Contents 7.1 Vectors in 2-Space 7.2 Vectors in 3-Space 7.3 Dot Product 7.4 Cross Product 7.5 Lines and Planes in 3-Space

7.1 Vectors in 2-Space Review of Vectors

Fig 7.1 (Geometric Vectors)

Fig 7.3 (Parallel vectors)

Fig 7.4 (sum)

Fig 7.5 (difference)

Fig 7.6 (position vectors)

Example 1 Fig 7.7

a – b = <a1− b1, a2 − b2> (4) Let a = <a1, a2>, b = <b1, b2> be vectors in R2 (i) Addition: a + b = <a1 + a2, b1 + b2> (1) (ii) Scalar multiplication: ka = <ka1, ka2>, k is a scalar (2) (iii) Equality: a = b if and only if a1 = b1, a2 = b2 (3) DEFINITION 7.1 Addition, Scalar Multiplication, Equality a – b = <a1− b1, a2 − b2> (4)

Graph Solution Fig 7.8 shows the graph solutions of the addition and subtraction of two vectors.

Example 2 If a = <1, 4>, b = <−6, 3>, find a + b, a − b, 2a + 3b. Solution Using (1), (2), (4), we have

Properties (i) a + b = b + a (ii) a + (b + c) = (a + b) + c (iii) a + 0 = a (iv) a + (−a) = 0 (v) k(a + b) = ka + kb k scalar (vi) (k1 + k2)a = k1a + k2a k1, k2 scalars (vii) k1(k2a) = (k1k2)a k1, k2 scalars (viii) 1a = a (ix) 0a = 0 = <0, 0> 0 = <0, 0>

Magnitude, Length, Norm a = <a1 , a2>, then Clearly, we have ||a||  0, ||0|| = 0

Unit Vector A vector that has magnitude 1 is called a unit vector. u = (1/||a||)a is a unit vector, since

Example 3 Given a = <2, −1>, then the unit vector in the same direction u is and

The i, j vectors If a = <a1, a2>, then (5) Let i = <1, 0>, j = <0, 1>, then (5) becomes a = a1i + a2j (6)

Example 4 (i) <4, 7> = 4i + 7j (ii) (2i – 5j) + (8i + 13j) = 10i + 8j (iii) (iv) 10(3i – j) = 30i – 10j (v) a = 6i + 4j, b = 9i + 6j are parallel because b = (3/2)a

Example 5 Let a = 4i + 2j, b = –2i + 5j. Graph a + b, a – b Solution Fig 7.11

7.2 Vectors in 3-Space Simple Review Please refer to Fig 7.22 through Fig 7.24. Fig 7.22

Fig 7.23

Fig 7.24

Example 1 Graph the points (4, 5, 6) and (−2, −2, 0). Solution See Fig 7.25.

Distance Formula (1) Fig 7.26

Example 2 Find the distance between (2, −3, 6) and (−1, −7, 4) Solution

Midpoint Formula (2)

Example 2 Find the midpoint of (2, −3, 6) and (−1, −7, 4) Solution From (2), we have

Vectors in 3-Space Fig 7.27.

Component Definitions in 3-Spaces Let a = <a1, a2 , a3>, b = <b1, b2, b3 > in R3 (i) a + b = <a1 + b1, a2 + b2, a3 + b3> (ii) ka = <ka1, ka2, ka3> (iii) a = b if and only if a1 = b1, a2 = b2, a3 = b3 (iv) –b = (−1)b = <− b1, − b2, − b3> (v) a – b = <a1 − b1, a2 − b2, a3 − b3> (vi) 0 = <0, 0 , 0> (vi) Component Definitions in 3-Spaces

Fig 7.28

Example 4 Find the vector from (4, 6, −2) to (1, 8, 3) Solution

Example 5 From Definition 7.2, we have

The i, j, k vectors i = <1, 0, 0>, j = <0, 1, 0>, k = <0, 0, 1> a = < a1, a2, a3> = a1i + a2j + a3k

Fig 7.29

Example 6 a = <7, −5, 13> = 7i − 5j + 13k Example 7 (a) a = 5i + 3k is in the xz-plance (b) Example 8 If a = 3i − 4j + 8k, b = i − 4k, find 5a − 2b Solution 5a − 2b = 13i − 20j + 48k

7.3 Dot Product Dot Product of Two Vectors DEFINITION 7.3 The dot product of a and b is the scalar (1) where  is the angle between the vectors 0    . Dot Product of Two Vectors

Fig 7.32

Example 1 From (1) we obtain i  i = 1, j  j = 1, k  k = 1 (2)

Component Form of Dot Product (3) (4) See Fig 7.33

Fig 7.33

Example 2 If a = 10i + 2j – 6k, b = (−1/2)i + 4j – 3k, then

Properties (i) a  b = 0 if a = 0 or b = 0 (ii) a  b = b  a (iii) a  (b + c) = a  b + a  c (iv) a  (kb) = (ka)  b = k(a  b) (v) a  a  0 (vi) a  a = ||a||2

