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# Vectors 2.pdf

Vector Product
Cross product in R3
Triple Product
Applications of Vector Products
Equation of lines
Equation of Planes

Vector Product
Cross product in R3
Triple Product
Applications of Vector Products
Equation of lines
Equation of Planes

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### Vectors 2.pdf

1. 1. VECTORS Dr. Gabriel Obed Fosu Department of Mathematics Kwame Nkrumah University of Science and Technology Google Scholar: https://scholar.google.com/citations?user=ZJfCMyQAAAAJ&hl=en&oi=ao ResearchGate ID: https://www.researchgate.net/profile/Gabriel_Fosu2 Dr. Gabby (KNUST-Maths) Vectors 1 / 37
2. 2. Lecture Outline 1 Vector Product Cross product in R3 Triple Product 2 Applications of Vector Products 3 Equation of Lines and Planes Equation of lines Equation of Planes Dr. Gabby (KNUST-Maths) Vectors 2 / 37
3. 3. Vector Product Outline of Presentation 1 Vector Product Cross product in R3 Triple Product 2 Applications of Vector Products 3 Equation of Lines and Planes Equation of lines Equation of Planes Dr. Gabby (KNUST-Maths) Vectors 3 / 37
4. 4. Vector Product Cross product in R3 Cross Product Definition The cross product of two vectors u = [u1,u2,u3] and v = [v1,v2,v3] in R3 is the vector denoted u×v and defined by u×v = [u2v3 −u3v2]i+[u3v1 −u1v3]j+[u1v2 −u2v1]k (1) u×v is also called the vector product or outer product. Dr. Gabby (KNUST-Maths) Vectors 4 / 37
5. 5. Vector Product Cross product in R3 Alternative 1 for Cross Product An easy way to remember the components of u×v is to consider the three vectors i,j,k and three consecutive elements of the sequence 1,2,3,1,2. 1 We have 123, 231 and 312. 2 The first element points to the position of a basis vector. The second and third elements indicate the indices of the components of u and v used to calculate the coefficients. u×v = [u2v3 −u3v2]i+[u3v1 −u1v3]j+[u1v2 −u2v1]k (2) For instance: 1 in 123, 1 points to i and 2,3 yield [u2v3 −u3v2]; 2 in 231, 2 points to j and 3,1 yield [u3v1 −u1v3]; and 3 in 312, 3 points to k and 1,2 yield [u1v2 −u2v1]. Dr. Gabby (KNUST-Maths) Vectors 5 / 37
6. 6. Vector Product Cross product in R3 Alternative 2 for Cross Product Definition The determinant of order two is defined by ¯ ¯ ¯ ¯ a b c d ¯ ¯ ¯ ¯ = ad −bc (3) and the cross product can be written as follows u×v = ¯ ¯ ¯ ¯ ¯ ¯ i j k u1 u2 u3 v1 v2 v3 ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ u2 u3 v2 v3 ¯ ¯ ¯ ¯i− ¯ ¯ ¯ ¯ u1 u3 v1 v3 ¯ ¯ ¯ ¯j+ ¯ ¯ ¯ ¯ u1 u2 v1 v2 ¯ ¯ ¯ ¯k (4) = (u2v3 −u3v2)i+(u3v1 −u1v3)j+(u1v2 −u2v1)k (5) Dr. Gabby (KNUST-Maths) Vectors 6 / 37
7. 7. Vector Product Cross product in R3 Example Find u×v where 1 u = [4,3,6] and v = [2,5,−3]. 2 u = 2i−j+3k and v = −i+2j+4k, Dr. Gabby (KNUST-Maths) Vectors 7 / 37
8. 8. Vector Product Cross product in R3 Example Find u×v where 1 u = [4,3,6] and v = [2,5,−3]. 2 u = 2i−j+3k and v = −i+2j+4k, 1 u × v = ¯ ¯ ¯ ¯ ¯ ¯ i j k 4 3 6 2 5 −3 ¯ ¯ ¯ ¯ ¯ ¯ (6) = −39i+24j+14k (7) = [−39,24,14] (8) Dr. Gabby (KNUST-Maths) Vectors 7 / 37
