Consider P2 with the inner product given by evaluation at -1, 0, and 1. That is < p, q >= p(−1)q(−1) + p(0)q(0) + p(1)q(1). Let p(t) = 3t − t2 and q(t) = 3 + 2t2 .
A) Compute < p, q >, ||p||, and ||q||.
B) Compute the orthogonal projection of q onto the subspace spanned by p.

A).

B).The orthogonal projection of q onto the subspace spanned by p is

We have

And < p, q >= p(−1)q(−1) + p(0)q(0) + p(1)q(1).

Now,

A).

B)Let V be the subspace spanned by p.

Now, an orthonormal basis for p is

 Then the projection of q onto the subspace spanned by p is

Learn more:link

(A)

Evaluate and at the given values of , then plug them into the inner product:

Now,

(B)

Let denote the subspace spanned by . We need an orthonormal basis for . Since

we have the basis vectors ; normalize these vectors to get the orthonormal basis,

Then the projection of onto is

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.