मराठी

Prove that the Points Having Position Vectors ^ I + 2 ^ J + 3 ^ K , 3 ^ I + 4 ^ J + 7 ^ K , − 3 ^ I − 2 ^ I − 5 ^ K Are Collinear. - Mathematics

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प्रश्न

Prove that the points having position vectors \[\hat{i} + 2 \hat{j} + 3 \hat{k} , 3 \hat{i} + 4 \hat{j} + 7 \hat{k} , - 3 \hat{i} - 2 \hat{i} - 5 \hat{k}\] are collinear.

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उत्तर

Let A, B, C  be the points with position vectors
\[\hat{i} + 2 \hat{j} + 3 \hat{k} , 3 \hat{i} + 4 \hat{j} + 7 \hat{k} , - 3 \hat{i} - 2 \hat{j} - 5 \hat{k} .\]
 Then,
\[\overrightarrow{AB} =\]  Position vector of B - Position vector of A 
\[= 3 \hat{i} + 4 \hat{j} + 7 \hat{k} - \hat{i} - 2 \hat{j} - 3 \hat{k} \]
\[ = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}\]
\[\overrightarrow{BC} =\] Position vector of C -  Position vector of B
\[= - 3 \hat{i} - 2 \hat{j} - 5 \hat{k} - 3 \hat{i} - 4 \hat{j} - 7 \hat{k} \]
\[ = - 6 \hat{i} - 6 \hat{j} - 12 \hat{k} \]
\[ = - 3\left( 2 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)\]
∴ \[\overrightarrow{BC} = - 3 \overrightarrow{AB}\]
\[\text{ So, }\overrightarrow{AB} \text{ and }\overrightarrow{BC}\]  are parallel vectors.
But B  is a point common to them.
So,
\[\overrightarrow{AB}\] and \[\overrightarrow{BC}\]  are collinear.
Hence, 
A, B, C are collinear.

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Position Vector of a Point Dividing a Line Segment in a Given Ratio
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पाठ 23: Algebra of Vectors - Exercise 23.7 [पृष्ठ ६०]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 23 Algebra of Vectors
Exercise 23.7 | Q 3 | पृष्ठ ६०

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