9. 9. Vector Product Cross product in R3 Example Find u×v where 1 u = [4,3,6] and v = [2,5,−3]. 2 u = 2i−j+3k and v = −i+2j+4k, 1 u × v = ¯ ¯ ¯ ¯ ¯ ¯ i j k 4 3 6 2 5 −3 ¯ ¯ ¯ ¯ ¯ ¯ (6) = −39i+24j+14k (7) = [−39,24,14] (8) 2 u × v = ¯ ¯ ¯ ¯ ¯ ¯ i j k 2 −1 3 −1 2 4 ¯ ¯ ¯ ¯ ¯ ¯ (9) = −10i−11j+3k. (10) Dr. Gabby (KNUST-Maths) Vectors 7 / 37
10. 10. Vector Product Cross product in R3 Properties of Cross Product 1 Cross product is not commutative u×v ̸= v×u rather u×v = −v×u (11) Dr. Gabby (KNUST-Maths) Vectors 8 / 37
11. 11. Vector Product Cross product in R3 Properties of Cross Product 1 Cross product is not commutative u×v ̸= v×u rather u×v = −v×u (11) 2 The associative law for multiplication does not usually hold; that is, [u×v]×w ̸= u×[v×w] (12) Dr. Gabby (KNUST-Maths) Vectors 8 / 37
12. 12. Vector Product Cross product in R3 Properties of Cross Product 1 Cross product is not commutative u×v ̸= v×u rather u×v = −v×u (11) 2 The associative law for multiplication does not usually hold; that is, [u×v]×w ̸= u×[v×w] (12) 3 u×[v+w] = u×v+u×w. Dr. Gabby (KNUST-Maths) Vectors 8 / 37
13. 13. Vector Product Cross product in R3 Properties of Cross Product 1 Cross product is not commutative u×v ̸= v×u rather u×v = −v×u (11) 2 The associative law for multiplication does not usually hold; that is, [u×v]×w ̸= u×[v×w] (12) 3 u×[v+w] = u×v+u×w. 4 u×[kv] = ku×v where k ∈ R. Dr. Gabby (KNUST-Maths) Vectors 8 / 37
14. 14. Vector Product Cross product in R3 Properties of Cross Product 1 Cross product is not commutative u×v ̸= v×u rather u×v = −v×u (11) 2 The associative law for multiplication does not usually hold; that is, [u×v]×w ̸= u×[v×w] (12) 3 u×[v+w] = u×v+u×w. 4 u×[kv] = ku×v where k ∈ R. 5 w·[u×v] = v·[w×u] = u·[v×w]. Dr. Gabby (KNUST-Maths) Vectors 8 / 37
15. 15. Vector Product Cross product in R3 Properties of Cross Product 1 Cross product is not commutative u×v ̸= v×u rather u×v = −v×u (11) 2 The associative law for multiplication does not usually hold; that is, [u×v]×w ̸= u×[v×w] (12) 3 u×[v+w] = u×v+u×w. 4 u×[kv] = ku×v where k ∈ R. 5 w·[u×v] = v·[w×u] = u·[v×w]. 6 ∥u×v∥ = ∥u∥∥v∥sinθ where θ is the internal angle between the directions of u and v. Dr. Gabby (KNUST-Maths) Vectors 8 / 37
16. 16. Vector Product Cross product in R3 Exercise 1 Show that i×i = 0, j×i = −k, k×i = j, i×j = k, j×j = 0, k×j = −i i×k = −j, j×k = i, k×k = 0. 2 Find u×v where 1 u = [1,−1,1] and v = [2,1,0], 2 u = i+jcosθ +ksinθ and v = i−jsinθ +kcosθ. Dr. Gabby (KNUST-Maths) Vectors 9 / 37
17. 17. Vector Product Triple Product Triple Product Definition (Triple Product) The product u·[v×w] is called the scalar triple product of the vectors u,v and w. We can write the scalar triple product as a determinant: u·[v×w] = ¯ ¯ ¯ ¯ ¯ ¯ u1 u2 u3 v1 v2 v3 w1 w2 w3 ¯ ¯ ¯ ¯ ¯ ¯ (13) Dr. Gabby (KNUST-Maths) Vectors 10 / 37
18. 18. Vector Product Triple Product Triple Product Definition (Triple Product) The product u·[v×w] is called the scalar triple product of the vectors u,v and w. We can write the scalar triple product as a determinant: u·[v×w] = ¯ ¯ ¯ ¯ ¯ ¯ u1 u2 u3 v1 v2 v3 w1 w2 w3 ¯ ¯ ¯ ¯ ¯ ¯ (13) Definition (Right-Handed System) Three vectors u,v and w [in the given order] are said to constitute a right-handed system if the scalar triple product u·[v×w] > 0 (14) Dr. Gabby (KNUST-Maths) Vectors 10 / 37
