Hello,
If a plane has ax+by+cz=d as equation, a normal vector of it has coordinates (a,b,c)
To find the unit vector, divide each coordinate by the norm of the vector, given by
A vector joining 2 points and has its coordinates as follow :
Hi guys, I have this question which I really need help ...
The equation of a plane P1 in a cartesian coordinate system is given by 3x+y-z-6=0. THe cordinates of two points A and B are [1,1,3] and [2,1,2] respectively.
i)Find the unit vector normal to the plane P1.
ii)Find the vector AB.
iii)Find the equation of a plane P2 that passes thru the points A and B and is perpendicular to the plane P1.
so far i only got AB=OB-OA ..i think so which is equals to i-k.
No, you have the hypothetic equation of P2. You will have to determine a, b, c and d
As A and B are in P2, the coordinates of A verify the equation of P2 and the coordinates of B verify the equation of P2.
This will give you two equations in the aim to solve for a, b, c and d. But you need 2 others to find the complete equation.
Oh, there can be a way of doing it... even 2. But i don't really like it because it doesn't use the previous questions...
I don't remember if there are other properties we can get from the equation of a plane... With orthogonality.
Perhaps you can try a thing :
The vector (a,b,c) is orthogonal (perpendicular) to P2 (same rule as for the 1st question).
As is in P2, you can say that the scalar product of with the vector (a,b,c) is 0...
I'm not sure... can you try ?
1. The normal vector of P1 is as Moo has posted.
2. The vector
3. That means the plane P2 contains the point A and the 2 vectors which span the plane:
which yield the equation of the plane in parametric form:
4. To get this equation in coordinate form(?) calculate the normal vector of the 2 direction vectors which span the plane:
The plane contains point A. Calculate the dot-product to get the constant summand. Therefore the equation of the plane P2 is:
let be the perpendicular plane
then its normal vector is
so since the planes are perpendicular the dot product of their normals must be zero. so
Now if we evaluate the plane at the two given points we get
(1,1,3) and (2,1,2)
This is an underdetermined system
solving the system
so d=2c c=c b=-2c a=c
plugging into our equation we get
divide by c and we get