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# Let Then If c 2 = –1 and c 3 = 1, show that no value of c 1 can make coplanar.

4(b)   Let $\vec a = \hat i + \hat j + \hat k , \vec b = \hat i \: \: and \: \: \vec c = c_1 \hat i + c_2 \hat j + c_3 \hat k$  Then
If $c_2 =- 1$ and $c_3 =1$ show that no value of $c_3$ can make $\vec a , \vec b , \vec c$ coplanar.

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Given

$\vec a = \hat i + \hat j + \hat k , \vec b = \hat i \: \: and \: \: \vec c = c_1 \hat i - \hat j + \hat k$

these will be coplanar when,

$\left [ \vec a ,\vec b, \vec c \right ]=\begin{vmatrix} 1 &1 &1 \\ 1& 0& 0\\ c_1&-1 &1 \end{vmatrix}=1(0)-1(1)+1(-1)=-2$

Hence the value of  $\left [ \vec a ,\vec b, \vec c \right ]$  is -2, irrespective of the value of $c_3$. Hence  no value of  $c_3$ satisfies the condition of three vectors being coplanar.

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