19. 19. Vector Product Triple Product Definition (Left-Handed System) Three vectors u,v and w [in the given order] are said to constitute a left-handed system if the scalar triple product u·[v×w] < 0 (15) Dr. Gabby (KNUST-Maths) Vectors 11 / 37
20. 20. Vector Product Triple Product Definition (Left-Handed System) Three vectors u,v and w [in the given order] are said to constitute a left-handed system if the scalar triple product u·[v×w] < 0 (15) Linearly Dependent Vectors The vectors u,v and w are linearly dependent or coplanar (lie in the same plane) if and only if u·[v×w] = 0 (16) These vectors are linearly independent or non-coplanar if and only if u·[v×w] ̸= 0 (17) Dr. Gabby (KNUST-Maths) Vectors 11 / 37
21. 21. Applications of Vector Products Outline of Presentation 1 Vector Product Cross product in R3 Triple Product 2 Applications of Vector Products 3 Equation of Lines and Planes Equation of lines Equation of Planes Dr. Gabby (KNUST-Maths) Vectors 12 / 37
22. 22. Applications of Vector Products Area and Volume Area of the parallelogram The area of the parallelogram formed by u and v is given by ∥u×v∥ (18) u and v are collinear (lie on the same line) or linearly dependent if u×v = 0. u v Two vectors or collinear if one is a scalar multiple of the other. Example [2, 1] and [6, 3] are collinear Dr. Gabby (KNUST-Maths) Vectors 13 / 37
23. 23. Applications of Vector Products Area and Volumes Paralellepiped is a three-dimensional shape whose faces are all parallelograms. Dr. Gabby (KNUST-Maths) Vectors 14 / 37
24. 24. Applications of Vector Products Area and Volumes Paralellepiped is a three-dimensional shape whose faces are all parallelograms. Volume of the paralellepiped The volume of the paralellepiped formed by u,v and w is Vp = ∥u·[v×w]∥ (19) Dr. Gabby (KNUST-Maths) Vectors 14 / 37
25. 25. Applications of Vector Products Area and Volumes Tetrahedron is a solid having four plane triangular faces; that is a triangular pyramid. Dr. Gabby (KNUST-Maths) Vectors 15 / 37
26. 26. Applications of Vector Products Area and Volumes Tetrahedron is a solid having four plane triangular faces; that is a triangular pyramid. Volume of a tetrahedron The volume of the tetrahedron formed by u,v and w is Vt = 1 6 ∥u·[v×w]∥ (20) Dr. Gabby (KNUST-Maths) Vectors 15 / 37
27. 27. Applications of Vector Products Example Use the scalar triple product to show that the vectors u = [1,4,−7], v = [2,−1,4],w = [0,−9,18] are coplanar. Dr. Gabby (KNUST-Maths) Vectors 16 / 37
28. 28. Applications of Vector Products Example Use the scalar triple product to show that the vectors u = [1,4,−7], v = [2,−1,4],w = [0,−9,18] are coplanar. u·[v×w] = ¯ ¯ ¯ ¯ ¯ ¯ 1 4 −7 2 −1 4 0 −9 18 ¯ ¯ ¯ ¯ ¯ ¯ (21) = 1 ¯ ¯ ¯ ¯ −1 4 −9 18 ¯ ¯ ¯ ¯−4 ¯ ¯ ¯ ¯ 2 4 0 18 ¯ ¯ ¯ ¯−7 ¯ ¯ ¯ ¯ 2 −1 0 −9 ¯ ¯ ¯ ¯ (22) = 1(18)−4(36)−7(−18) (23) = 0 (24) This means that u,v and w are coplanar. Dr. Gabby (KNUST-Maths) Vectors 16 / 37
29. 29. Applications of Vector Products Exercise 1 Let u = [u1,u2,u3], v = [v1,v2,v3] and w = [w1,w2,w3] be three nonzero vectors. a. Express w·[u×v] in terms of their coordinates. b. Show that u×v is perpendicular to u and v. 2 Find a unit vector perpendicular to the plane that passes through the points P = (1,4,6),Q = (−2,5,−1) and R = (1,−1,1) and compute the area of the triangle PQR. Dr. Gabby (KNUST-Maths) Vectors 17 / 37
