4.1 Vectors and Lines

1. 4.1 Vectors and Lines

2. Vectors • Vectors -- relay information about magnitude and direction. • Ex. Velocity: 12 mph to NE • Head of vector: terminal point • Tail of vector: initial point • Vectors will be bold: u,v • Magnitude is ||u||, will be illustrated by length of vector • Vectors are equal iff they have same direction and same magnitude • 0 is zero vector iff magnitude is 0

3. Vectors (continued) • v and u are parallel if they have the same or opposite direction uvw • U = -v since same magnitude and opp. direction

4. Vector Addition • If I ride my bike to school, I have 2 routes. One takes me 2 miles East and then 1/2 mile north. The other takes me 1/2 mile north, then 2 miles east. Draw diagrams representing the total displacement in each case.

5. Vector Addition (continued) • Parallelogram Rule: To find u + v, we draw u and v starting at the same point P. Then draw the parallelogram of which these two vectors are two sides. The diagonal from P will be the resultant vector. • Note that u + v = v + u

6. Vector subtraction • u - v = u + (-v) • v • -v u • u - v • Also, x + v = u • Implies u-v = x • So we find x such u x = u-v • That x+v = u • v

7. Vector Properties • u+ v = v + u • u + (v + w) = (u + v) + w (illustrate them) • v + 0 = v • v + (-v) = 0

8. Proving properties in Geometry • Show that diagonals of parallelogram bisect each other. • ABCD will be parallelogram. B C • E is the intersection of the diagonals M,E • M is the midpoint of AC A D • Just need to show that BM=MD (vectors) • same direction means M is on BD so M = E • same magnitude means that M = E is midpoint of MD

9. continued • AM = MC (vectors) since M is midpoint • BA = CD (vectors) since parallelogram • BM = BA + AM (by par rule) • = CD + MC (from above) • = MC + CD • = MD (by par rule) • B C • M,E • A D

10. Scalar Multiplication • a is a real number. av is a vector such that: • ||av|| = |a| ||v|| • Direction of av is: • The same as v if a > 0 and v≠ 0 • No direction if a = 0 or v = 0 • Opposite of v if a < 0 and v ≠ 0 • A vector of magnitude 1 is called a unit vector

11. Helpful property • Often, we will need to be able to find a unit vector in the direction of a given vector, v: • show that this is true. • this is clearly a scalar times v, so same direction

12. Theorem 1-Properties of vectors • u+ v = v + u • u + (v + w) = (u + v) + w (illustrate them) • v + 0 = v • v + (-v) = 0 • 1u = u • a(bu) = (ab)u • (a + b)u = au + au • a(u + v) = au + av

13. Coordinates • Position vector of a point, P, in R3 is the vector p = OP from the origin to P. If P = P(x,y,z) then p = (x,y,z) • X,y,z are the X,Y,Z components of p. • If P = P(x,y) is in R2, then p = (x,y) • Coordinate Vectors -- unit vectors along the axes i = (1,0,0) j = (0,1,0) k = (0,0,1)

14. Expressing vectors w/ coordinate vectors • P (x,0,0) ==> p = xi ||p|| = ||xi|| = |x| ||i|| = |x| Direction is clearly along x axis as in point P • xi = (x,0,0) • yj = (0,y,0) • zk = (0,0,z) • Show v = xi + yj + zk

15. Theorem 2 • u = (x,y,z) and u1 = (x1,y1,z1) • u = u1 iff x = x1, y = y1, and z = z1 • u + u1 = (x+x1, y+y1,z+z1) • au = (ax,ay,az) for a scalar a • u - u1 = (x-x1,y-y1,z-z1) To prove 2,3,4, start as sum,diff,scalar mult of unit vectors Properties in R2 are analogous

16. Example u = (2,3,-1), v = (0,3,2) Find 3u - 2v = 3(2,3,-1) - 2(0,3,2) = (6,9,-3) - (0,6,4) = (6,3,-7)

17. Theorem 3 • P1 = (x1,y1,z1), P2 = (x2,y2,z2) P1 • the vector from P1 to P2 is: p2-p1 • P1P2 = (x2 - x1, y2 - y1, z2 - z1) p1 • p2 P2 • Proofp1 = (x1,y1,z1), p2 = (x2,y2,z2) • P1P2 = p2- p1 based on the figure • = (x2-x1,y2-y1,z2-z1) based on Theorem 2

18. Example Proof Prove that the midpoint of P1=(x1,y1,z1) and P2=(x2,y2,z2) is P1 p1 P p P2 p2 P1P = 1/2(P1P2) P1P2 = p2 - p1 p = p1 + P1P = p1 + 1/2(p2 - p1) = 1/2(p1 + p2) = 1/2 (x1+x2,y1+y2,z1+z2)

19. Lines in space • Direction vector -- any nonzero vector that is parallel to a given line. For any given line, there is an infinite number of direction vectors

20. Lines (cont) • Vector equation of a line-- the line parallel to d ≠ 0 through the point with position vector p0 is given by p = p0 + td for some scalar t. So, the point with position vector p is on the line iff a real number t exists such that p = p0 + td P0 P d p0 p

21. Lines (cont) • In component form: • (x,y,z) = (x0,y0,z0) + t(a,b,c) If we then set components equal we get: • Parametric equations The line through P0 (x0,y0,z0) w/ direction vector d=(a,b,c) ≠ 0 is x = x0 + ta y = y0 + tb z = z0 + tc

22. Linear Equations • Recall that if AX = B, and if A is invertible (det A ≠ 0), then • X = A-1B So...

23. Examples • Find the equations of the line through the points P0(1,2,3) and P1(-1,3,-2) • Find d = P0P1 • Use one of the points for P0 in par equations • Find the equations of the line through P0(2,1,-2) parallel to line w/ equations x = 2 +3t , y = -2 - t, z = 1 - 2t • Find d from par equations given • Use d for d in new case (since parallel)

24. Examples • Find the point of intersection of the following lines (if one exists). x = 2-2t x=1-s • y= 2-3t y = 2+s • z = 1+2t z = 3 -s • Set x’s equal, y’s equal and z’s equal • Determine whether any point satisfies all 3 equations

25. Note Interesting example proof on p. 153 proving the point slope form of an equation of a line.