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Show that the Points Whose Position Vectors Are as Given Below Are Collinear: 3 ^ I − 2 ^ J + 4 ^ K , ^ I + ^ J + ^ K and − ^ I + 4 ^ J − 2 ^ K

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Question

Show that the points whose position vectors are as given below are collinear: \[3 \hat{i} - 2 \hat{j} + 4 \hat{k}, \hat{i} + \hat{j} + \hat{k}\text{ and }- \hat{i} + 4 \hat{j} - 2 \hat{k}\]

Sum
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Solution

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

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Position Vector of a Point Dividing a Line Segment in a Given Ratio
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Chapter 22: Algebra of Vectors - Exercise 23.8 [Page 65]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 22 Algebra of Vectors
Exercise 23.8 | Q 1.2 | Page 65
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