30. 30. Equation of Lines and Planes Outline of Presentation 1 Vector Product Cross product in R3 Triple Product 2 Applications of Vector Products 3 Equation of Lines and Planes Equation of lines Equation of Planes Dr. Gabby (KNUST-Maths) Vectors 18 / 37
31. 31. Equation of Lines and Planes Equation of lines Vector equation of lines 1 A line in the 2D plane can be determined by a point on the line and slope (direction of the line). Similarly, a line L in the 3D space is determined a point Po(xo, yo,zo) on L and the direction of L. 2 In the 3D space the direction of a line is conveniently described by a vector, so we let v = [a,b,c] be a vector parallel to L. 1 Let P(x, y,z) be an arbitrary point on L and let ro = [xo, yo,zo] and r = [x, y,z] be the position vectors of Po and P 2 If a is the vector − − → PoP, then the Triangle Law for vector addition r = ro +a =⇒ r = ro + tv (25) Eqn. (25) is the vector equation of L Dr. Gabby (KNUST-Maths) Vectors 19 / 37
32. 32. Equation of Lines and Planes Equation of lines Parametric Equation of L 1 The vector equation of L in component form is r = ro + tv (26) [x, y,z] = [xo, yo,zo]+[ta,tb,tc] (27) [x, y,z] = [xo + ta, yo + tb,zo + tc] (28) Dr. Gabby (KNUST-Maths) Vectors 20 / 37
33. 33. Equation of Lines and Planes Equation of lines Parametric Equation of L 1 The vector equation of L in component form is r = ro + tv (26) [x, y,z] = [xo, yo,zo]+[ta,tb,tc] (27) [x, y,z] = [xo + ta, yo + tb,zo + tc] (28) 2 Two vectors are equal if and only if corresponding components are equal. Thus x = xo + ta, y = yo + tb, z = zo + tc (29) 3 These equations (29) are called parametric equations of the line L through the point Po and parallel to the vector v. 4 Each value of the parameter t gives a point (x, y,z) on L. Dr. Gabby (KNUST-Maths) Vectors 20 / 37
34. 34. Equation of Lines and Planes Equation of lines Example 1 Find a vector equation and parametric equations for the line that passes through the point (5,1,3) and is parallel to the vector i +4j −2k. 2 Find two other points on the line. Dr. Gabby (KNUST-Maths) Vectors 21 / 37
35. 35. Equation of Lines and Planes Equation of lines Example 1 Find a vector equation and parametric equations for the line that passes through the point (5,1,3) and is parallel to the vector i +4j −2k. 2 Find two other points on the line. 1 ro = [5,1,3] = 5i + j +3k and v = i +4j −2k, so the vector equation is r = [5i + j +3k]+ t[i +4j −2k] The parametric equations are x = 5+ t, y = 1+4t, z = 3−2t Dr. Gabby (KNUST-Maths) Vectors 21 / 37
36. 36. Equation of Lines and Planes Equation of lines Example 1 Find a vector equation and parametric equations for the line that passes through the point (5,1,3) and is parallel to the vector i +4j −2k. 2 Find two other points on the line. 1 ro = [5,1,3] = 5i + j +3k and v = i +4j −2k, so the vector equation is r = [5i + j +3k]+ t[i +4j −2k] The parametric equations are x = 5+ t, y = 1+4t, z = 3−2t 2 Choosing the parameter value t = 1 gives the point (6,5,1) is a point on the line. Similarly, t = −1 gives the point (4,−3,5). Dr. Gabby (KNUST-Maths) Vectors 21 / 37
