# What is parametric equation of the line created by the intersecting planes x = 2 and z = 2?

Aug 2, 2016

$\vec{r} = \left(\begin{matrix}2 \\ 0 \\ 2\end{matrix}\right) + \lambda \left(\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right)$

#### Explanation:

no need to do much here because every point on the line of intersection will have values x = 2 and z = 2, and any value of y is possible.

a fixed point on the line is (2,0,2) so we can say that the line is this....

$\vec{r} \left(\lambda\right) = \left(\begin{matrix}2 \\ 0 \\ 2\end{matrix}\right) + \lambda \left(\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right)$

Aug 3, 2016

$\vec{r} = \left(2 , 0 , 2\right) + t \left(0 , 1 , 0\right) , t \in \mathbb{R}$, or,

in the usual Cartesian Form $\frac{x - 2}{0} = \frac{y}{1} = \frac{z - 2}{0}$.

#### Explanation:

Let the given planes be ${\pi}_{1} : x - 2 = 0 , \mathmr{and} {\pi}_{2} : z - 2 = 0$.

Let the reqd. line be $L = {\pi}_{1} \cap {\pi}_{2}$.

To determine the eqn. of $L$, we need, $\left(1\right)$ a pt., say, $A \in L$, &

$\left(2\right)$ the direction $\vec{l}$ of $L$.

(1) : The point A :-.

$A \left(x , y , z\right) \in L = {\pi}_{1} \cap {\pi}_{2} \Rightarrow A \in {\pi}_{1} , \mathmr{and} , A \in {\pi}_{2}$

$\Rightarrow x = 2 , z = 2$. As regards, $y \in \mathbb{R}$, y is arbitrary, and, by our

choice, $y = 0$. Hence, $A = A \left(2 , 0 , 2\right)$

(2) : The Direction $\vec{l}$ of $L$ :-

Let vecn_1, &, vecn_2 be the normals to pi_1, &, pi_2, resp.

:. vecn_1=(1,0,0)=hati, &, vecn_2=(0,0,1)=hatk

L=pi_1nnpi_2 rArr L sub pi_1, &, L sub pi_2 rArr vecl bot vecn_1, &, vecl bot vecn_2.

$\Rightarrow \vec{l}$ is along ${\vec{n}}_{1} \times {\vec{n}}_{2} = \hat{i} \times \hat{k} = - \hat{j} = \left(0 , - 1 , 0\right)$.

we take, $\vec{l} = \hat{j} = \left(0 , 1 , 0\right)$

Using (1), &, (2), the vector eqn., or, parametric eqn. of $L :$

$\vec{r} = \vec{a} + t \vec{l} , t \in \mathbb{R}$, where, $\vec{a}$ is the position vector of

the pt.$A$.

Hence, $L : \vec{r} = \left(2 , 0 , 2\right) + t \left(0 , 1 , 0\right) , t \in \mathbb{R}$, or, in the usual

Cartesian Form $L : \frac{x - 2}{0} = \frac{y}{1} = \frac{z - 2}{0}$.

Hope, this will be helpful! Enjoy Maths!