Orthogonal Vectors (i) a  b > 0 if and only if  is acute (ii) a  b < 0 if and only if  is obtuse (iii) a  b = 0 if cos  = 0,  = /2 Note: Since 0  b = 0, we say the zero vector is orthogonal to every vector. Two nonzero vectors a and b are orthogonal if and only if a  b = 0. THEOREM 7.1 Criterion for an Orthogonal Vectors

Example 3. i, j, k are orthogonal vectors Example 3 i, j, k are orthogonal vectors. i  j = j  i = 0, j  k = k  j = 0, k  i = i  k = 0 (5) Example 4 If a = −3i − j + 4k, b = 2i + 14j + 5k, then a  b = –6 – 14 + 20 = 0 They are orthogonal.

Angle between Two Vectors (6)

Example 5 Find the angle between a = 2i + 3j + k, b = −i + 5j + k. Solution

Direction Cosines Referring to Fig 7.34, the angles , ,  are called the direction angles. Now by (6) We say cos , cos , cos  are direction cosines, and implies cos2 + cos2 + cos2 = 1

Fig 7.34

Example 6 Find the direction cosines and the direction angles of a = 2i + 5j + 4k. Solution

Component of a on b Since a = a1i + a2j + a3k, then (7) We write the components of a as (8) See Fig 7.35. The component of a on any vector b is compba = ||a|| cos  (9) Rewrite (9) as (10)

Fig 7.35

Example 7 Let a = 2i + 3j – 4k, b = i + j + 2k. Find compba and compab. Solution Form (10), a  b = −3

Physical Interpretation See Fig 7.36. If F causes a displacement d of a body, then the work done is W = F  d (11)

Fig 7.36

Example 8 Let F = 2i + 4j. If the block moves from (1, 1) to (4, 6), find the work done by F. Solution d = 3i + 5j W = F  d = 26 N-m

Projection of a onto b See Fig 7.37. the projection of a onto i is See Fig 7.38. the projection of a onto b is (12)

Fig 7.37

Fig 7.38

Example 9 Find the projection of a = 4i + j onto b = 2i + 3j. Solution

Fig 7.39

7.4 Cross Product Cross Product of Two Vectors The cross product of two vectors a and b is (1) where  is the angle between them, 0    , and n is a unit vector perpendicular to the plane of a and b with direction given by right-hand rule. DEFINITION 7.4 Cross Product of Two Vectors

Fig 7.46

Example 1 To understand the physical meaning of the cross product, please see Fig 7.47 and 7.48. The torque  done by a force F acting at the end of position vector r is given by  = r  F. Fig 7.47 Fig 7.48

Properties (i) a  b = 0, if a = 0 or b = 0 (ii) a  b = −b  a (iii) a  (b + c) = (a  b) + (a  c) (iv) (a + b)  c = (a  c) + (b  c) (v) a  (kb) = (ka)  b = k(a  b) (vi) a  a = 0 (vii) a  (a  b) = 0 (viii) b  (a  b) = 0 Two nonzero vectors a and b are parallel, if and only if a  b = 0. THEOREM 7.2 Criterion for Parallel Vectors

Example 2 (a) From properties (vi) i  i = 0, j  j = 0, k  k = 0 (2) (b) If a = 2i + 3j – k, b = –6i – 3j + 3k = –3a, then a and b are parallel. Thus by Theorem 7.2, a  b = 0 If a = i, b = j, then (3) According to the right-hand rule, n = k. So i  j = k

Example 3 See Fig 7.49, we have (4)

Fig 7.49

Alternative Definition Since (5) we have (6)

We also can write (6) as (7) In turn, (7) becomes (8)

Example 4 Let a = 4i – 2j + 5k, b = 3i + j – k, Find a  b. Solution From (8), we have

Special Products We have (9) is called the triple vector product. The following results are left as an exercise. (10)

Area and Volume Area of a parallelogram A = || a  b|| (11) Area of a triangle A = ½||a  b|| (12) Volume of the parallelepiped V = |a  (b  c)| (13) See Fig 7.50 and Fig 7.51

Fig 7.50

Fig 7.51

Example 5 Find the area of the triangle determined by the points (1, 1, 1), (2, 3, 4), (3, 0, –1). Solution Using (1, 1, 1) as the base point, we have two vectors a = <1, 2, 3>, b = <2, –1, –2>

Coplanar Vectors a  (b  c) = 0 if and only if a, b, c are coplanar.

7.5 Lines and Planes in 3-Space Lines: Vector Equation See Fig 7.55. We find r2 – r1 is parallel to r – r2, then r – r2 = t(r2 – r1) (1) If we write a = r2 – r1 = <x2 – x1, y2 – y1, z2 – z1> = <a1, a2, a3> then (1) implies a vector equation for the line is r = r2 + ta (2) where a is called the direction vector.

