Question

The points $(0,-1,-1),(-4,4,4),(4,5,1)$ and $(3,9,4)$ are

• A. collinear
• B. coplanar
• C. forming a square
• D. forming a triangle

Answer 1

The correct option is B coplanar

We know that:

The equation of the plane passing through $(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)$ is

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

The equation of the plane passing through (0 , -1 , -1) , (-4 , 4 , 4) , ( 4 , 5 , 1 ) is

$\Rightarrow \left|\begin{array}{ccc}x-0& y-(-1)& z-(-1)\\ (-4)-0& 4-(-1)& 4-(-1)\\ 4-0& 5-(-1)& 1-(-1)\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x& y+1& z+1\\ -4& 4+1& 4+1\\ 4& 5+1& 1+1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x& y+1& z+1\\ -4& 5& 5\\ 4& 6& 2\end{array}\right|=0$

$\Rightarrow (5\cdot 2-6\cdot 5)x-\left(\right(-4)\cdot 2-5\cdot 4)(y+1)+\left(\right(-4)\cdot 6-5\cdot 4)(z+1)=0$

$\Rightarrow (10-30)x-\left(\right(-8)-20)(y+1)+\left(\right(-24)-20)(z+1)=0$

$\Rightarrow (-20)x-(-28)(y+1)+(-44)(z+1)=0$

$\Rightarrow -20x+28y+28-44z-44=0$

$\Rightarrow -20x+28y-44z-16=0$

$\Rightarrow -5x+7y-11z-4=\mathrm{0.................}\left(1\right)$

Substituting the last point (3 , 9 , 4) in eq(1), we get

$=-5\cdot 3+7\cdot 9-11\cdot 4-4$

$=-15+63-44-4=0$

The equation of plane is satisfied by these points.

Therefore, these points are coplanar.