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# Vectors and geometry

## Equation of a line in space

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 $\displaystyle \frac{x-x_0}{u_x}= \frac{y-y_0}{u_y}$.

Parametric and Cartesian equation of a line

## Equation of a plane in space

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 $\nvec = \uvec\times\vvec$.

Parametric and Cartesian equation of a plane

## Finding a plane passing through three points

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 $$x +6y + z = 1 +12 + 6 = 19.$$