How can I find the value of k for the intersection of 3 planes to form a line?

[choob]

Homework Statement

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

Homework Equations

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.

 

[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)

 

[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?

 

[rock.freak667]

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

 

[choob]

line is normal to the plane?

 

[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?

 

[rock.freak667]

line is normal to the plane?

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?

 

[choob]

absolutely

 

[rock.freak667]

absolutely

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).

 

[choob]

what is the other r in the question?

 

[rock.freak667]

what is the other r in the question?

that is just <x,y,z>

so that for the point <x₀,y₀,z₀>

\vec{r}- \vec{r_0} = &lt;(x-x_0),(y-y_0),(z-z_0)&gt;

 

[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 don't 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?

 

[rock.freak667]

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 don't even know where to start on this one.

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?

 

[choob]

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

 

[rock.freak667]

-a, -b, -c is a point

Good

p, q, r are components in the x, y, z axis

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

 

[choob]

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