1. Jul 19, 2005

### ziddy83

hey guys...stuck on a hw problem here... very simple but...im not sure how to start...

Here is the question: Find a vector equation and parametric equations for:

The line through the point (1,0,6) and perpendicular to the plane x + 3y +z =5
can i just..use that equation to find parametric equations and then find vector equations from that ? Thanks for any help...

2. Jul 19, 2005

### Imo

do you remember what the meaning of the co-efficients of the plane are? I suggest you look that up, and then remember that all you need to make the equation of a line is a directional vector and a point. That's the absolute simplest way to do it. The parametric equations come straight from that.

3. Jul 20, 2005

### geosonel

try here:
http://www.usd.edu/~jflores/MultiCalc02/WebBook/Chapter_13/Graphics/Chapter13_5/DemoHtml13_5/13.5%20LinesAndPlanes.htm [Broken]
the first equation indicates the vector equation of a line thru point r0 (which is given to be <1,0,6> here) and parallel to vector v. if the line is normal to the plane given above, then this line is parallel to the plane's normal vector (which will be used for v). scroll down the page of the above URL to determine a vector v normal to the plane (which can be determined from the coefficients of the plane equation given above). the parametric line equation can easily be found from the vector equation.

Last edited by a moderator: May 2, 2017
4. Jul 20, 2005

### Dr.Brain

What do you need for the parametric equation of a line? , You need one point on it and the direction cosines of the line.Now because the line is prependicular to the above given plane.So the direction ratios of the line will be same as direction ratios of a normal vector to the plane.From the equation of the plane , you can make out that the vector prependicular to the plane given is a vector: i + 3j + k , So the direction ratios of the required line are 1,3,1 and it passes through point 1,0,6 . Find the direction cosines from the direction ratios and form the parametric equation of the line.

BJ