Thread: Equation of a plane in normal form ( with 2 lines)

Hi
Problem: a plane has got a straight line
(x,y,z)= (5+t,6-t,7+2t) and is parallell with
The line (x,y,z)= (3+t,5+t,6+t) determine the equation
Of the plane in normal form

We got two lines

I can write line1 as
(5,6,7) + t*(1,-1,2)

Means a point 1 and vector 1

The same applies for line2

(3,6,5) + t(1,1,1)

Point 2 and vector 2

I have a feeling I have to use Sarrus rule on this
Can someone show me how solve it?

Hey Riazy.

You can use the equations to get two direction vectors (remember the direction is important, not the location) to get the normal vector to the plane, and you also have a point on the plane (or several given your line equation).

Do you know the normal equation that relates a plane normal with a general point on the plane with a specific known point on the plane?

Do you mean I should cross multiply those two vectors
With the basis vectors e1,e2,e3?

Yes, if you multiply the two direction vectors (note the word direction), you'll get a normal vector to the plane and you can do this because one lies on the plane and the other is parrallel.

The multiplication I refer to is the cross product.

Ok so when I have got the normal, what's the next step?
Ihave two points, how can I deal with the problem now?

The plane equation in linear form is n . (r - r0) = 0 where r is any point on the plane, r0 is a specific point and n is the normal. Have you gone throught this in class? If not I think that if they haven't mentioned it then you should go ask your lecturer/teacher.