Geometry of lines and planes II

Equation of a line through a line with position vector $\vec{a}$ parallel to $\vec{b}$



Vector form
$\vec{r}=\vec{a}+\lambda\vec{b},\lambda \in \mathbb{R}$

Cartesian form
$\vec{r}=(x,y,z), \vec{a}=(a_1,a_2,a_3), \vec{b}=(b_1,b_2,b_3)$
$(x,y,z)=(a_1+\lambda b_1,a_2+\lambda b_2,a_3+\lambda b_3)$
$\Rightarrow x=a_1+\lambda b_1, y=a_2+\lambda b_2, z=a_3+\lambda b_3$

If $b_1,b_2,b_3 \neq 0$, then eliminating $\lambda$ from these equations yields
$\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}(=\lambda)$

Example:
Equation of line through $(-2,0,5)$ parallel to $(1,2,-3)$
$\frac{x+2}{1}=\frac{y}{2}=\frac{z-5}{-3} \Rightarrow 3y=6x+12=10-2z$

Example:
Equation of line through $(1,2,3)$ parallel to $(-2,0,5)$
Note that $b_2=0$ in this case.
$x=1-2\lambda, y=2+0\lambda, z=3+5\lambda$
Eliminating $\lambda \Rightarrow y=2, \frac{1-x}{2}=\frac{z-3}{5}$
Thus, the equations are $y=2, 5x+2z=11$.

Remark: Both equations are needed to describe the line. Each equation on its own describes a plane. The required line is the intersection of these two planes.

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Equation of a line through two points



$\vec{AB}=\vec{b}-\vec{a}$

Vector form
$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})=(1-\lambda)\vec{a}+\lambda\vec{b}, \lambda \in \mathbb{R}$

Cartesian form
$x=a_1+\lambda(b_1-a_1), y=a_2+\lambda(b_2-a_2), z=a_3+\lambda(b_3-a_3)$
or $\Large \frac{x-a_1}{b_1-a_1}=\frac{y-a_2}{b_2-a_2}=\frac{z-a_3}{b_3-a_3}$ if the denominators are non-zero.

Example:
Prove that medians of a triangle are concurrent.



Proof:
Let $\vec{a}$ be the position vector from the origin to A, $\vec{b}$ be the position vector from the origin to B, and so on.
Then $\vec{f}=\frac{1}{2}(\vec{a}+\vec{b}),\vec{e}=\frac{1}{2}(\vec{b}+\vec{c}),\vec{e}=\frac{1}{2}(\vec{a}+\vec{c})$
Any point on the line BE is given by $(1-\lambda)\vec{b}+\lambda\frac{1}{2}(\vec{a}+\vec{c})$.
Similarly, any point on AD is $(1-\mu)\vec{a}+\mu\frac{1}{2}(\vec{b}+\vec{c})$.
Since $G$ lies on both of these lines, we have $(1-\lambda)\vec{b}+\lambda\frac{1}{2}(\vec{a}+\vec{c})=(1-\mu)\vec{a}+\mu\frac{1}{2}(\vec{b}+\vec{c})$.
$\Rightarrow 1-\lambda=\frac{\mu}{2}$ and $\frac{\lambda}{2}=1-\mu$
$\Rightarrow \lambda=\mu=\frac{2}{3}$
Therefore $G=\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$.
Finally, check that G lies on CF.

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Equation of plane through the origin and parallel to $\vec{a}$ and $\vec{b}$



$\vec{r}=\lambda\vec{a}+\mu\vec{b}, \lambda, \mu \in \mathbb{R}$

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Equation of plane through C parallel to $\vec{a}$ and $\vec{b}$

A general plane can be specified by giving two vectors which lie on the plane and the position vector of a point lying on the plane.



$\vec{r}=\vec{OP}=\vec{OC}+\vec{CP}=\vec{c}+\lambda\vec{a}+\mu\vec{b}, \lambda, \mu \in \mathbb{R}$

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Equation of plane through points $\vec{a},\vec{b},\vec{c}$

We can also specify a plane uniquely by giving 3 non-collinear points which lie on it.


$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})+\mu(\vec{c}-\vec{a})=(1-\lambda-\mu)\vec{a}+\lambda\vec{b}+\mu\vec{c}, \lambda, \mu \in \mathbb{R}$

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Equation of plane in terms of normal to the plane


Projection of $OP$ onto $ON=\vec{r}\cdot\hat{n}=p$
Let $\hat{n}=(a,b,c), \vec{r}=(x,y,z)$
$ax+by+cz=p$

Key point:
One parameter is needed to write the equation of a line and two parameters are needed for that of a plane.

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