Algebra Examples

Given points ${\displaystyle C=(2,9,7)}$ and ${\displaystyle D=(3,3,3)}$, find a plane containing points ${\displaystyle A=(1,2,3)}$ and ${\displaystyle B=(2,5,6)}$ that is parallel to line ${\displaystyle \mathrm{CD}}$.

First, calculate the direction vector of the line through points ${\displaystyle C}$ and ${\displaystyle D}$. This can be done by taking the coordinate values of point ${\displaystyle C}$ and subtracting them from point ${\displaystyle D}$.

Replace the ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ values and then simplify to get the direction vector ${\displaystyle V_{\mathrm{CD}}}$ for line ${\displaystyle \mathrm{CD}}$.

Calculate the direction vector of a line through points ${\displaystyle A}$ and ${\displaystyle B}$ using the same method.

Replace the ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ values and then simplify to get the direction vector ${\displaystyle V_{\mathrm{AB}}}$ for line ${\displaystyle \mathrm{AB}}$.

The solution plane will contain a line that contains points ${\displaystyle A}$ and ${\displaystyle B}$ and with the direction vector ${\displaystyle V_{AB}}$. For this plane to be parallel to the line ${\displaystyle CD}$, find the normal vector of the plane which is also orthogonal to the direction vector of the line ${\displaystyle CD}$. Calculate the normal vector by finding the cross product ${\displaystyle V_{AB}}$x${\displaystyle V_{CD}}$ by finding the determinant of the matrix ${\displaystyle \left[\begin{array}{ccc}i& j& k\\ x_B-x_A& y_B-y_A& z_B-z_A\\ x_D-x_C& y_D-y_C& z_D-z_C\end{array}\right]}$.

Add the constant to find the equation of the plane to be ${\displaystyle \left(6\right)x+\left(7\right)y+(-9)z=-7}$.