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 ? ??(Headbang)

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- Jun 7th 2008, 08:34 PMCrakavector 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 ? ??(Headbang) - Jun 7th 2008, 08:43 PMTheEmptySet
The Normal Vector to the plane is $\displaystyle \vec n =1 \vec i + 3 \vec j + 1 \vec k$ You can get this by taking the gradient of the plane.

To find the vector equation of the plane

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

so we get

$\displaystyle \vec r(t)=(\vec i +6 \vec k) +t(1 \vec i + 3 \vec j + 1 \vec k)=(1+t)\vec i +(3t)\vec j +(6+t) \vec k$ - Jun 7th 2008, 08:44 PMJhevon
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? - Jun 7th 2008, 08:44 PMearboth
- Jun 7th 2008, 08:45 PMo_O
Given a line of the form $\displaystyle ax + by + cz + d = 0$, the vector $\displaystyle (a,b,c)$ is perpendicular to the line. This is the vector component that is parallel to your line, i.e. the $\displaystyle \vec{v}$ of the vector equation: $\displaystyle \vec{r} = \vec{r}_{0} + \lambda \vec{v}$. You have a point on the line and thus you can find $\displaystyle \vec{r}_{0}$ to complete your equation.

EDIT: WOW BEaTEN BY 3 PEOPLE???? - Jun 7th 2008, 08:49 PMTheEmptySet
- Jun 7th 2008, 09:00 PMCrakaQuote:

Quote:

Originally Posted by**Craka**http://www.mathhelpforum.com/math-he...s/viewpost.gif

*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 ? ??(Headbang)

The Normal Vector to the plane is http://www.mathhelpforum.com/math-he...a93880f4-1.gif You can get this by taking the gradient of the plane.

To find the vector equation of the plane

http://www.mathhelpforum.com/math-he...069ae4df-1.gif

so we get

http://www.mathhelpforum.com/math-he...eead6920-1.gif

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) ? - Jun 7th 2008, 09:08 PMo_O
Given the plane $\displaystyle ax + by + cz + d = 0$, the vector $\displaystyle (a,b,c)$ is orthogonal to the plane.

The j component is present as 0j. - Jun 7th 2008, 09:09 PMTheEmptySet
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

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

Then the equation of the line would be

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

They also could have been written in parametric form.

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