Question

To find an equation of the plane that passes through the points \((1, - 1,1)\) and contains the line with symmetric equations \(x = 2y = 3z\).

Short answer

The equation of the plane that passes through the point \((1, - 1,1)\) and contains the line with symmetric equations \(x = 2y = 3z\) is \(5x - 4y - 9z = 0\).

Step-by-step solution

Concept of Equation of the Plane

Formula used:

1. The equation of the plane through the point\({P_0}\left( {{x_0},{y_0},{z_0}} \right)\)with normal vector\(n = \langle a,b,c\rangle \)is\(a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0.\)

2. The parametric equations for a line through the point\(\left( {{x_0},{y_0},{z_0}} \right)\)and parallel to the direction vector\(\langle a,b,c\rangle \)are\(\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\).

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

The given points on the plane is v and the symmetric equations of line are \(x = 2y = 3z\).

Rewrite the symmetric equations as standard form \(\frac{{(x - 0)}}{6} = \frac{{(y - 0)}}{3} = \frac{{(z - 0)}}{2}\).

By the formula (2), obtain the direction vector from the line equations as \(a = \langle 1,3,2\rangle \) and position vector as \(\left( {{x_0},{y_0},{z_0}} \right) = (0,0,0)\).

Calculate the vector from the point \((0,0,0)\) to \((1, - 1,1)\) and named as,

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

Since 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\(n\)

The normal vector \({\rm{n}} = {\rm{a}} \times {\rm{b}}\) with \({\rm{a}} = \langle 1,3,2\rangle \) and \({\rm{b}} = \langle 1, - 1,1\rangle \) as follows.

\(\begin{array}{l}{\rm{n}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\6&3&2\\1&{ - 1}&1\end{array}} \right|\\{\rm{n}} = {\rm{i}}\left| {\begin{array}{*{20}{c}}3&2\\{ - 1}&1\end{array}} \right| - {\rm{j}}\left| {\begin{array}{*{20}{c}}6&2\\1&1\end{array}} \right| + {\rm{k}}\left| {\begin{array}{*{20}{c}}6&3\\1&{ - 1}\end{array}} \right|\\{\rm{n}} = {\rm{i}}((3)(1) - (2)( - 1)) - {\rm{j}}((6)(1) - (2)(1)) + {\rm{k}}((6)( - 1) - (3)(1))\\{\rm{n}} = {\rm{i}}(3 + 2) - {\rm{j}}(6 - 2) + {\rm{k}}( - 6 - 3)\\{\rm{n}} = 5{\rm{i}} - 4{\rm{j}} - 9{\rm{k}}\end{array}\)

Hence, the normal vector is \({\rm{n}} = 5{\rm{i}} - 4{\rm{j}} - 9{\rm{k}}\) whose component form is \({\rm{n}} = \langle 5, - 4, - 9\rangle \).

Calculation of the equation of the plane

Consider the point \((0,0,0)\) as \({P_0}\left( {{x_0},{y_0},{z_0}} \right)\) and the normal vector \({\rm{n}} = \langle 5, - 4, - 9\rangle \).

Apply the formula \(\left( 1 \right)\)and obtain the equation of plane as follows.

That is, substitute \({x_0} = 0,{y_0} = 0,{z_0} = 0,a = 5,b = - 4\), and \(c = - 9\) in the formula \(\left( 1 \right)\)

\(\begin{array}{l}(5)(x - 0) + ( - 4)(y - 0) + ( - 9)(z - 0) = 0\\5x - 4y - 9z = 0\end{array}\)

Thus, the equation of the plane that passes through the point \((1, - 1,1)\) and contains the line with symmetric equations \(x = 2y = 3z\) is \(5x - 4y - 9z = 0\).