37. 37. Equation of Lines and Planes Equation of lines 1 The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. Dr. Gabby (KNUST-Maths) Vectors 22 / 37
38. 38. Equation of Lines and Planes Equation of lines 1 The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. 2 For instance, if, instead of (5,1,3), we choose the point (6,5,1) in above, then the parametric equations of the line become x = 6+ t, y = 5+4t, z = 1−2t (30) Dr. Gabby (KNUST-Maths) Vectors 22 / 37
39. 39. Equation of Lines and Planes Equation of lines 1 The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. 2 For instance, if, instead of (5,1,3), we choose the point (6,5,1) in above, then the parametric equations of the line become x = 6+ t, y = 5+4t, z = 1−2t (30) 3 Or, if we stay with the point (5,1,3) but choose the parallel vector 2i +8j −4k, we arrive at the equations x = 5+2t, y = 1+8t, z = 3−4t (31) Dr. Gabby (KNUST-Maths) Vectors 22 / 37
40. 40. Equation of Lines and Planes Equation of lines 1 The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. 2 For instance, if, instead of (5,1,3), we choose the point (6,5,1) in above, then the parametric equations of the line become x = 6+ t, y = 5+4t, z = 1−2t (30) 3 Or, if we stay with the point (5,1,3) but choose the parallel vector 2i +8j −4k, we arrive at the equations x = 5+2t, y = 1+8t, z = 3−4t (31) 4 In general, if a vector v = [a,b,c] is used to describe the direction of a line L, then the numbers a,b, and c are called direction numbers of L. Dr. Gabby (KNUST-Maths) Vectors 22 / 37
41. 41. Equation of Lines and Planes Equation of lines 1 The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. 2 For instance, if, instead of (5,1,3), we choose the point (6,5,1) in above, then the parametric equations of the line become x = 6+ t, y = 5+4t, z = 1−2t (30) 3 Or, if we stay with the point (5,1,3) but choose the parallel vector 2i +8j −4k, we arrive at the equations x = 5+2t, y = 1+8t, z = 3−4t (31) 4 In general, if a vector v = [a,b,c] is used to describe the direction of a line L, then the numbers a,b, and c are called direction numbers of L. 5 Since any vector parallel to v could also be used, we see that any three numbers proportional to a,b, and c could also be used as a set of direction numbers for L. Dr. Gabby (KNUST-Maths) Vectors 22 / 37
42. 42. Equation of Lines and Planes Equation of lines Symmetric Equations of L 1 Another way of describing a line L is to eliminate the parameter t from x = xo +ta, y = yo + tb, z = zo + tc. If none of a,b, or c is 0, we can solve each of these equations for t: t = x − xo a , t = y − yo b , t = z − zo c (32) Equating the results, we obtain x − xo a = y − yo b = z − zo c (33) These equations (33) are called symmetric equations of L. Dr. Gabby (KNUST-Maths) Vectors 23 / 37
43. 43. Equation of Lines and Planes Equation of lines Symmetric Equations of L 1 Another way of describing a line L is to eliminate the parameter t from x = xo +ta, y = yo + tb, z = zo + tc. If none of a,b, or c is 0, we can solve each of these equations for t: t = x − xo a , t = y − yo b , t = z − zo c (32) Equating the results, we obtain x − xo a = y − yo b = z − zo c (33) These equations (33) are called symmetric equations of L. 2 If one of a,b, or c is 0, we can still eliminate t. For instance, if a = 0 we could write the equations of L as x = xo, y − yo b = z − zo c (34) This means that L lies in the vertical plane x = xo. Dr. Gabby (KNUST-Maths) Vectors 23 / 37
