# Thread: vector equation of line through point perpendicular to plane

1. ## vector equation of line through point perpendicular to plane

Hi,

How do I find the vector equation of this.
given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

I'm guessing here by do I do something like this
(x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??

2. Originally Posted by Craka
Hi,

How do I find the vector equation of this.
given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

I'm guessing here by do I do something like this
(x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??
The Normal Vector to the plane is $\vec n =1 \vec i + 3 \vec j + 1 \vec k$ You can get this by taking the gradient of the plane.

To find the vector equation of the plane

$\vec r(t)=P_0+t\cdot \vec v$

so we get

$\vec r(t)=(\vec i +6 \vec k) +t(1 \vec i + 3 \vec j + 1 \vec k)=(1+t)\vec i +(3t)\vec j +(6+t) \vec k$

3. Originally Posted by Craka
Hi,

How do I find the vector equation of this.
given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

I'm guessing here by do I do something like this
(x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??
the direction vector of the line is the normal vector to the plane. so we have a line passing through (1,0,6) in the direction <1,3,1>. how would you write the equation for this line?

You know why the normal vector of the plane is the direction vector of the line, right?

EDIT: well, if i'm going to be beaten by anyone, it might as well be TES... you mind me calling you TES, TheEmptySet?

4. Originally Posted by Craka
Hi,

How do I find the vector equation of this.
given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

I'm guessing here by do I do something like this
(x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??
From the equation of the plane you already know it's normal vector: $\vec n = (1,3,1)$

Therefore the equation of the line is:

$l: \vec r=(1,0,6)+t \cdot (1,3,1)$

Beaten again ...

5. Given a line of the form $ax + by + cz + d = 0$, the vector $(a,b,c)$ is perpendicular to the line. This is the vector component that is parallel to your line, i.e. the $\vec{v}$ of the vector equation: $\vec{r} = \vec{r}_{0} + \lambda \vec{v}$. You have a point on the line and thus you can find $\vec{r}_{0}$ to complete your equation.

EDIT: WOW BEaTEN BY 3 PEOPLE????

6. Originally Posted by Jhevon

EDIT: well, if i'm going to be beaten by anyone, it might as well be TES... you mind me calling you TES, TheEmptySet?
Fine by be I have been called ALOT worse

7. Quote:
Originally Posted by Craka
Hi,

How do I find the vector equation of this.
given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

I'm guessing here by do I do something like this
(x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??

The Normal Vector to the plane is You can get this by taking the gradient of the plane.

To find the vector equation of the plane

so we get

It is a normal vector as the plane equation is not equal to zero, right?

sorry didn't realise needed to use the equation of that format. Obviously if there were a j component that would be included in (i+6k) ?

8. Given the plane $ax + by + cz + d = 0$, the vector $(a,b,c)$ is orthogonal to the plane.

The j component is present as 0j.

9. Originally Posted by Craka
It is a normal vector as the plane equation is not equal to zero, right?

sorry didn't realise needed to use the equation of that format. Obviously if there were a j component that would be included in (i+6k) ?
Yes it would need to be included. The notation I used is what I am used to. There are many other ways this could have been written for example

$n=<1,3,1> \\\ P=<1,0,6>$

Then the equation of the line would be

$r(t)=<1,0,6>+t<1,3,1>=<1+t,3t,6+t>$

They also could have been written in parametric form.

So I would use whatever form your professor or book uses.

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# find the vector equation for the line that goes through point and is perpendiculat to the x-axis and vector u

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