Show that the points A\left ( -2\hat{i} +3\hat{j}+5\hat{k}\right ),B\left ( \hat{i} \right +2\hat{j}+3\hat{k})and\: C\left ( 7\hat{i} -\hat{k}\right ) \right ) are collinear.

 

 

 

 
 
 
 
 

Answers (1)

given: \vec{A}= -2\hat{i}+3\hat{j}+5\hat{k}
           \vec{B}= \hat{i}+2\hat{j}+3\hat{k}
          \vec{C}= 7\hat{i}+0\hat{j}+\hat{k}
Now \overrightarrow{AB}=  position vector of B p.r of A
= \hat{i} +2\hat{j} +3\hat{k}-\left ( -2\hat{i} +3\hat{j}+5\hat{k}\right )
= \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{BC}=p.r of C-p.r of B
        = 7\hat{i}+0\hat{j}-0\hat{k}-\left ( \hat{i} +2\hat{j}+3\hat{k}\right )
       = 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}
       = 2\left ( 3\hat{i}-\hat{j} -2\hat{k}\right )  
clearly,\overrightarrow{BC}= 2\overrightarrow{AB}\Rightarrow AB\parallel BC but B is a common point to \overrightarrow{AB}  and\overrightarrow{BC}.
\therefore \overrightarrow{AB} \: and\: \: \overrightarrow{BC} are collinear vector
Hence, points A, B, and C are collinear. hence proved.
       
       

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