IB Mathematics Analysis and Approaches (SL/HL) Lecture Notes - Topic 3 Geometry and Trigonometry PDF
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Uploaded by WarmerGauss
2022
International Baccalaureate
Christos Nikolaidis
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These lecture notes cover Topic 3: Geometry and Trigonometry for International Baccalaureate (IB) Mathematics Analysis and Approaches (SL and HL). The notes detail topics like 3D coordinate geometry, the sine rule, the cosine rule, and trigonometric identities, along with examples and solutions. The content is part of a larger set of lecture notes.
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International Baccalaureate MATHEMATICS Analysis and Approaches (SL and HL) Lecture Notes Christos Nikolaidis TOPIC 3 GEOMETRY AND TRIGONOME...
International Baccalaureate MATHEMATICS Analysis and Approaches (SL and HL) Lecture Notes Christos Nikolaidis TOPIC 3 GEOMETRY AND TRIGONOMETRY 3.1 THREE-DIMENSIONAL GEOMETRY ……………………………………………………… 1 3.2 TRIANGLES – THE SINE RULE - THE COSINE RULE...………………… 5 3.3 APPLICATIONS IN 3D GEOMETRY – NAVIGATION ……..…………………….. 16 3.4 THE TRIGONOMETRIC CIRCLE – ARCS AND SECTORS …………………….. 22 3.5 SIN, COS, TAN ON THE UNIT CIRCLE – IDENTITIES ………………………… 29 3.6 TRIGONOMETRIC EQUATIONS ………………………………………………………….….. 38 3.7 TRIGONOMETRIC FUNCTIONS ………………………………………………………………. 48 Only for HL 3.8 MORE TRIGONOMETRIC IDENTITIES AND EQUATIONS..….………………. 61 3.9 INVERSE TRIGONOMETRIC FUNCTIONS ………………………………………………. 67 VECTORS 3.10 VECTORS: GEOMETRIC REPRESENTATION …………………………………………. 73 3.11 VECTORS: ALGEBRAIC REPRESENTATION ………..……………………………….. 80 3.12 SCALAR (or DOT) PRODUCT – ANGLE BETWEEN VECTORS …………… 87 3.13 VECTOR EQUATION OF A LINE IN 2D …..…………………………………………….. 92 3.14 VECTOR EQUATION OF A LINE IN 3D ………………………………………………… 98 3.15 KINEMATICS …………………………………………………………………………………………. 105 3.16 VECTOR (or CROSS) PRODUCT ………………………………………………………….… 108 3.17 PLANES ………………………………………………………………………………………………….. 113 3.18 INTERSECTIONS AMONG LINES AND PLANES ………………………………….. 120 3.19 DISTANCES …………………………………………………………………………………………….. 126 December 2022 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.1 THREE DIMENSIONAL GEOMETRY 3D COORDINATE GEOMETRY We know that a point in the Cartesian plane has the form P(x,y). In 3D space we add one more coordinate, thus a point has the form P(x,y,z). The distance between two points A(x1,y1,z1) and B(x2,y2,z2) is given by dAB (x 1 x 2 ) 2 (y1 y2 ) 2 (z 1 z 2 ) 2 while the midpoint of the line segment AB is given by x1 x 2 y1 y2 z1 z 2 M( , , ) 2 2 2 EXAMPLE 1 Let A(1,0,5) and B(2,3,1). Find (a) the distance between A and B (b) the distance between O and B (c) the coordinates of the midpoint M of the line segment [AB] (d) the coordinates of point C given that B is the midpoint of [AC] Solution (a) dAB (1 2) 2 (0 3) 2 (5 1) 2 1 9 16 26 (b) d OB 2 2 3 2 1 2 14 1 2 0 3 5 1 3 3 (c) M( , , ) i.e. M( , ,3 ) 2 2 2 2 2 (d) C(3,6,-3) Notice: the coordinates of A,B,C (B midpoint) form arithmetic sequences x: 1,2,3 y: 0,3,6 z: 5,1,-3 1 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis VOLUMES AND SURFACE AREAS OF KNOWN SOLIDS The volumes and the surface areas of 5 known solids are given below: Solid Volume Surface area Cuboid V xyz S 2xy 2yz 2zx Pyramid 1 S (sum of areas V (area of base) (height) 3 of the faces) Cylinder 2 V πr 2 h S 2π rh 2π r Cone S πrL πr2 1 V πr 2 h where 3 L r 2 h2 Sphere 4 V πr 3 S 4π r 2 3 Notation x, y, z : length-width-height r: radius of circular base h : vertical height 2 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 2 The volume and the surface area for the following solids Cube of side x Cuboid of square base x y x x x Cube: V xxx x 3 S 6x 2 Cuboid of square base: V x 2y S 2x 2 4xy EXAMPLE 3 Given that the volume of a cylinder is 25, (a) express h in terms of r (b) hence express the surface area in terms of r Solution 25 (a) V πr 2 h πr 2 h 25 h π r2 25 50 (b) 2 S 2π rh 2π r 2π r 2 2π r 2 2π r 2 πr r EXAMPLE 4 Given that the surface area of a cylinder is 100π, (a) express h in terms of r (b) hence express the volume in terms of r Solution 50 - r 2 (a) S 2π rh 2π r 2 2π rh 2π r 2 100π h r 50 - r 2 (b) V πr 2 h πr 2 πr(50 - r 2 ) 50π r - 50r 3 r 3 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 5 Find the volume and the surface area of a right pyramid of square base of side 6 and vertical height 4. 