Geometry of perpendicular vectors

I'm trying to self-study linear algebra but some topics seems incomprehensible to me, such as this:

It should be evident (says the book), that Perpendicular vectors to $V$ $= [1,1,1]$ lie on a plane while perpendicular vectors to $[1,1,1]$ and $[1,2,3]$ lie on a line.

I don't understand why. I mean, if I have a vector with 3 coordinates isn't that vector on $\mathbb{R}^3$ and not $\mathbb{R}^2$ and then it's perpendicular vectors will lie also on $\mathbb{R}^3$. And obviously the second premise seems even more strange. Can anyone please explain me these in a simple way (to a newbie), please.

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Well, from context I'll assume we are dealing with the vector space $\mathbb{R}^3$ with usual inner product, i.e.: given two vectors $x = (x_1, x_2, x_3)$ and $y=(y_1, y_2, y_3)$ their inner product is the real number obtained by:

$$\left\langle x, y \right\rangle=\sum_{i=1}^3x_iy_i$$

Well, why is this important? Simply because we can't talk about orthogonal vectors without one inner product defined (it's one extra layer of structure we add to a vector space to allows us to do that). For the moment, what's of interest is: why the hell the vectors perpendicular to $V=(1,1,1)$ lie on a plane? Well, let's try it out. Take some arbitrary vector $x =(x_1, x_2, x_3)$ and say that it's orthogonal to $V$, which by definition means that the inner product of $x$ and $V$ is zero. We must have:

$$\left\langle V, x\right\rangle=x_1+x_2+x_3=0$$

However, what is this? The equation of a plane through the origin. In particular, we can parametrize this plane by setting $x_1 = \lambda_1$ and $x_2=\lambda_2$ so that we have that a vector is orhtogonal to $V$ if and only if he is of the form:

$$(x_1, x_2, x_3)=(\lambda_1, \lambda_2, -\lambda_1 - \lambda_2)$$

A little more work and we see that this is equivalent to saying that:

$$(x_1,x_2,x_3) = \lambda_1(1, 0, -1)+ \lambda_2(0, 1,-1)$$

Hence, any vector orthogonal to $V$ must be on the plane of equation $x+y+z=0$ which is equivalent to say that it should be on the plane generated by the vectors $(1, 0, -1)$ and $(0,1,-1)$ as we've shown above.

What's the intuition behind this? This vector $V$ determines a line in space, so given a line in three spaces, we can find a plane which has this line as it's normal, in other words, it should really be the case the those things orthogonal to a line in three space lives on planes (remember, this is intuition, the proof we already gave, we are jus trying to understand), so that you can imagine the line and imagine that if you plug a vector orthogonal to the line, there are many others you can plug orhtogonal to it, it's just a question of getting round the line (try to imagine this). This is equivalent to say that those vectors should be in a plane.

Now, if you give me two vectors: $v = (1,1,1)$ and $w = (1,2,3)$ the vector $x$ is orthogonal to them at the same time if and only if:

$$\left \langle v, x \right \rangle = 0 \quad \quad \text{and} \quad \quad \left\langle w, x \right\rangle=0$$

If you use the definition of the inner product we gave you'll get a linear system of two equations and two unknowns. If the results can be paremetrized by just one parameter (just one degree of freedom), you'll see that the set of solutions is a line and the geometric interpretation is that the normal to those two vectors must be a line. If the results needs two parameters it means that the original vectors were colinear, so that in the end it's the same working with both or just one of them.

Try for yourself to give the same argument I gave in the first case to show that it is really a line, and try to imagine it geometrically. It always helps!

I hope this helps you somehow. Good luck with your studies!

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Thanks for your vey comprehensive answer. I do the exercise as you suggest, it seems from the results that the orthogonal vectors to both will be the span of precisely {1, 0, -1} and such they'll be a line. Also your definition helps me a lot. Thanks! – Randolf Rincón Fadul Apr 16 '13 at 3:00

All of the vectors are still in $\mathbb{R}^3$. The issue is the dimensionality of the spaces: a 2-dimensional subspace (defined by the span of two vectors) will be orthogonal to a subspace that is 1-dimensional (which can be described by a line or vector). A 1-dimensional subspace will then be orthogonal to a 2-D subspace.

When I say a space is orthogonal to another, I am referring to all vectors in one space being orthogonal to all vectors in the other.

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Yes, those vectors all live in $\mathbb{R}^3$, but one can be more specific. It's like saying "all points 5 units away from the origin lie in a sphere". Yeah, they're in $\mathbb{R}^3$ too, but one can describe it completely by using 'sphere'.

Vectors perpendicular to a given vector are exactly a plane. Think about standing a pencil up on a table. Every vector lying in the table is at a right angle to the pencil.

For the second question, think about two of these planes intersecting. They intersect in a line. (As long as those two vectors are not parallel, which would mean the planes are too.)

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You can probably find in your book, that the vector $[A,B,C]$ and the point $(x_0,y_0,z_0)$ defines the plane $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$. The vector $[A,B,C]$ is said normal, i.e. perpendicular to the defined plane.

You can check that the vector product of [1,1,1] and [1,2,3] is a nonzero vector, from properties of the vector product perpendicular to them, hence it generates a line.

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You should always believe in the "The Book"!! :):)(Silly Joke!).

A plane doesn't exactly mean all points in it are in $\mathbb{R}^2$. Consider the $xy$ plane in 3-D, the $z$ co-ordinates of all the points in the plane are zero. A point in it will look like $(x,y,0)$, but still it lies in 3-dimensions, since you considered three dimensions in the first place. Consider a plane which is parallel to $xy$ plane and is at a unit-distance from it. It will have $3$ co-ordinates, but the last one will always be fixed. If you are considering $\mathbb{R}^3$, a plane is a set of points such that all points $(x,y,z)$ in that set satisfy a equation of the type $ax+by+cz=d$ for some constants $a,b,c,d$.

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