Question

For this question to display nicely, you may need to close the menu on the left side of the eClass page. Let B = {x2 + 3x +1,2x2 + 5x + 2, 3x2 + 8x + 4} CP2 . (a) Prove that B is a basis for P2.

(i) Show that B is linearly independent. Suppose a(x2 + 3x + 1) + b(2x2 + 5x + 2) + c(3x2 + 8x + 4) = 0 for some a, b, c ER. We need to show that a = 0, b = 0,c = 0. Now a(x2 + 3x + 1) + b(2x2 + 5x + 2) + c(3x2 + 8x + 4) = 0 PO ) x2 + ( )x+ )1 = 0 = (0)x2 + (0)x + (0)1 1 = 0 (by equating coefficients of x?, x and 1, respectively) = 0 = 0 This gives us the augmented matrix H Therefore, the only solution is a = 0, b = 0 and c = 0. Thus a(x² + 3x + 1) + b(2x² + 5x + 2) + c(3x² + 8x + 4) = 0 → a = b = c = 0, Therefore, the only solution is a = 0, b = 0 and c = 0. Thus a(x² + 3x + 1) + b(2x² + 5x + 2) + c(3x² + 8x + 4) = 0 a = b = c = 0, so B is linearly independent.

(ii) To show that the set B spans P2 we must show that for any fx2 + gx +he P2 there exist coefficients a, b and c such that a(x2 + 3x + 1) + b(2x² + 5x + 2) + c(3x2 + 8x + 4) = fx2 + gx +h. The augmented matrix of this system is rref Since the rref of the coefficient matrix of this system contains a leading entry in every row, this system will be consistent for any augmenting column, so the system has a solution regardless of what f, g and h are. Thus, for any f, g and h the polynomial fx2 + gx + h e Span(B). Therefore, B spans P2. (Note: To answer part

(ii), it would have been sufficient to have obtained simply a row echelon form of the augmented matrix.) Since B is a linearly independent spanning set, B is a basis for P2. (b) Find the coordinate vector [p]b of the polynomial p= 12x2 + 34x + 17 with respect to B. (plb = w 9 (C) Let q E P2 such that [9]B = -8 . Then q= (To enter x2 type x^2.)

Answer 1

The distance from a point to the origin is invariant under rotation because the sum of the squares of the coordinates does not change with rotation. The cross product of two vectors is a vector perpendicular to the plane of the original vectors, while a vanishing dot product indicates orthogonality.

Step-by-step explanation:

Proof of Invariance of Distance Under Rotation

To prove that the distance of point P to the origin is invariant under rotations of the coordinate system, consider that the distance is given by the formula d = √(x² + y²). When we rotate the coordinate system, the position of P changes in terms of coordinates, yet the actual distance to the origin does not. No matter how we rotate the system, the sum of the squares of the coordinates, and therefore the distance, remains the same due to the rotational symmetry of the circle equation representing the distance from the origin.

Cross and Dot Product Relations

The cross product of two vectors results in a vector perpendicular to the plane containing the original vectors, as shown with A × B. If the cross product vanishes, the vectors are either parallel or one is the zero vector. For the dot product, if it vanishes (A ⋅ B = 0), then the vectors A and B are orthogonal. Finally, the dot product of a vector A with the cross product of A and another vector B is zero because they are orthogonal by definition of the cross product.