I need to find the normal vector for the following 3d vector presented in the vectorial equation because I need to find a plane that is orthogonal to the following line:
$(x,y,z)=(1,0,0)+k(1,2,3)$
I understand that two normal vectors will have a dot product of $0$, however, I am not sure what to do in 3D space?
In 2D space, if I have an equation of $(x,y) = (p1,p2) +k(a,b)$, I know that the normal vector would be $(-b,a)$.
Furthermore, I would like to ask if I have an equation for plan: $2x+3y+4z=0$, in which $(2,3,4)$ is the normal vector. How can I obtain the direction vector for this plan?
Thanks!
Best Answer
In $3$ dimensions, there are infinitely many vectors perpendicular to a given vector.
As you said $(x,y,z)\perp(1,2,3)\iff x+2y+3z=0$.
One solution is $(x,y,z)=(1,1,-1)$ by inspection.
One way to find a vector perpendicular to a given vector in $3$ dimensions is to take the cross-product with another (non-collinear) vector.
For example, $(1,0,0)\times(1,2,3)=(0,-3,2)$ is perpendicular to both $(1,0,0)$ and $(1,2,3)$, as you can verify by showing their dot product is $0$.
Now that we have two vectors perpendicular to $(1,2,3)$, any linear combination of those two vectors $\alpha(1,1,-1)+\beta(0,-3,2)$ with $\alpha,\beta\in\mathbb R$ will also be perpendicular to $(1,2,3)$.