# How do you find parametric equations for the line through the point (0,1,2) that is perpendicular to the line x =1 + t , y = 1 – t , z = 2t and intersects this line?

Dec 11, 2016

See below.

#### Explanation:

Given the line $L$ and ${p}_{1} = \left(0 , 1 , 2\right)$ where

$L \to p = {p}_{0} + t \vec{v}$

where $p = \left(x , y , z\right)$, ${p}_{0} = \left(1 , 1 , 0\right)$ and $\vec{v} = \left(1 , - 1 , 2\right)$

The elements ${p}_{1}$ and $L$ define a plane $\Pi$ with normal vector $\vec{n}$ given by

$\vec{n} = {\lambda}_{1} \left({p}_{0} - {p}_{1}\right) \times \vec{v}$ where ${\lambda}_{1} \in \mathbb{R}$

The sought line ${L}_{1} \in \Pi$ and is orthogonal to $L$ so

${L}_{1} \to p = {p}_{1} + {t}_{1} {\vec{v}}_{1}$ where ${\vec{v}}_{1} = {\lambda}_{2} \vec{v} \times \vec{n}$ with ${\lambda}_{2} \in \mathbb{R}$

Dec 11, 2016

Please see the helpful video and the explanation for my solution to the problem.

#### Explanation:

Here is a video that helped me to know how to do this problem. Helpful Video

Lets write the vector equation of the line:

$\left(x , y , z\right) = \left(1 , 1 , 0\right) + t \left(\hat{i} - \hat{j} + 2 \hat{k}\right)$

A plane that is perpendicular to this line will have the general equation:

$x - y + 2 z = c$

We make it contain the point by substituting in the point and solving for c:

$0 - 1 + 2 \left(2\right) = c$

$c = 3$

The plane $x - y + 2 z = 3$ contains the point $\left(0 , 1 , 2\right)$ and is perpendicular to the line.

To find the point where the line intersects the plane, substitute the parametric equations of the line into the equation of the plane:

$x - y + 2 z = 3$

$\left(1 + t\right) - \left(1 - t\right) + 2 \left(2 t\right) = 3$

$1 + t - 1 + t + 4 t = 3$

$6 t = 3$

$t = \frac{1}{2}$

$x = 1 + \frac{1}{2} = \frac{3}{2}$

$y = 1 - \frac{1}{2} = \frac{1}{2}$

$z = 2 \left(\frac{1}{2}\right) = 1$

The line intersects the plane at the point (3/2, 1/2, 1)

check $x - y + 2 z = 3$:

$\frac{3}{2} - \frac{1}{2} + 2 = 3$

$3 = 3$

This checks.

The vector, $\overline{v}$, from the given point to the intersection point is:

$\overline{v} = \left(\frac{3}{2} - 0\right) \hat{i} + \left(\frac{1}{2} - 1\right) \hat{j} + \left(1 - 2\right) \hat{k}$

$\overline{v} = \frac{3}{2} \hat{i} - \frac{1}{2} \hat{j} - \hat{k}$

The vector equation of the line is:

$\left(x , y , z\right) = \left(0 , 1 , 2\right) + t \left(\frac{3}{2} \hat{i} - \frac{1}{2} \hat{j} - \hat{k}\right)$

The parametric equations are:

$x = \frac{3}{2} t$

$y = 1 - \frac{1}{2} t$

$z = 2 - t$