44. 44. Equation of Lines and Planes Equation of Planes Equation of Planes 1 Although a line in space is determined by a point and a direction, a plane in space is more difficult to describe. 2 A single vector parallel to a plane is not enough to convey the ‘direction’ of the plane, but a vector perpendicular to the plane does completely specify its direction. 1 Thus a plane in space is determined by a point Po(xo, yo,zo) in the plane and a vector n that is orthogonal to the plane. 2 This orthogonal vector n is called a normal vector. 3 Let P(x, y,z) be an arbitrary point in the plane, and let ro and r be the position vectors of Po and P. Then the vector r−ro is represented by − − → PoP Dr. Gabby (KNUST-Maths) Vectors 24 / 37
45. 45. Equation of Lines and Planes Equation of Planes 1 The normal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to r−ro and so we have n·[r−ro] = 0 =⇒ n·r = n·ro (35) 2 Expanded as n·[r−ro] = 0 (36) [a,b,c]·[x − xo, y − yo,z − zo] = 0 (37) The scalar equation of the plane through the point Po(xo, yo,zo) with the normal vector n = [a,b,c] is a(x − xo)+b(y − yo)+c(z − zo) = 0 (38) Dr. Gabby (KNUST-Maths) Vectors 25 / 37
46. 46. Equation of Lines and Planes Equation of Planes Example Find an equation of the plane through the point (2,4,−1) with normal vector n = [2,3,4]. Find the intercepts and sketch the plane. Dr. Gabby (KNUST-Maths) Vectors 26 / 37
47. 47. Equation of Lines and Planes Equation of Planes Example Find an equation of the plane through the point (2,4,−1) with normal vector n = [2,3,4]. Find the intercepts and sketch the plane. We first compute the scalar equation of the plane: a(x − xo)+b(y − yo)+c(z − zo)= 0 (39) 2(x −2)+3(y −4)+4(z +1) = 0 (40) 2x +3y +4z= 12 (41) Dr. Gabby (KNUST-Maths) Vectors 26 / 37
48. 48. Equation of Lines and Planes Equation of Planes Example Find an equation of the plane through the point (2,4,−1) with normal vector n = [2,3,4]. Find the intercepts and sketch the plane. We first compute the scalar equation of the plane: a(x − xo)+b(y − yo)+c(z − zo)= 0 (39) 2(x −2)+3(y −4)+4(z +1) = 0 (40) 2x +3y +4z= 12 (41) To find the x-intercept we set y = 0,z = 0 in this equation and obtain x = 6. Similarly, the y-intercept is 4 and the z- intercept is 3. Dr. Gabby (KNUST-Maths) Vectors 26 / 37
49. 49. Equation of Lines and Planes Equation of Planes Example Find an equation of the plane through the point (2,4,−1) with normal vector n = [2,3,4]. Find the intercepts and sketch the plane. We first compute the scalar equation of the plane: a(x − xo)+b(y − yo)+c(z − zo)= 0 (39) 2(x −2)+3(y −4)+4(z +1) = 0 (40) 2x +3y +4z= 12 (41) To find the x-intercept we set y = 0,z = 0 in this equation and obtain x = 6. Similarly, the y-intercept is 4 and the z- intercept is 3. Dr. Gabby (KNUST-Maths) Vectors 26 / 37
50. 50. Equation of Lines and Planes Equation of Planes Angle Between Two Planes Two planes are parallel if their normal vectors are parallel. For instance, the planes x+2y − 3z = 4 and 2x +4y −6z = 3 are parallel because their normal vectors are n1 = [1,2,−3] and n2 = [2,4,−6] and n2 = 2n1. Dr. Gabby (KNUST-Maths) Vectors 27 / 37
