Show that an equation of a plane passing through three noncolinear points p 1 = ( x 1 , y 1 , z 1 ), p 2 = ( x 2 , y 2 , z 2 ), p 3 = ( x 3 , y 3 , z 3 ) is given by ( p − p 1 ) × ( p − p 2 ) ⋅ ( p − p 3 ) = 0, where p=(x,y,z) is an arbitrary point of the plane and p − p 1 , for instance, means the vector ( x − x 1 , y − y 1 , z − z 1 )I have the following reasoning:"Explicitly, the plane P through the point P 0 = ( x 0 , y 0 , z 0 ) is uniquely determined (up to a scalar multiple) by a normal vector n=⟨a,b,c⟩ according to the following: a point P lies on P if and only if n and P 0 P → are orthogonal if and only if n ⋅ P 0 P → = 0 if and only if a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0.By setting d = a x 0 + b y 0 + c z 0 , we have a x + b y + c z = d .But I have not been able to finish the problem because I must get the equation of the plane of the form ( p − p 1 ) × ( p − p 2 ) ⋅ ( p − p 3 ) = 0. I need help to do this.

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Show that an equation of a plane passing through three noncolinear points       p    1    =  (      x    1    ,      y    1    ,      z    1    ),       p    2    =  (      x    2    ,      y    2    ,      z    2    ),       p    3    =  (      x    3    ,      y    3    ,      z    3    ) is given by   (  p  −      p    1    )  ×  (  p  −      p    2    )  ⋅  (  p  −      p    3    )  =  0, where p=(x,y,z) is an arbitrary point of the plane and   p  −      p    1  , for instance, means the vector   (  x  −      x    1    ,  y  −      y    1    ,  z  −      z    1    )I have the following reasoning:&quot;Explicitly, the plane       P   through the point       P    0    =  (      x    0    ,      y    0    ,      z    0    ) is uniquely determined (up to a scalar multiple) by a normal vector n=⟨a,b,c⟩ according to the following: a point P lies on       P   if and only if n and                     P        0            P        →   are orthogonal if and only if   n  ⋅                    P        0            P        →    =  0 if and only if  a  (  x  −      x    0    )  +  b  (  y  −      y    0    )  +  c  (  z  −      z    0    )  =  0.By setting   d  =  a      x    0    +  b      y    0    +  c      z    0    , we have   a  x  +  b  y  +  c  z  =  d  .But I have not been able to finish the problem because I must get the equation of the plane of the form   (  p  −      p    1    )  ×  (  p  −      p    2    )  ⋅  (  p  −      p    3    )  =  0. I need help to do this.

kunstboom8w · Accepted Answer

As noticed, we should have a       p    3   instead of a   p for the cross product used to determine the normal, then we can use the triple product determinant      [    (          p      3        −          p      1        )    ×    (          p      3        −          p      2        )    ]    ⋅  (  p  −      p    3    )  =  det      |                                        x            3                    −                      x            1                                                y            3                    −                      y            1                                                z            3                    −                      z            1                                                            x            3                    −                      x            2                                                y            3                    −                      y            2                                                z            3                    −                      z            2                                                x          −                      x            3                                    y          −                      y            3                                    z          −                      z            3                                |    =  0to get the plane equation.

Show that an equation of a plane passing through three noncolinear points p_1=(x_1,y_1,z_1), p_2=(x_2,y_2,z_2), p_3=(x_3,y_3,z_3) is given by (p−p_1) xx (p−p_2) * (p−p_3)=0, where p=(x,y,z) is an arbitrary point of the plane and p−p_1, for instance, means the vector (x−x_1,y−y_1,z−z_1)

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