have, A, B ,C with position vectors \[\vec{a} - 2 \vec{b} + 3 \vec{c} , 2 \vec{a} + 3 \vec{b} - 4 \vec{c} , - 7 \vec{b} + 10 \vec{c}\]
Then,
\[\overrightarrow{AB} =\] Position Vector of B - Position Vector of A 
\[= 2 \vec{a} + 3 \vec{b} - 4 \vec{c} - \vec{a} + 2 \vec{b} - 3 \vec{c} \]
\[ = \vec{a} + 5 \vec{b} - 7 \vec{c}\]

\[\overrightarrow{BC} =\] Position Vector of C - Position Vector of B
\[= - 7 \vec{b} + 10 \vec{c} - 2 \vec{a} - 3 \vec{b} + 4 \vec{c} \]
\[ = - 2 \vec{a} - 10 \vec{b} + 14 \vec{c} \]
\[ = - 2 \left( \vec{a} + 5 \vec{b} - 7 \vec{c} \right)\]
∴ \[\overrightarrow{BC} = - 2 \overrightarrow{AB}\]
Hence, 
\[\overrightarrow{AB} \text{ and }\overrightarrow{BC}\] are parallel vectors.
But B is a point common to them.
So,
\[\overrightarrow{AB}\text{ and }\overrightarrow{BC}\] are collinear.
Hence, points A, B and C  are collinear.
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Position Vector of a Point Dividing a Line Segment in a Given Ratio
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Algebra of Vectors - Exercise 23.7 [पृष्ठ ६०]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 22 Algebra of Vectors
Exercise 23.7 | Q 1 | पृष्ठ ६०
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