Calculate a normal to vector lying on a plane formed by $2$ vectors

Question

Let's presume I have two vectors $V_1$ and $V_2$. As far as I understand normal to a vector is all vectors lying on a plane perpendicular to it. What I need is a normal to $V_1$ that lies on a plane formed by $V_1$ and $V_2$.

Example

Answer 1

You can apply the cross product twice. First find $\vec{v_3} = \vec{v_1} \times \vec{v_2}$, then find $\vec{v_4} = \vec{v_1} \times \vec{v_3}$. You can use any vector $\lambda \vec{v_4}$ where $\lambda$ is an arbitrary nonzero scalar as your answer.

With the first cross product, you're finding a normal vector to the plane defined by $\vec{v_1}$ and $\vec{v_2}$. With the second cross product, you're finding a vector normal to both that vector and $\vec{v_1}$, which, in $3$-dimensional space, has to lie in the same plane as $\vec{v_1}$ and $\vec{v_2}$.

Answer 2

Any vector on this plane is of the form $v = av_1 + bv_2$. For a vector to be normal to $v_1$ it must satisfy $0 = v_1 \cdot v = v_1 \cdot (a v_1 + b v_2) = a(v_1 \cdot v_1) + b(v_1 \cdot v_2)$. Choose any $b \neq 0$, then solve for $a$. The resulting $v$ is your answer.