1. Aug 12, 2008

### terryfields

i think i get this now but just checking as i have the exam tomorrow and wont do this part if im getting it wrong.

(2) (a) find a vector perpendicular to a=i+2j-2k and b=-2i+3j+5k
just use the cross product? and get 15i+j+7k

(b) (i) find the equation of the plane through position vector (2,1,1) and perpendicular to the vector(3,-1,2)

is this Plane II=T(3,-1,2) + (2,1,1)

determin the perpendicular distance of plane II from the orogin

no idea here??

(ii) Find the equation of the plane III through three points (2,-2,1)=a (4,-1,6)=b and (3,-3,2)=c i use a-b and a-c and then do the cross product of these two answers to get (6i,-3j,-3k) am i right in thinking that it doesnt matter which 2 i use to get this cross product e.g i could use b-c b-a etc to find the normal then i have the answer planeIII=T(6,-3,-3)+(2,-2,1) (plus any point)

(iii) find the angles between the 2 planes, i'll wait for some feedback before i attempt this, thanks

2. Aug 12, 2008

### Defennder

Yes correct.

The equation of any plane is given by $$\vec{n} \cdot \vec{P_0 P} = 0$$ where n is the normal vector to the plane, OP is the position vector (x,y,z) and P0 is any reference point on the plane. The perpendicular distance from the plane can be found by assuming that the normal vector can be scaled by some constant so that it "touches" the origin.

You're right it does not matter, but make sure the two vectors you use are not parallel or anti-parallel (why?).

3. Aug 12, 2008

### terryfields

sorry i dont understand this second part what is the vector P0P? dont understand what i have to do with the normal vector here to find the equation of the plane?

4. Aug 12, 2008

### terryfields

and so ive got the 2 plane equations wrong right?

5. Aug 12, 2008

### Defennder

The P0 simply refers to any point on the plane taken as a reference. For example if it is given that the point (1,2,3) lies on the plane, you can let P0 be the reference point.

That is equation of a line, if I interpret your T to mean the parameter which varies. The vector equation of a plane has two parameters, not one. You're not required to express the equation of a plane in vector form, are you? If not just stick to the x,y,z notation alone.

6. Aug 12, 2008

### terryfields

so on this b(i) how do i go about finding (a) any point on the plane and (b) the normal vector as the only information i am given is a position vector (2,1,1,) and a perpendicular vector (3,-1,2)

7. Aug 12, 2008

### Defennder

The position vector given is a point on the plane. As for the normal vector, what is definition of "normal"?

8. Aug 12, 2008

### terryfields

but i have to do N.P0P doesnt that mean N dot product with the distance between the position vector and any point on the plane? if so isnt the distance between the position vector and itself going to be zero? as for the normal vector isnt that the opposite of the perpendicular? but i have no clue how to find it

9. Aug 12, 2008

### Defennder

Yes it will equate to zero. That is how you get an equation. The normal vector to the plane is also perpendicular to the plane.