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# Visualizing a column space as a plane in R3

Determining the planar equation for a column space in R3. Created by Sal Khan.

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• At why are you dotting the normal vector with the difference of two vectors? •   If you have two non-zero vectors, a & b, and their dot product equals zero, then the two vectors are perpendicular to each other. HOWEVER, just because these two vectors are perpendicular to each other DOES NOT necessarily mean that one of those vectors is also perpendicular to a specified plane that the other happens to lie on. So, if you have three non-zero/non-collinear vectors (let's call them a, b, & c) and the vectors b & c lie on the same specified plane, the only way that vector a is perpendicular to the entire plane is if a*b = a*c = 0 (where: the "*" is being used to indicate the dot product operation).
Next, subtracting a*c from both sides gives us: a*b - a*c = 0.
Lastly, because the dot product is distributive, we can write this as: a*(b - c) = 0, which is the equation that Sal uses (except in the video "n" replaces "a", "b" is an arbitrary vector [x y z], and "c" is one of our two Basis vectors).

The point is that, given two non-zero vectors, b & c, (that AREN'T Collinear but ARE Coplanar) the only way a third vector, a, is perpendicular to the entire plane is if the dot product of a & b AND the dot product of a&c BOTH equal zero.

Does this make sense?

Another way of looking at it is to first think of the xy-plane. The x-axis is obviously on the xy-plane and the y-axis is perpendicular to the x-axis but it, too, is on the xy-plane. However, the z-axis is perpendicular to the xy-plane because it is perpendicular to BOTH the x-axis AND the y-axis.

Hope this helps...
• Sorry, I don't have a particular spot in the video to refer to, but here goes:
I have been watching the Linear Algebra sessions from the beginning and thought I had a good grip on what was going on then, when I got here and we had been working with columns for a few lectures, the question came to mind:
How does what we are doing in column spaces relate to the origin of the original matrix. I mean the rows of the matrix represent a linear equation, but the column of a matrix only references the same variable in each of the linear equations.
Hoping for an explanation
Mark • If you have Ax = b, then b must be a linear combination of the columns of A. Say it is 1 times the first column, plus 2 times the second column... then x is the vector (1,2). For, say, a 2x2 matrix A there is a row way of looking at it, and the lines represented by each row intersect at x, plus a column way of looking at it where each column is a vector and the numbers in x tell you how much to multiply each column by so that their sum results in the b vector. It is 2 ways to do the same problem or think about it.
• I don't get it. Is it always a plane? The column space is only a plane if there are 2 basis vectors. What if there are 3? • For the second approach, what about the constraints in the top 2 rows?

e.g., that x_2 - x_3 - x_4 = 2x - y.

Is it not possible that this would alter the plane constraint?

We already heuristically know that the solution should be a plane. Is this why these cases were not (or did not have to be) considered? • So, since spanning subspaces must always contain the zero vector, then all spanning subspaces intersect the origin? • Since the rref of the A only has two identity vectors, e1 and e2, doesn't the mean that the column space spans R^2 and not R^3? meaning the plane would be R^2? This kinda goes for the whole video • @ sal ,cross products the two positions vectors and those vectors are not "ON" the plane they are just touching the plane . so how can the cross product be a normal to the plane ?

for those having the same question
at 13.34 , Sal clearly explains why he was able to get the normal vector by doing a cross product. ie the two vectors are also part of the plane. now why are they part of the plane because all the points of the vectors can be obtained from their linear combination. eg the equation av1 + bv2 where v1,v2 are vectors and a,b are scalars a can be zeroed out to get the end point of vector v2 and then b can be incremented from zero to get all the points in vector v2. the same thing can be done for v1 • At you said the plane has to intersect (0,0,0) because Subspace always contain the zero vector. But i thinks it a little mistake right? Because you said vector doesn't contain position information? Is the real reason is you draw the vectors at the standard point? I also think any other parallel plane with the one that you found can contain the same set of vector. • Why do we pronounce the word echolon as esholon rather than ekolon (just like we pronounce echo) ? • At , it is said that the normal vector is the result of the cross product of the two basis vectors. Just to confirm: Is possible then to generalize this by saying that the cross product of any two vectors that are a "basis" (therefore lie on the plane they describe) generate a normal vector? Thus, this wouldn't work with "position vectors" (which do not lie on the plane), right? Finally, is this the main difference between "position vectors" and "basis vectors"? 