Question #198762

Find the vector equation of the plane determined by the points (-1,-2,1),(1,0,1) and (1,-1,1). Also check whether (1/2,1/2,1/2) lies on it.


1
Expert's answer
2021-05-26T16:36:56-0400

Solution:-

points  (-1,-2,1),(1,0,1) and (1,-1,1).

we will use this formula

xxAyyAzzAxBxAyByAzBzAxCxAyCyAzCzA=0\begin{vmatrix} x - x_A & y - y_A&z - z_A \\ x_B - x_A & y_B - y_A & z_B - z_A\\ x_C - x_A &y_C - y_A & z_C - z_A \end{vmatrix}=0


on putting the values


x(1)y(2)z11(1)0(2)111(1)(1)(2)11=0\begin{vmatrix} x - (-1) & y - (-2)&z - 1 \\ 1 - (-1) & 0 - (-2) &1 - 1\\ 1 - (-1) &(-1) - (-2) & 1 - 1 \end{vmatrix}=0

    \implies

[x(1)][2001][y(2)][2002]+[z1][2122]=0[x - (-1)][2·0-0·1]-[y - (-2)][2·0-0·2]+[z - 1][2·1-2·2]=0

    \implies

2z+2=0\boxed{- 2z + 2 = 0}

this is the required equation

Now we check the points (12,12,12)(\frac{1}{2},\frac{1}{2},\frac{1}{2})

=2×12+2=1+2=10=-2\times\frac{1}{2}+2\\ =-1+2\\ =1 \ne0

    \implies points does not lie on plane.


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