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 ? ??
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?
Given a line of the form , the vector is perpendicular to the line. This is the vector component that is parallel to your line, i.e. the of the vector equation: . You have a point on the line and thus you can find to complete your equation.
EDIT: WOW BEaTEN BY 3 PEOPLE????
It is a normal vector as the plane equation is not equal to zero, right?
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
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
Then the equation of the line would be
They also could have been written in parametric form.
So I would use whatever form your professor or book uses.