हिंदी

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

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

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.

shaalaa.com
Position Vector of a Point Dividing a Line Segment in a Given Ratio
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 22: Algebra of Vectors - Exercise 23.8 [पृष्ठ ६५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 22 Algebra of Vectors
Exercise 23.8 | Q 1.2 | पृष्ठ ६५
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×