# Thread: vector equation of a plane

1. ## vector equation of a plane

for this question (photo 1), i am not sure whether this is type 1 (as the type in photo 2) or type 2 ( as in photo 3 ). the question didnt provide a diagram, this is making me confused. so i did it another way on the right , (using pencil ). is my working acceptable ?

2. ## Re: vector equation of a plane

Originally Posted by delso

for this question (photo 1), i am not sure whether this is type 1 (as the type in photo 2) or type 2 ( as in photo 3 ). the question didnt provide a diagram, this is making me confused. so i did it another way on the right , (using pencil ). is my working acceptable ?
I have absolutely no idea what your post means.
In general if $A,~B,~\&~C$ are three non-colinear then the plane determined is:
Let $R=<x,y,z>$ then the plane is $\left( {\overrightarrow {AB} \times \overrightarrow {AC} } \right) \cdot \left[ {\overrightarrow R - \overrightarrow A } \right] = 0$

3. ## Re: vector equation of a plane

how do u know a b and c are not collinear?

4. ## Re: vector equation of a plane

my question is how would we know whether A is connected to C or B is connected to C ? If A is connected to C , then the solution is just like case 1 (photo 2 ), and vice versa

5. ## Re: vector equation of a plane

Originally Posted by delso
my question is how would we know whether A is connected to C or B is connected to C ? If A is connected to C , then the solution is just like case 1 (photo 2 ), and vice versa
They are not col-linear if $\overrightarrow {AB} \times \overrightarrow {AC} \ne 0$

6. ## Re: vector equation of a plane

in order to satisfy the condition r dot n = a dot n . the n which is the normal to the plane must originate from point a am i right? so the point a must be connected DIRECTLY to point B and point P , SO the angle between BAP can be found, but for photo 3, the point A is connected to B , but point P is connected to point B , but not A !. please correct me if my concept is wrong. VECTOR almost making me crazy.