# Reta orthogonal to the plan

• Jun 19th 2008, 07:08 PM
Apprentice123
Reta orthogonal to the plan
To determine the equations of a line that passes through M r (2.1,-1) and is orthogonal to the plan, ret, and r: $\displaystyle \frac{x-2}{3}=\frac{y+1}{2}=z-1$ and t: $\displaystyle \frac{x}{6}=\frac{y}{4}=\frac{z}{2}$

$\displaystyle \frac{x-2}{-3}=y-1=\frac{z+1}{7}$
• Jun 21st 2008, 09:02 AM
TKHunny

Ax + By + Cz + D = 0

Orthogonal to r

(1) 3A + 2B + C = 0

Orthogonal to t

6A + 4B + 2C = 0 ==> 3A + 2B + C = 0

Ah! They played a little trick on us. r and t are parallel. Fortunately, we can generate more direction numbers form the two most obvious points on the lines.

2-0 = 2, -1-0 = -1, and 1-0 = 1

Orthogonal to a line with those dirction numbers.

(2) 2A - B + C = 0

Contains the point (2,1,-1)

(3) 2A + B - C + D = 0

That's enough. With three constraints and four variables, we'll have to pick one to parameterize the others. I chose A.

B = -A/3, C = -7A/3, and D = -4A

Substituting into the original general form and dividing by -A/3 gives:

3x - y - 7z - 12 = 0

There are the values we seek. The line is then trivial.

$\displaystyle \frac{x-2}{3}\;=\;\frac{y-1}{-1}\;=\;\frac{z+1}{-7}\;\implies\;\frac{x-2}{-3}\;=\;y-1\;=\;\frac{z+1}{7}$
• Jun 21st 2008, 03:02 PM
Apprentice123
Thank you very much. I am Brazilian and I'm trying to communicate better. You have MSN?