Question

To find an equation of the plane that passes through the points \((0,1,1),(1,0,1)\) and \((1,1,0)\).

Short answer

The equation of the plane that passes through the points \((0,1,1),(1,0,1)\) and \((1,1,0)\) is \(x + y + z = 2\).

Step-by-step solution

Concept of Equation of the Plane

Formula:

Write the expression to find equation of the plane through the point\({P_0}\left( {{x_0},{y_0},{z_0}} \right)\)with normal vector\({\rm{n}} = \langle a,b,c\rangle \)as follows.

\(a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0(1)\)

Write the expression to find vector from the point\(P\left( {{x_1},{y_1},{z_1}} \right)\)to\(Q\left( {{x_2},{y_2},{z_2}} \right)\).

\(\overrightarrow {PQ} = \left\langle {\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)} \right\rangle (2)\)

Calculation of the vector\(a\)and \(b\)

Consider the vector from the point \((0,1,1)\) to \((1,0,1)\) is \(\left( a \right)\)and the vector from the point \((0,1,1)\) to \((1,1,0)\) is \(\left( b \right).\)

For vector\(a\)

Substitute \(0\) for \({x_1},1\) for \({y_1},1\) for \({z_1},1\) for \({x_2},0\) for \({y_2}\), and \(1\) for \({z_2}\) in equation \(\left( 2 \right),\)

\(\begin{array}{l}a = \langle (1 - 0),(0 - 1),(1 - 1)\rangle \\a = \langle 1, - 1,0\rangle \end{array}\)

For vector\(b\)

Substitute \(0\) for \({x_1},1\) for \({y_1},1\) for \({z_1},1\) for \({x_2},1\) for \({y_2}\), and\(0\) for \({z_2}\) in equation \(\left( 2 \right),\)

\(\begin{array}{l}{\rm{b}} = \langle (1 - 0),(1 - 1),(0 - 1)\rangle \\{\rm{b}} = \langle 1,0, - 1\rangle \end{array}\)

As both the vectors \(a\) and \(b\) lie on the plane, the cross product of \(a\)and \(b\)is the orthogonal of the plane and it is considered as normal vector.

Calculation of the normal vector\(n\)

We know the normal vector \(\left( n \right).\)

\({\rm{n}} = {\rm{a}} \times {\rm{b}}\)

Substitute \(\langle 1, - 1,0\rangle \) for \({\bf{a}}\) and \(\langle 1,0, - 1\rangle \) for \({\bf{b}}\),

\({\rm{n}} = \langle 1, - 1,0\rangle \times \langle 1,0, - 1\rangle \)

The cross product as follows.

\(\begin{array}{l}{\rm{n}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&{ - 1}&0\\1&0&{ - 1}\end{array}} \right|\\{\rm{n}} = {\rm{i}}(( - 1)( - 1) - (0)(0)) - {\rm{j}}((1)( - 1) - (1)(0)) + {\rm{k}}((1)(0) - (1)( - 1))\\{\rm{n}} = {\rm{i}}(1 - 0) - {\rm{j}}( - 1 - 0) + {\rm{k}}(0 + 1)\\{\rm{n}} = {\rm{i}} + {\rm{j}} + {\rm{k}}\end{array}\)

Calculation of the equation of the plane

Consider the point \((0,1,1)\) as \({P_0}\left( {{x_0},{y_0},{z_0}} \right)\) and the normal vector is \(\langle 1,1,1\rangle \).

Calculation of equation of the plane:

Substitute \(0\) for \({x_0},1\) for v for \({z_0},1\) for \(a\), \(1\)for \(b\), and \(1\) for \(c\) in equation \(\left( 1 \right),\)

\(\begin{array}{l}(1)(x - 0) + (1)(y - 1) + (1)(z - 1) = 0\\x + y - 1 + z - 1 = 0\\x + y + z = 2\end{array}\)

Thus, the equation of the plane that passes through the points \((0,1,1),(1,0,1)\) and \((1,1,0)\) is \(x + y + z = 2\).