Describing vector equation geometrically

Question

How would I describe geometrically the vector equation:

$$\vec{x} = s(0,2,1) + t(1,1,-1) ,\qquad s,t \in \Bbb R$$

Answer 1

this is equation of plane so you can using cross product and obtain normal of plane and write ax+by+cz+d=0 that (a,b,c) is normal vector and d obtain by an optional point.

Answer 2

If you let $\vec{x}=(x,y,z)$, this gives the equations $x=t, y=2s+t, z=s-t$;

so $y-2z=3t$ and therefore $3x-y+2z=0$.

This shows that the given vector equation represents a plane in $\mathbb{R^3}$ passing through the points

$(0,0,0)$, $(0,2,1)$, and $(1,1,-1)$.

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More generally, the vector equation $\vec{x}=s(a,b,c)+t(d,e,f)$ represents a plane passing through

the points $(0,0,0)$, $(a,b,c)$, and $(d,e,f)$.