Question

show that the set of all vectors in R3 that are orthogonal to (1,-2,2) is a subspace of R3. (use standard inner product for R3); find a basis for the subspace;

Answer 1

how do we find all the vectors that are perpendicular to a given vector?

Answer 2

or, how do we know when 2 vectors are perped? what property can we use?

Answer 3

To find perpendicular, do the inner product of two vector and if u get 0, then they are orthogoanl

Answer 4

so lets take a set of vectors (x,y,x) and the stated vector (1,-2,2) and work the dot product that defines them to be perp. wht do we get?

Answer 5

*** set of vectors (x,y,z) that is ....

Answer 6

.

Answer 7

I get x-2y+2z=0, if I let y=t, z=r, then I get the vector to be (2t-2r, t, r).

Answer 8

x - 2y + 2z = 0 defines all the vectors in R^3 that are perp to (1,-2,2). its also known as the equation of a plane

now, what are the properties we need to check to determine its a subspace?

Answer 9

contains the zero vector (the origin in in the plane), well ... 0,0,0 satisfies the equation so it checks out

what else we got for subspace properties?

Answer 10

Closed under addition and scalar multiplication, but I don't know how to show addition & multi

Answer 11

let v1 = (a1,a2,a3) and let v2=(b1,b2,b3) be any vectors that satisfy x - 2y +2z = 0

when kv1 and jv2 for some constants k and j will also satisfy the equation (be contained in the plane)

Answer 12

right?

Answer 13

thats for scalar multiplication

and (kv1 + jv2) for addition would have to satisfy the plane equation as well

Answer 14

kv1 is any vector in the plane scaled, since v1 is defined to be in the plane then kv1 outta satisfy

test it out

ka1 -2(ka2) +2k(a3) = 0

k (a1 -2a2 +2a3) = 0

since we know that a1,a2,a3 is in the plane by definition,
then we know that a1 -2a2 +2a3=0

k (0) = 0 for all k scalar multiples of any vector in the plane

Answer 15

test for addition

(ka1+jb1) -2(ka2+jb2)+2(ka3+jb3)=0

ka1+jb1 -2ka2 -2jb2 +2ka3 +2jb3 = 0

k (a1 -2a2 +2a3) +j (b1 -2b2 +2b3) = 0

we now a1,a2,a3 and b1,b2,b3 satisfy the equation by definition of being in the plane already soo

a1 -2a2 +2a3=0
b1 -2b2 +2b3=0

k (0) +j (0) = 0 is true for all vector additions in the plane.

Answer 16

the next step is just pulling 2 linearly independant vectors to numerically define v1 and v2 and those would form the basis of the plane

Answer 17

thank you so muchhhhhhh, i got it!!!! XD