How do you find a normal of a direction vector in 3 space?

[thomasrules]

Question #1. How do you find a normal of a direction vector in 3 space?
Question #2.:

What is the scalar equation of the plane that contains the x-axis and the point (4,-2,1)?

 

[HallsofIvy]

There is no single normal to a vector in 3-space. There exist an entire plane normal to any vector and any vector in that plane is normal to the original plane.

If a plane includes the entire x-axis, then it also includes the vector from (0,0,0) to (1, 0, 0): i. Since this plane includes the point (4, -2, 1) as well as (0,0,0) it includes the vector 4i- 2j+ k and so the cross product of those two vectors is perpendicular to the plane: -j- 2k is perpendicular to the entire plane. If (x, y, z) is any point in the plane, then the vector from (0,0,0) to (x,y,z), xi+ yj+ zk, is perpendicular to that vector: (xi+ yj+ zk).(-j- 2k)= 0 or -y- z= 0. The desired plane is z= -y.

 

[thomasrules]

hallsofivy...that is not the answer...I think you are wrong...can you please make it a bit more clear on the steps made to get the answer.

for my first question ok sorry maybe I worded it wrong...how do i find a normal plane of a direction vector in 3 space?

 

[thomasrules]

or maybe ur right ...thats the plane...but not the answer

I'm not sure how to find the normal...(1,0,0)cross what? to get a normal...then after that put the (x,y,z) values in the equation Ax+By+Cz+D=0
Then find D by inserting the point (4,-2,1) and you've got the equation?

Is this process correct