Surely you can find a normal vector to a plane if you know the equation of the plane >_<
Given cartesian coordinates x=(z-1)/2, y=1. And p defined as 2x+y-z=1.
1. The plane p divides R3 into two parts. Find the unit vector perpendicular to p and pointing into the part containing the origin O.
2. Let P(x,y,z) be the point on the line l given by x=(z-1)/2 and y=1. Letting x=t for some constant t, find the y and z coordinates of P. Calculate the distance from P to the plane p.
I am so confused as to how to do question 1, can you show me how to work the answers out??
Also for question 2, I get P(x,y,z)=(1,0,2). Is this right? and if so how do I work out the next part of the question, to find the distance from the point to the plane.
Thanks.
I agree with this unit vector, the other of course being $\displaystyle \begin{align*} \frac{1}{\sqrt{6}} \left( -2, -1, 1 \right) \end{align*}$. Now don't guess which one points towards the origin. Choose a point on the plane. See what happens when you move from that point to the point specified by the normal unit vectors. Do you end up closer to the origin or further away?
You don't substitute anything. You know that one of your vectors is $\displaystyle \begin{align*} \left( -\frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right) \end{align*}$, this tells you how far to move in each of the x, y and z directions from your starting point.
Not sure if I'm correct, but if the plane intercepts the z axis at (0,0,-1) then the plane lies below the origin. That means the normal vector with a positive k value is the normal which points to the side of the plane containing the origin (1/root6)(-2,-1,1)
OH I SEE, thank you, i get it this one now.
Sorry, to be asking a lot but can you also help me with Question 2, was the point (1,0,2) correct? What do I do next? do i use the formula : |n.PQ|/|n| . If so the normal would be (2,1,-1) but how would I find PQ?