Find the value of x such that the four points with position vectors, A\left ( 3\hat{i}+2\hat{j}+\hat{k} \right ),B\left ( 4\hat{i}+x\hat{j}+5\hat{k} \right ),C\left ( 4\hat{i}+2\hat{j}-2\hat{k} \right )\; and\: \: D\left ( 6\hat{i}+5\hat{j}-\hat{k} \right ) are  coplanar.

 

 

 

 
 
 
 
 

Answers (1)

Here \overline{AB}= \hat{i}+\left ( x-2 \right )\hat{j}+4\hat{k}
        \overrightarrow{}{AC}= \hat{i}-3\hat{k}
        \overrightarrow{AD}= 3\hat{i}+3\hat{j}-2\hat{k}   
As A,B,C and D are coplanar so \left [ \overrightarrow{AB}\, \overrightarrow{AC} \overrightarrow{AD} \right ]= 0
\Rightarrow \begin{vmatrix} 1 & x-2&4 \\ 1& 0&-3 \\ 3 & 3 &-2 \end{vmatrix}= 0
\Rightarrow 9-7\left ( x-2 \right ) +4\times 3= 0
\therefore x= 5 

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