1. 6.5 – Vectors in R² and R³ - In R², the position vector OP has its tail located at the origin O (0, 0), and its head at any arbitrary point P (a, b). Therefore, OP = (a, b). - Similarly in R³, the position vector OP has its tail located at the origin O(0, 0, 0), and its head at P (a, b, c). Therefore, OP = (a, b, c). - R³ can also be visualized by forming the right hand system, where the index finger is the positive x – axis, the middle finger is the positive y – axis, and the thumb represents the positive z – axis.

2. Example Determine the coordinates of points : A, B, C, D, E, F, given P(-2, 4, -7).

3. Solution A (-2, 0, 0) B (-2, 4, 0) C (0, 4, 0) D (0, 0, -7) E (0, 4, -7) F (-2, 0, 7)

4. 6.6 – Operations with Algebraic Vectors in R² Vectors can be simplified by factoring a constant out. For example, vector (3, 12) can be simplified to 3 (1, 4). Thus, m(x, y) = (mx, my). OP = (a, b) = ai + bj, and |OP| = sqrt ((a² + b²)). Given : A = (3, 1), and B = (4, 2), then AB = (4-3, 2-1) = (1, 1), (x2-x1, y2-y1) |AB| = sqrt (( x2 – x1 )² + (y2 – y1)² ))

5. Examples 1.Find a single vector that is equivalent to: -3 (4, -9) – 9 (2, 3). 2.If a = ( -60, 11 ) and b = ( -40 -9 ), Calculate |a – b|.

6. Solutions 1. 2.

7. 6.7 – Operations with Algebraic Vectors in R³ OP = (a, b, c) = ai + bj + ck, and |OP| = sqrt ((a² + b² + c² )). A vector between two points : A (x1, y1, z1) and B (x2, y2, z2), then AB is (x2 - x1, y2 - y1, z2 - z1) |AB| = sqrt ( ( x2 – x1 )² + (y2 – y1)² ) + (z2 - z1)² ) Two vectors are parallel if they are scalar multiples of each other.

8. 6.8 – Linear Combinations and Spanning Sets When two vectors u and v are collinear (u = mv) then they don’t span R², however, if u does not equal to mv, which means vector u is not collinear to v, then they span R². When three vectors are not coplanar (u does not equal to nv + mw), then the those three vectors span R³. To check if three vectors span R³ 1.Start by equating the three vectors : u = nv + mw 2.Solve for m, and n, and check for consistency. (LS = RS) 3.If LS = RS, there is a unique solution for m and n, which means the three vectors are coplanar and do not span R³.

9. Example Solve for a, b, and c in the equation : 2(a, 3, c) + 3(c, 7, c) = (5, b + c, 15)

10. Solution 2c + 3c = 15 5c = 15 c = 3, Subst. into 2a + 3c = 5 a = -2 27 = b + c b = 24, therefore a = -2, b = 24, c = 3