Re: Find parametric equation

Quote:

Originally Posted by

**icelated** I have an odd problem in the book where i have no solution to and i am not sure if i am doing it correctly.

Let L be the line given by $\displaystyle x = 3-t, y=2+t, z= -4 +2t$

L intersects the plane $\displaystyle 3x-2y+z=1$ at the point $\displaystyle P=(3, 2, -4)$

Find parametric equations for the line through P which lies in the plane and is perpendicular to L

**Attempt:**

Direction Vector v = $\displaystyle <-1,1,2>$

Normal Vector n = $\displaystyle <3, -2, 1>$

$\displaystyle v \times n = <5, 7, -1>$

$\displaystyle X = P + vt$

Point: $\displaystyle P=(3, 2, -4)$

v = $\displaystyle <-1,1,2>$

putting it together

$\displaystyle (3, 2, -4) + <-1,1,2>t$

Parametric equation: $\displaystyle x = 3-t, y = 2+t , z = -4 +2t$

Somehow, i dont think this is correct?

You have a mistake in concept. Your line is

$\displaystyle \\x=3+5t\\y=2+7t\\z=-4-t$

**WHY?**

Re: Find parametric equation

Are you saying i should use

$\displaystyle x = P + (v \times n) t$

instead of

$\displaystyle X = P + vt$

Re: Find parametric equation

Quote:

Originally Posted by

**icelated** I dont know what you mean. What you are trying to get.Are you saying i should use v X n instead of v?

Can you read?

You posted: Parametric equation: $\displaystyle \color{red}x = 3-t, y = 2+t , z = -4 +2t$

**THAT IS INCORRECT!**

THIS IS CORRECT: Parametric equation: $\displaystyle x = 3+5t,~ y = 2+7t ,~ z = -4 -t$.

Then I asked you why that is correct?