Co-ordinate of Point on Plane & Perpendicular Line

[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 ..

 

[rock.freak667]

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 ..

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> ?

 

[noooffence]

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> ?

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 don't know how to use the dot product to find the vector of that equation...

 

[rock.freak667]

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 don't know how to use the dot product to find the vector of that equation...

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)?

 

[Susanne217]

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 ..

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

 

[HallsofIvy]

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

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

 

[Susanne217]

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

Okay my bad,

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

 

[Rasalhague]

i know the direction vector is <1,-2,-5>

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

and i got a random point <5,8,11>.

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

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

but i don't know how to use the dot product to find the vector of that equation...

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.