Find the equation of a hyperplane in $\mathbb R^4$ that contains $3$ given vectors. The vectors are:
$$v_1 = (1,0,-1,0),\quad
v_2 = (0,1,1,1),\quad
v_3 = (1,1,-1,0) $$
I've found the equation in parametric form ($s$ and $t$ are scalars):
$$x=-s+1,\quad
y=s+t,\quad
z=2s-1,\quad
w=s$$
The answer in the back of the book says the answer is: $x_1+x_3-x_4=0$.
How would I convert my answer into an equation for the hyper plane?
Best Answer
Your parametric form isn't right: the hyperplane is $3$-dimensional (the given vectors are linearly independent), so it would need three parameters: $$(r,s,t) \mapsto rv_1+sv_2+tv_3$$ But this is not what you want anyway. You want a non-parametric equation: $$c_1x_1+c_2x_2+c_3x_3+c_4x_4=0 \tag{1}$$ It must be such that the given vectors are contained in the hyperplane, meaning $$ \begin{cases} c_1-c_3&=0\\ c_2+c_3+c_4&=0\\ c_1+c_2-c_3&=0 \end{cases} $$ Solve this system, and you'll have the coefficients for (1).