# Vectors problem

1. Apr 8, 2010

### noooffence

A plane has cartesian equation x-2y-5z=6. Give the co-ordinates of any point on this plane . Give in, parametric form, the equation of the straight line which is perpendicular to the plane and passes through your chosen point on the plane..

pls help ..

2. Apr 8, 2010

### rock.freak667

How do you think you will do this part?

For a plane ax+by+cz=d what is significant about the vector <a,b,c> ?

3. Apr 8, 2010

### noooffence

i know the direction vector is <1,-2,-5> and i got a random point <5,8,11>.

i also know that for an equation that is perpendicular to the plane a.b = 0

but i dont know how to use the dot product to find the vector of that equation...

4. Apr 8, 2010

### rock.freak667

If you know that is the direction vector and it is the same as the normal to the plane then what are those two vectors (if one happened to be a scalar multiple of the other)?

5. Apr 12, 2010

### Susanne217

Isn't what the professor of nooooffence is asking for him/her to write the normal-vector of that plan in parametric form?

6. Apr 12, 2010

### HallsofIvy

No, he is asked to write the equation of the normal line, not vector.

7. Apr 12, 2010

### Susanne217

If I remember correctly the eqn of normal or normal line is

$$y = -\frac{(x-x_0)}{f'(x_0) }+ f(x_0)$$ for y = f(x) if it has a skew tangent at point $$(x_0,f(x_0))$$.

Susanne

8. Apr 12, 2010

### Rasalhague

That's right, $<1,-2,-5>$ is perpendicular to the plane, and will be parallel to whichever line you chose.

To test whether this is a point in the plane, try substituting its components, $<x,y,z>$, into the equation $x-2y-5z=6$.

A parametric equation for a line looks like this:

$$\textbf{R}(t)=\textbf{R}(0)+t\textbf{V}$$

where $\textbf{R}(t)$ for all values of $t$ are position vectors representing each of the points on the line, $\textbf{R}(0)$ is the position vector for one particular point on the line where your parameter $t$ happens to be equal to 0, and $\textbf{V}$ is a vector parallel to the line. Oh, and the parameter $t$ takes the value of each of the real numbers.

One way to describe a plane is in terms of which vectors are perpendicular to a certain vector, $\textbf{N}$.

$$\textbf{N} \cdot (\textbf{R}-\textbf{R}_0)=0$$

$$\textbf{N} \cdot \textbf{R}= \textbf{N} \cdot \textbf{R}_0$$

Where $\textbf{R}_0$ is a constant position vector indicating some particular point in the plane, and position vectors of the form $\textbf{R}$ stand for each of the other points in the plane. In your case,

$$\textbf{N} \cdot \textbf{R}= 6$$

with $\textbf{R} = <x,y,z>$ and $\textbf{N} = <1,-2,-5>$, as you worked out.