# Angle between a plane and a line

Gold Member

## Homework Statement

If the plane 3x+y+2z+6=0 is parallel to the line (3x-1)/2b = 3-y = (z-1)/a then 3a+3b is?

## Homework Equations

angle between two lines=90 degree when l1l2+m1m2+n1n2=0 where l,m,n are direction ratios...(i)

## The Attempt at a Solution

First I divided x/3 and 3-y by -1 to get the direction ratios 2b/3; -1 and a. Since the plane is parallel the normal to the plane is perpendicular to the line. Using 3,1,2 for the plane's dr's and putting it in (i) we get b+3a=1, I'm stuck here. I tried to use AM>=GM but that gave me ab<=1/12 which isn't very useful. How do I get a second equation?

Homework Helper

## Homework Statement

If the plane 3x+y+2z+6=0 is parallel to the line (3x-1)/2b = 3-y = (z-1)/a then 3a+3b is?

## Homework Equations

angle between two lines=90 degree when l1l2+m1m2+n1n2=0 where l,m,n are direction ratios...(i)

## The Attempt at a Solution

First I divided x/3 and 3-y by -1 to get the direction ratios 2b/3; -1 and a. Since the plane is parallel the normal to the plane is perpendicular to the line. Using 3,1,2 for the plane's dr's ? and putting it in (i) we get b+3a=1, I'm stuck here. I tried to use AM>=GM but that gave me ab<=1/12 which isn't very useful. How do I get a second equation?
What is the normal of the plane? What is the directional vector of the line? And they are perpendicular...
I think you have some mistake when calculating the scalar product.

Homework Helper
Dearly Missed

## Homework Statement

If the plane 3x+y+2z+6=0 is parallel to the line (3x-1)/2b = 3-y = (z-1)/a then 3a+3b is?

## Homework Equations

angle between two lines=90 degree when l1l2+m1m2+n1n2=0 where l,m,n are direction ratios...(i)

## The Attempt at a Solution

First I divided x/3 and 3-y by -1 to get the direction ratios 2b/3; -1 and a. Since the plane is parallel the normal to the plane is perpendicular to the line. Using 3,1,2 for the plane's dr's and putting it in (i) we get b+3a=1, I'm stuck here. I tried to use AM>=GM but that gave me ab<=1/12 which isn't very useful. How do I get a second equation?
You need ##(2b/3,-1,a) \perp (3,1,2).## How do you express that algebraically? What condition do you get on ##a## and ##b##?

Note: if you think about the problem geometrically you will see that you were not given enough information to determine ##a## and ##b## separately, but can at least get a relationship between them.