# Vectors and geometry

The parametric equation, or vector equation, of a line through $$\xvec_0=(x_0, y_0, z_0)$$ and with direction $$\uvec = (u_x, u_y, u_z)\ne \vecmath{0}$$ is $$$ \xvec = \xvec_0 + \lambda\uvec,\qquad \lambda\in\R$$$

The equation of the line going through $$\xvec_0\ne\xvec_1$$ is found by using the equation above, with the direction vector $$\uvec=\xvec_1-\xvec_0.$$

The line going through $$(1,-3)$$ and $$(7,-1)$$ is $$(1,-3) + \lambda(6,2)$$. We can find points on the line by changing $$\lambda$$: when $$\lambda = 2$$, we find the point $$(13,1)$$.

The Cartesian equation of the line is $$$ \frac{x-x_0}{u_x}= \frac{y-y_0}{u_y}= \frac{z-z_0}{u_z}.$$$ If $$u_z=0$$, this means $$z=z_0$$ and $$\displaystyle \frac{x-x_0}{u_x}= \frac{y-y_0}{u_y}$$.

The parametric equation of a plane in space through $$\xvec_0=(x_0, y_0, z_0)$$ which contains two non-collinear vectors $$\uvec$$ and $$\vvec$$ is $$$ \xvec = \xvec_0 + \lambda\uvec+\mu\vvec,\qquad \lambda,\mu\in\R.$$$ This is also called the vector equation. $$\uvec$$ and $$\vvec$$ are non unique.

The Cartesian equation of the plane going through $$\xvec_0$$ and orthogonal to the vector $$\nvec\ne0$$ is $$$(\xvec-\xvec_0)\cdot\nvec = 0 .$$$ This reads $$$ n_x x + n_x y +n_x z = d,\quad d= \nvec\cdot\xvec_0 = n_x x_0 + n_y y_0 +n_z z_0.$$$

The formula gives the parametric equation of a plane through three non-collinear points $$\xvec_0$$, $$\xvec_1$$, $$\xvec_2$$, with $$\uvec =\xvec_1-\xvec_0$$ and $$\vvec =\xvec_2-\xvec_0$$.

To find the Cartesian equation, we apply the formula with $$\nvec = \uvec\times\vvec$$.

The parametric equation of a plane is $$$ \xvec = \xvec_0 + \lambda\uvec+\mu\vvec,\qquad \lambda,\mu\in\R.$$$

The Cartesian equation is $$$(\xvec-\xvec_0)\cdot\nvec = 0$$$

The parametric equation of a plane through $$(1,2,6)$$, $$(4,2,3)$$ and $$(-1,3,2)$$ is $$$(1,2,6) + \lambda(3,0,-3) + \mu(-2,1,-4)$$$ or $$$(1,2,6) + \lambda(1,0,-1) + \mu(-2,1,-4)$$$

For the Cartesian equation, we form a system of equations \begin{align*} &x = 1 + \lambda -2\mu\\ &y = 2 + \mu\\ &z = 6 - \lambda - 4\mu\\ \end{align*} We solve these to get $$x+6y+z=19$$

Alternatively, the plane goes through $$(1,2,6)$$ and is orthogonal to $$(1,0,-1)\times(-2,1,-4)=(1,6,1)$$, so we have $$$x +6y + z = 1 +12 + 6 = 19.$$$