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4th edition, linear algebra and its application, gilbert strang exercise 2.1 question 17

  1. Let P be the plane in R^3 with equation x+y-2z = 4. The origin (0,0,0) is not in P! Find two vectors in P and check that their sum is not in P Answer at the back --> (4,0,0) is on the plane, (0,4,0) is on the plane but their sum (4,4,0) is not on the plane.

Questions: What does it mean for a vector to be in a plane? Does the end point of the vector being on the plane mean that the vector is in the plane?

This is a central theme in second chapter that strang has used wherein he called the necessity of passing through origin for a subspace as a consequence of closure under addition and closure under scalar multiplication. to quote something - "the distinction between a subset and a subspace is made clear by example. In each case can you add vectors and multiply by scalars, without leaving the space?" and he goes on to demonstrate with a few examples

(Can you ever add two vectors and leave the space, or perform scalar multiplication to the same result?)

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    There is no royal decree stating that affine planes have to go through the origin. $4$ points in $\mathbb{R}^3$ are coplanar if the determinant of the matrix whose rows (or colums) are the coordinates of the given points augmented with a $1$ is zero. That follows from the fact that a determinant is more or less an oriented volume, hence four points are coplanar if they are vertices of a pyramid with volume zero. – 2017-01-24
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    There is very little conceptual difference between a vector with its tail at the origin and the point that lies at the head of said vector. In fact, I would argue that keeping the two concepts separate leads to confusions like this one in exchange for no immediately apparent gain. Mathematics are not and should not be compartmentalised more than absolutely necessary. – 2017-01-24
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    Let P be the plane in R^3 with equation x+y-2z = 4. The origin (0,0,0) is not in P! Find two vectors in P and check that their sum is not in P Answer at the back --> (4,0,0) is on the plane, (0,4,0) is on the plane but their sum (4,4,0) is not on the plane. Does this make sense? statements that (4,0,0) is on the plane and (0,4,0) is on the plane – 2017-01-24
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    @Arthur : Are you not asking to keep a scalar and a vector in the same bucket? – 2017-01-24
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    No. Vectors and scalars are different. What I'm saying is that the vector $(1,2,3)$ and the point $(1,2,3)$ are really the same thing. Thus, asking for vectors in a given plane is the same as asking about points in that same plane. – 2017-01-24
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    @Arthur : So vector (4,0,0) is on x+y-2z = 4 according to you? I really fail to see how this can be. What am i missing? – 2017-01-24

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To say a vector belongs to a plane is to say that it is "lies" on it. Imagine that you draw an arrow on a sheet of paper. That arrow is a vector, the paper a plane. It lies on the paper.

Now, the neat thing about vector addition and scalar multiplication is that, given an underlying vector space, these operations are closed. You'll soon learn about what a vector space is, but you'll get to linear combinations and spans soon - these are important.

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    Let P be the plane in R^3 with equation x+y-2z = 4. The origin (0,0,0) is not in P! Find two vectors in P and check that their sum is not in P Answer at the back --> (4,0,0) is on the plane, (0,4,0) is on the plane but their sum (4,4,0) is not on the plane. This does not make sense then? – 2017-01-24
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    That plane does not define a vector space. Why? It doesn't admit an additive identity, namely (0, 0, 0) – 2017-01-24
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    Let P be the plane in R^3 with equation x+y-2z = 4. The origin (0,0,0) is not in P! Find two vectors in P and check that their sum is not in P Answer at the back --> **(4,0,0) is on the plane, (0,4,0) is on the plane** but their sum (4,4,0) is not on the plane Is the explanation, given in the book, correct for it to not be a space? (I also argued the lack of additive identity to be the reason for it not being a space) excuse my persistence to have miute details delineated. I just dont see why gilbert strang would say vector (4,0,0) is on the plane neither do i expect the guy to be wrong – 2017-01-24
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Take the $x,y$ and $z$ coordinates of the vector, insert them for $x,y$ and $z$ in the equation for the plane, and see what you end up with. In this case we get $4+0-2\cdot0=4$. Is that equality correct? If so, the vector is on the plane. If not, then the vector is not in the plane.

Do the same thing for $(0,4,0)$ and $(4,4,0)$ and check whether those two are on the plane too.