Show that the four points A, B, C and D with position vectors \hat{i}+2\hat{j}-\hat{k},3\hat{i}-\hat{j},\: 2\hat{i}+3\hat{j}+2\hat{k} and 4\hat{i}+3\hat{k} respectively are coplanar.

 

 

 

 
 
 
 
 

Answers (1)

given points are
A= \hat{i}+2\hat{j}-\hat{k}    B= 3\hat{i}-\hat{j}, C= 2\hat{i}+3\hat{j}+2\hat{k}\, and\: D= 4\hat{i}+3\hat{k}
we know that the four points A,B,C and D will be coplanar, if there three vectors \overrightarrow{AB},\overrightarrow{AC}\: and\: \overrightarrow{AD}  are coplanar ie if \left [ \overrightarrow{AB},\overrightarrow{AC}\: and\: \overrightarrow{AD} \right ]= 0
\therefore \overrightarrow{AB}= \left ( 3\hat{i}-\hat{j} \right )-\left (\hat{i}+2\hat{j} -\hat{k} \right )= 2\hat{i}-3\hat{j}+\hat{k}
\therefore \overrightarrow{AC}= \left ( 2\hat{i}+3\hat{j}+2\hat{k} \right )-\left (\hat{i}+2\hat{j} -\hat{k} \right )= \hat{i}+\hat{j}+3\hat{k}
\therefore \overrightarrow{AD}= \left ( 4\hat{i}+3\hat{k} \right )-\left (\hat{i}+2\hat{j} -\hat{k} \right )= 3\hat{i}-2\hat{j}+4\hat{k}
now \left [ \overrightarrow{AB}\, \overrightarrow{AC}\, \overrightarrow{AD} \right ]= \begin{vmatrix} 2 &-3 &1 \\ 1& 1 &3 \\ 3 & -2 & 4 \end{vmatrix}
= 2\left ( 4+6 \right )+3\left ( 4-9 \right )+1\left ( -2-3 \right )
= 2\left ( 10 \right )+3\left ( -5 \right )+1\left ( -5 \right )
= 20-15-5= 0

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