(1 point) Find an orthonormal basis of the plane x1 + 2x2 – x3 = 0. -
(1 point) Find an orthonormal basis of the plane X1 + 4x2 – x3 = 0. Answer: To enter a basis into WebWork, place the entries of each vector inside of brackets, and enter a list of these vectors, separated by commas. For instance, if your basis is 2 then you would enter [1,2,3], 3 [1,1,1) into the answer blank.
Find an orthonormal basis for the plane viewed as a subspace of R3. Z (-1,0,2) (0,-1,0) (0,1,0)
Find an orthonormal basis of the plane in R3 defined by the equation 2a yz0
Please attempt both questions. 5. Find an orthonormal basis for the plane viewed as a subspace of R3. Z (-1,0,2) (0,-1,0) (0,1,0) X 6. Determine if each basis is orthogonal. Further, is the basis orthonormal? (a) In the vector space R3 (i.e. column vectors in 3-space): 1 2 5 -3 (b) In the vector space that consists of polynomial functions of degree less than or equal to 2: {f(x) = 22 - 3, 9() = 4, h(x) = 2² +2}...
D1. If a and b are nonzero, then an orthonormal basis for the plane z = ax + by is
Find an ONB (orthonormal basis) for the following plane in R3 2 + y + 3z = 0 First, solve the system, then assign parameters s and t to the free variables (in this order), and write the solution in vector form as su + tv. Now normalize u to have norm 1 and call it ū. Then find the component of v orthogonal to the line spanned by u and normalize it, call it ū. Below, enter the components...
(a) Find an orthonormal basis for the subspace U = span ((1, −1, 0, 1, 1),(3, −3, 2, 5, 5),(5, 1, 3, 2, 8)) of R 5 . (b) Express the vectors (0, −6, −1, 5, −1) as linear combinations of the orthonormal basis obtained in part (a). (c) Which of the standard basis vectors lie in U?
Find an orthonormal basis for the subspace of R3 spanned by Extend the basis you found to an orthonormal basis for R 3 (by adding a new vector or vectors). Is there a unique way to extend the basis you found to an orthonormal basis of R3 ? Explain.
4. (10+10pts.) Consider the homogeneous system 21 +22+ (3 - 2a).x3 = 0 2x1 + 12 + 7.03 - 14 = 0 -22 + 20.73 +2.04 = 0 21 +22 + 4x3 = 0 where a is a real constant. a. Find the value of a for which the dimension of the solution space of the system is 1. b. Find a basis of the solution space of the system for the value of a found in part (a).