# Homework Help: Intersection of 3 planes

1. Jan 21, 2009

### choob

1. The problem statement, all variables and given/known data

Find value of k so that
x+2y-z=0
x+9y-5z=0
kx-y+z=0
intersect in a line

2. Relevant equations

3. The attempt at a solution
multiply l1 by 5
subtract l2 from 5l1
end up with:
4x+y=0

subtract l1 from l2
end up with:
7y-4z=0

i have no idea what to do from here, or even if what i did was correct.
thanks for the help.

2. Jan 21, 2009

### rock.freak667

Rewrite it in maxtrix form and row-reduce until echelon form is obtained. For the planes to intersect in a line, then when you reduce in echelon form, the rank of the augmented matrix should be less than 3 (the number of variables)

3. Jan 21, 2009

### choob

ahh thank you.

i have another problem:

Find the vector equation of the plane through A(1,-7,-2) and perpendicular to line (5,0,0)+t(2,0,7)

i have parametric equations of the line
x=5+t
y=0
z=7t

what do i do from here?

4. Jan 21, 2009

### rock.freak667

If the line is perpendicular to the plane, what can you say about the normal to the plane and the line?

5. Jan 21, 2009

### choob

line is normal to the plane?

6. Jan 21, 2009

### choob

ok, i found took the vector from 5,0,0 from 1,-7,-2

i made the z component of the vector 0

then knowing that the dot product of the line and a vector of the plane is 0, i found the component of z of a vector of the plane which would make it perpendicular to the line

is that possible?

7. Jan 21, 2009

### rock.freak667

Yes, so for a vector line in the for r=a+tu, where u is the direction of the line.

In your example u=<2,0,7>, if the lines is perpendicular to the plane, then so is the direction.

So if the direction (u) is normal to the plane, doesn't that mean that the vector u is parallel to the normal?

8. Jan 21, 2009

### choob

absolutely

9. Jan 21, 2009

### rock.freak667

so if it is parellel to the plane then

Q<2,0,7> would be a normal to this plane. Pick any non-zero value for Q, and you'll get the normal to the plane,N. When you get that, use the formule

$$(\vec{r}-\vec{r_0})\cdot \vec{N}=0$$

Where $\vec{N}$ is the normal vector and $r_0$ is a point on the plane (which you were given in the question).

Last edited: Jan 21, 2009
10. Jan 21, 2009

### choob

what is the other r in the question?

11. Jan 21, 2009

### rock.freak667

that is just <x,y,z>

so that for the point <x0,y0,z0>

$$\vec{r}- \vec{r_0} = <(x-x_0),(y-y_0),(z-z_0)>$$

12. Jan 21, 2009

### choob

thank you.

i have another question:

find a vector equation of the line through the point 4,5,5

that meets the line: (x-11)/3=y+8=z-4

i dont even know where to start on this one.

perhaps i do the same thing as before?

find the vector between 4,5,5 and the point on the line when t=0? (starting point?)
then set one of the components to 0 and use the dot product formula?

Last edited: Jan 21, 2009
13. Jan 21, 2009

### rock.freak667

For a vector line written in the form

$$\frac{x-a}{p} = \frac{y-b}{q}= \frac{z-c}{r}$$

What does <a,b,c> represent and what does <p,q,r> represent?

14. Jan 21, 2009

### choob

-a, -b, -c is a point
p, q, r are components in the x, y, z axis

15. Jan 21, 2009

### rock.freak667

Good
<p,q,r> would be the direction of the line.

16. Jan 21, 2009

### choob

haha where do i go from there then? btw is my process that i edited in at 2:27 am correct?