51. 51. Equation of Lines and Planes Equation of Planes Angle Between Two Planes Two planes are parallel if their normal vectors are parallel. For instance, the planes x+2y − 3z = 4 and 2x +4y −6z = 3 are parallel because their normal vectors are n1 = [1,2,−3] and n2 = [2,4,−6] and n2 = 2n1. If two planes are not parallel, then they intersect in a straight line and the angle between the two planes is defined as the acute angle between their normal vectors. Thus cosθ = n1 ·n2 ||n1||||n2|| (42) Dr. Gabby (KNUST-Maths) Vectors 27 / 37
52. 52. Equation of Lines and Planes Equation of Planes Distance from a Point to a Plane The distance D from a point P1 = (x1, y1,z1) to the plane ax +by +cz +d = 0 is D = |ax1 +by1 +cz1 +d| p a2 +b2 +c2 (43) Dr. Gabby (KNUST-Maths) Vectors 28 / 37
53. 53. Equation of Lines and Planes Equation of Planes Example 1 Find parametric equations and symmetric equations of the line that passes through the points A = (2,4,−3) and B = (3,−1,1). 2 At what point does this line intersect the xy-plane? Dr. Gabby (KNUST-Maths) Vectors 29 / 37
54. 54. Equation of Lines and Planes Equation of Planes Example 1 Find parametric equations and symmetric equations of the line that passes through the points A = (2,4,−3) and B = (3,−1,1). 2 At what point does this line intersect the xy-plane? Dr. Gabby (KNUST-Maths) Vectors 29 / 37
55. 55. Equation of Lines and Planes Equation of Planes Example 1 Find parametric equations and symmetric equations of the line that passes through the points A = (2,4,−3) and B = (3,−1,1). 2 At what point does this line intersect the xy-plane? We first find the vector v. Here, we are not explicitly given a vector parallel to the line, but observe that the vector v with representation − → AB is parallel to the line. v = − → AB = − − → OB − − − → OA (44) = [3,−1,1]−[2,4,−3] (45) = [1,−5,4] (46) Dr. Gabby (KNUST-Maths) Vectors 29 / 37
56. 56. Equation of Lines and Planes Equation of Planes So v = [a,b,c] = [1,−5,4] and taking the point Po as (2,4,−3), then parametric equations are x = xo + ta, y = yo + tb, z = zo + tc (47) x = 2+ t, y = 4−5t, z = −3+4t (48) Dr. Gabby (KNUST-Maths) Vectors 30 / 37
57. 57. Equation of Lines and Planes Equation of Planes So v = [a,b,c] = [1,−5,4] and taking the point Po as (2,4,−3), then parametric equations are x = xo + ta, y = yo + tb, z = zo + tc (47) x = 2+ t, y = 4−5t, z = −3+4t (48) symmetric equations x − xo a = y − yo b = z − zo c (49) x −2 1 = y −4 −5 = z +3 4 (50) Dr. Gabby (KNUST-Maths) Vectors 30 / 37
58. 58. Equation of Lines and Planes Equation of Planes 2. The line intersects the xy-plane when z = 0 in the symmetric equations and obtain x −2 1 = y −4 −5 = 3 4 (51) Thus x −2 = 3 4 =⇒ x = 11 4 (52) y −4 −5 = 3 4 =⇒ y = 1 4 (53) so the line intersects the xy-plane at the point µ 11 4 , 1 4 ,0 ¶ Dr. Gabby (KNUST-Maths) Vectors 31 / 37
59. 59. Equation of Lines and Planes Equation of Planes Example Find an equation of the plane that passes through the points P = (1,3,2), Q = (3,−1,6), R = (5,2,0) Dr. Gabby (KNUST-Maths) Vectors 32 / 37