Fig 7.55

Example 1 Find a vector equation for the line through (2, –1, 8) and (5, 6, –3). Solution Define a = <2 – 5, –1 – 6, 8 – (– 3)> = <–3, –7, 11>. The following are three possible vector equations: (3) (4) (5)

Parametric equation We can also write (2) as (6) The equations (6) are called parametric equations.

Example 2 Find the parametric equations for the line in Example 1. Solution From (3), it follows x = 2 – 3t, y = –1 – 7t, z = 8 + 11t (7) From (5), x = 5 + 3t, y = 6 + 7t, z = –3 – 11t (8)

Example 3 Find a vector a that is parallel to the line: x = 4 + 9t, y = –14 + 5t, z = 1 – 3t Solution a = 9i + 5j – 3k

Symmetric Equations From (6) provided ai are nonzero. Then (9) are said to be symmetric equation.

Example 4 Find the symmetric equations for the line through (4, 10, −6) and (7, 9, 2) Solution Define a1 = 7 – 4 = 3, a2 = 9 – 10 = –1, a3 = 2 – (–6) = 8, then

Example 5 Find the symmetric equations for the line through (5, 3, 1) and (2, 1, 1) Solution Define a1 = 5 – 2 = 3, a2 = 3 – 1 = 2, a3 = 1 – 1 = 0, then

Fig 7.56

Example 6 Write vector, parametric and symmetric equations for the line through (4, 6, –3) and parallel to a = 5i – 10j + 2k. Solution Vector: <x, y, z> = < 4, 6, –3> + t(5, –10, 2) Parametric: x = 4 + 5t, y = 6 – 10t, z = –3 + 2t, Symmetric:

Planes: Vector Equations Fig 7.57(a)-(c) shows the concept of the normal vector to a plane. Any vector in the plane should be perpendicular to the normal vector, that is n  (r – r1) = 0 (10)

Fig 7.57

Cartesian Equations If the normal vector is ai + bj + ck , then the Cartesian equation of the plane containing P1(x1, y1, z1) is a(x – x1) + b(y – y1) + c(z – z1) = 0 (11)

Example 7 Find the plane contains (4, −1, 3) and is perpendicular to n = 2i + 8j − 5k Solution From (11): 2(x – 4) + 8(y + 1) – 5(z – 3) = 0 or 2x + 8y – 5z + 15 = 0

Equation (11) can always be written as ax + by + cz + d = 0 (12) The graph of any ax + by + cz + d = 0, a, b, c not all zero, is a plane with the normal vector n = ai + bj + ck THEOREM 7.3 Plane with Normal Vector

Example 8 A vector normal to the plane 3x – 4y + 10z – 8 = 0 is n = 3i – 4j + 10k.

Given three noncollinear points, P1, P2, P3, we arbitrarily choose P1 as the base point. See Fig 7.58, Then we can obtain (13)

Fig 7.58

Example 9 Find an equation of the plane contains (1, 0 −1), (3, 1, 4) and (2, −2, 0). Solution We arbitrarily construct two vectors from these three points, say, u = <2, 1, 5> and v = <1, 3, 4>.

Example 9 (2) If we choose (2, −2, 0) as the base point, then <x – 2, y + 2, z – 0>  <−11, −3, 5> = 0

Graphs The graph of (12) with one or two variables missing is still a plane.

Example 10 Graph 2x + 3y + 6z = 18 Solution Setting: y = z = 0 gives x = 9 x = z = 0 gives y = 6 x = y = 0 gives z = 3 See Fig 7.59.

Fig 7.59

Example 11 Graph 6x + 4y = 12 Solution This equation misses the variable z, so the plane is parallel to the z-axis. Since x = 0 gives y = 3 y = 0 gives x = 2 See Fig 7.60.

Fig 7.60

Two planes that are not parallel must intersect in a line. See Fig 7 Two planes that are not parallel must intersect in a line. See Fig 7.62. Fig 7.63 shows the intersection of a line and a plane.

Fig 7.62

Fig 7.63

Example 13 Find the parametric equation of the line of the intersection of 2x – 3y + 4z = 1 x – y – z = 5 Solution First we let z = t, 2x – 3y = 1 – 4t x – y = 5 + t then x = 14 + 7t, y = 9 + 6t, z = t.

Example 14 Find the point of intersection of the plane 3x – 2y + z = −5 and the line x = 1 + t, y = −2 + 2t, z = 4t. Solution Assume (x0, y0, z0) is the intersection point. 3x0 – 2y0 + z0 = −5 and x0 = 1 + t0, y0 = −2 + 2t0, z0 = 4t0 then 3(1 + t0) – 2(−2 + 2t0) + 4t0 = −5, t0 = −4 Thus, (x0, y0, z0) = (−3, −10, −16)