Question

(a) Find the equation of the plane passing through the points (2,1,0), (5,0,1) and (4,1,1).

Answer 1

vectors between the points should lie in the plane; find two such vectors and their cross product will give you a plane normal.

Answer 2

How ?

Answer 3

find the gradient between two points and pick a point and sub into y-y1=m(x-x1)?

Answer 4

I got 3a-b+c=0 is it correct ??

Answer 5

which point did you choose to find the gradient and which point did you choose for the equation?

Answer 6

(2,1,0) as point

Answer 7

you need two points to find the gradient. remember y2-y1 divided by x2-x1?

Answer 8

oh ok i see now. dont do what i just told you. Sorry i read the question wrong eep!

Answer 9

(a) Find the equation of the plane passing through the points P(2,1,0), Q(5,0,1) and R(4,1,1).

consider: \(\vec{PR}=(2,0,1),\vec{PQ}=(3,-1,1)\) lie in the plane and their cross-product gives us a normal:$$\vec{PR}\times\vec{PQ}=(1,-1,-2)$$any vector lying in the plane should be orthogonal to our normal vector:$$\vec{X}\cdot(\vec{PR}\times\vec{PQ})=0$$to get a vector lying in the plane from a position vector \(\vec{x}\) we take its difference with a position vector pointing to any point in our plane e.g. \(\vec{P}\):$$(\vec{x}-\vec{P})\cdot(\vec{PR}\times\vec{PQ})=0\\((x,y,z)-(2,1,0))\cdot(1,-1,-2)=0\\(x-2,y-1,z)\cdot(1,-1,-2)=0\\(x-2)-(y-1)-2z=0\\x-y-2z-2+1=0\\x-y-2z-1=0$$

Answer 10

wait, it appears I made some dumb error... probably while computing my normal vector. it should be:$$\vec{PR}\times\vec{PQ}=(1,1,-2)$$hence:$$(x-2,y-1,z)\cdot(1,1-2)=0\\(x-2)+(y-1)-2z=0\\x+y-2z-3=0$$

Answer 11

Thanks @oldrin.bataku