60. 60. Equation of Lines and Planes Equation of Planes Example Find an equation of the plane that passes through the points P = (1,3,2), Q = (3,−1,6), R = (5,2,0) Let the vectors a and b corresponding to − − → PQ and − → PR that is a = [2,−4,4], b = [4,−1,−2] Since both a and b lie in the plane, their cross product a × b is orthogonal to the plane and can be taken as the normal vector. Dr. Gabby (KNUST-Maths) Vectors 32 / 37
61. 61. Equation of Lines and Planes Equation of Planes Example Find an equation of the plane that passes through the points P = (1,3,2), Q = (3,−1,6), R = (5,2,0) Let the vectors a and b corresponding to − − → PQ and − → PR that is a = [2,−4,4], b = [4,−1,−2] Since both a and b lie in the plane, their cross product a × b is orthogonal to the plane and can be taken as the normal vector. Thus n = a×b = ¯ ¯ ¯ ¯ ¯ ¯ i j k 2 −4 4 4 −1 −2 ¯ ¯ ¯ ¯ ¯ ¯ = 20i +20j +14k (54) Dr. Gabby (KNUST-Maths) Vectors 32 / 37
62. 62. Equation of Lines and Planes Equation of Planes With the point P = (1,3,2) and the normal vector n, an equation of the plane is 12(x −1)+20(y −3)+14(z −2) = 0 (55) or 6x +10y +7z = 50 (56) Dr. Gabby (KNUST-Maths) Vectors 33 / 37
63. 63. Equation of Lines and Planes Equation of Planes Example Find the point at which the line with parametric equations x = 2 + 3t, y = −4t, z = 5 + t intersect the plane 4x +5y −2z = 18. Dr. Gabby (KNUST-Maths) Vectors 34 / 37
64. 64. Equation of Lines and Planes Equation of Planes Example Find the point at which the line with parametric equations x = 2 + 3t, y = −4t, z = 5 + t intersect the plane 4x +5y −2z = 18. We substitute the expressions for x, y, and z from the parametric equations into the equation of the plane: 4(2+3t)+5(−4t)−2(5+ t) = 18 =⇒ t = −2 (57) Therefore the point of intersection occurs when the parameter value is t = −2. Then x = 2+3(−2) = −4, y = −4(−2) = 8, z = 5−2 = 7 (58) so the point of intersection is (−4,8,3). Dr. Gabby (KNUST-Maths) Vectors 34 / 37
65. 65. Equation of Lines and Planes Equation of Planes Example Find the angle between the planes x + y + z = 1 and x −2y +3z = 1 Dr. Gabby (KNUST-Maths) Vectors 35 / 37
66. 66. Equation of Lines and Planes Equation of Planes Example Find the angle between the planes x + y + z = 1 and x −2y +3z = 1 The normal vectors of these planes are n1 = [1,1,1] and n2 = [1,−2,3] so cosθ = n1 ·n2 ||n1||||n2|| (59) = 1[1]+1[−2]+1[3] p 3 p 14 (60) = 2 p 42 (61) θ = cos−1 · 2 p 42 ¸ (62) = 72ř (63) Dr. Gabby (KNUST-Maths) Vectors 35 / 37
67. 67. Equation of Lines and Planes Equation of Planes Exercise 1 Find the parametric equations of the line L that passes through the point A = (−1,2,1) and parallel to the vector i+j−k. 2 Find out if the following points belong to L : B = (0,3,0), C = (1,1,1), D = (−2,1,2). 3 If L ′ is a line with parametric equation x = 1+2s, y = 3,z = −2+ s, show that L and L ′ are not parallel and do not intercept. (They are skew lines.) 4 Show that the planes P : 2x+2y − z −10 = 0 and P ′ : 3 2 x − y + z = 0 are perpendicular. 5 Find the angle of intersection between the two planes P : 2x + 3y + 4z = 5 and P ′ : 2x −6y −3z = 0. 6 Find the component of the force F = 2i−j+2k in the direction of n = i+j+k p 2. Dr. Gabby (KNUST-Maths) Vectors 36 / 37
68. 68. END OF LECTURE THANK YOU Dr. Gabby (KNUST-Maths) Vectors 37 / 37