I took this lecture from Sir Denis at OCW about multivariable calculus where he explains vectors and he gives this question at the end of the lecture?
$x+2y+3z = 0$ is a plane.
But I don't really understand why is it a plane, granted the vectors $(x, y, z)$ and $(1, 2, 3)$ are perpendicular as their dot product is $0$ which implies the angle between them is $90^{\circ}$.
But if the angle is $90^{\circ}$, it should be a perpendicular, not a plane.
What am I missing?
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Since $x=-2y-3z$ each point vector $(x,y,z) = y(-2,1,0)+z(-3,0,1)$ in that set is linear combination of vectors $a= (-2,1,0)$ and $b=(-3,0,1)$. So every point in that set is in a plane spanned with these two vectors.
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You say "it should be a perpendicular". But a perpendicular to what ?
The answer is a perpendicular to the vector $(1,2,3)$, which is also a perpendicular to the line of direction $(1,2,3)$, and this is a plane.
To prove that the locus is not a single line, one can exhibit three points that are not aligned, such as $(0,0,0), (1,1,-1)$ and $(2,-1,0)$. Clearly, the vectors that they form two by two are not proportional to each other.
It is the set of all vectors (in all directions) which are orthogonal to $(1,2,3)$. Therefore, it's a plane.
For instance, the set of all vectors orthogonal to $(0,0,1)$ is the plane $z=0$.