Find a homogeneous linear system of two equations in three...

Find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in IR³ that are orthogonal to $\vec{a} = (-3, 2, -1)$ and $\vec{b} = (0, -2, -2)$. What kind of geometric object is the solution space? Find a general solution of the system obtained in part i., and confirm that Theorem 3.4.3 of the textbook holds. THEOREM 3.4.3 If A is an m x n matrix, then the solution set of the homogeneous linear system $Ax = 0$ consists of all vectors in $R^n$ that are orthogonal to every row vector of A.

Transcript

00:01 So in this problem, we're working in r3, and we're given these two vectors, a is 111, and b is negative 2, 3 ,0.

00:12 And we're asked to find a system, first we're asked to find a system of two equations, three unknowns, that are orthogonal to a and b.

00:22 So let's first talk about what orthogonal means.

00:26 So orthogonal means they're perpendicular to, right? which means that the dot product is zero.

00:43 So with x, y, and z, orthogonal to a, well, then we have the equation x plus y plus z is going to be zero.

01:13 Right because i take xyz times one one one and add them all together on the dot product and get zero okay so then orthogonal to b well that's negative 2x plus 3y and 0 z so i don't need to write it is equal to zero so there's my two equations with three unknowns that are orthogonal to these two vectors a and b.

01:59 So next we're asked, what kind of space do these two equations defined? well, we can think of these as two linear equations in three dimensions, right, in r3.

02:15 So what we have is two lines in r3.

02:33 And so then those two lines, let me say it this way.

02:51 We start out with a and b are two lines in r3, which intersect at one point.

03:09 So if we're orthogonal to both, then we are a line in three -dimensional space.

03:38 Next we're asked to find a general solution for this.

03:41 So let's set up the matrix.

03:45 1110 and negative 2 -3 -0.

03:54 We'll row reduce this.

03:55 So we'll take r2 plus 2 times r1, 111 -1 -1 -0.

04:05 I get a 0 here, right? then i have 3 plus 2 is 5 and 0 plus 1 is 2 and 0.

04:21 So then we'll take r2 divided by 5.

04:28 1 -1 -1 -0, 0, 1, 2 -5s, 0.