4 Μ΄ 6 Solution The vertical height is h=4. For the slant height AM we use the Pythagoras theorem on ANM. AM 2 AN 2 NM 2 AM 2 4 2 3 2 AM 5 The area of the triangle AED (and any side triangle) is 1 1 A ED AM 6 5 15 2 2 1 1 The volume is V (area of base) (height) = 6 2 4 48 3 3 The surface area is S (area of square base)+4A = 62+4×( 15 ) =96 Notice about the angles between lines and planes: ˆ Angle between line AM and plane BCDE = angle AMN Angle between line AD and plane BCDE = angle ADN ˆ ˆ Angle between the planes ADE and BCDE = angle AMN ˆ = 2× MAN Angle between the planes ACB and ADE = angle MAM ˆ 4 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.2 TRIANGLES – THE SINE RULE - THE COSINE RULE BASIC NOTIONS For any right-angled triangle C a b θ B c A we define the sine, the cosine and the tangent of angle θ by: b opposite c adjacent sinθ = = cosθ = = a hypotenuse a hypotenuse b opposite tanθ = = c adjacent Clearly sinθ tanθ = cosθ It also holds Pythagoras’ theorem a 2 b 2 c2 We can easily derive the so-called Pythagorean identity sin 2θ cos 2 θ 1 Indeed, 2 2 b c b 2 c2 a 2 sin 2 θ cos 2 θ 2 1 a a a2 a 5 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 1 C 4 sinB = 5 5 4 3 cosB = 5 θ 4 tanB = B 3 A 3 We can also confirm Pythagoras’ theorem: 5 2 = 32 4 2 Every angle has a fixed sine, cosine and tangent. For example 1 3 1 3 sin30o = , cos30o = , tan30o = 2 2 3 3 If we know the sine, the cosine or the tangent of an acute angle θ (i.e. θ 90ο). At the moment, it is enough to know that supplementary angles have equal sines but opposite cosines: e.g. sin30o = 0.5, sin 150o = 0.5 cos30o = 3/2 cos 150o = - 3/2 The values of sinθ and cosθ range between -1 and 1. 7 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis SINE RULE - COSINE RULE For any triangle two rules always hold: A c b B a C a b c SINE RULE = = sinA sinB sinC COSINE RULE a 2 b 2 c2 2b c cos A Notice: There are two more versions of the cosine rule: b 2 c2 a 2 2 ca cos B c2 a 2 b 2 2 ab cos C Consider, for example, the following triangle A 104.5o 2 3 46.60 28.9o B 4 C We confirm by GDC that the SINE RULE holds: 4 3 2 4.13 4.13 4.13 sin104.5 sin46.6 sin28.9 We can also confirm the three versions of the COSINE RULE: 42 = 32+22-2(3)(2)cos 104.5 (LHS = 16 RHS = 16) 32= 22+42-2(2)(4)cos 46.6 (LHS = 9 RHS = 9) 22= 42+32-2(4)(3)cos 28.9 (LHS = 4 RHS = 4) 8 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 2 Consider the following right-angled triangle A 90o c b B a C Then a b c b c o = = a = = sin90 sinB sinC sinB sinC and so b c sinB = and sinC = a a as expected by the definition of sinθ Also, a 2 b 2 c 2 2bc cos90 o implies a 2 b 2 c2 that is the Pythagoras’ theorem, since cos90o = 0. Moreover b 2 c 2 a 2 2ca cosB b 2 c 2 (b 2 c 2 ) 2ca cosB - 2c 2 2ca cosB c cosB = a b as expected by the definition of cosθ. Similarly we get cosC = a Consequently, SINE RULE generalizes the definition of sinθ COSINE RULE generalizes of the definition of cosθ and Pythagoras’ theorem 9 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis THE SOLUTION OF A TRIANGLE Any triangle has 6 basic elements: 3 sides and 3 angles. If we are given any 3 among those 6 elements (except 3 angles!) we are able to find the remaining 3 elements by using the sine rule or the cosine rule appropriately. Roughly speaking If we know we use (three sides) OR (two sides and an included angle) COSINE RULE otherwise SINE RULE In other words we use the SINE RULE when we know an angle-opposite side pair. EXAMPLE 3 (given three sides) A 2 3 B 4 C We use COSINE RULE 42 = 22 + 32 - 12 cosA 32 = 22 + 42 - 16 cosB 3 = -12cosA -11 = -16cosB cosA = - 0.25 cosB = 0.6875 A = 104.5o B = 46.6o Finally, C = 180o-A-B = 180o-104.5o-46.6o, Thus C = 28.9o 10 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis Notice: We may sometimes have no solutions at all. For example, if a=10, b=3, c=2 it is not possible to construct such a triangle! Indeed, the cosine rule gives us cosA = -7.25 which is not possible! EXAMPLE 4 (given two sides and an included angle) A 104.5o 2 3 B C We use COSINE RULE: BC 2 = 22 + 32 - 12 cos 104.5o = 16 Thus BC = 4 Then we know all the three sides and hence B and C can be found as above: B = 46.6o and C = 28.9o EXAMPLE 5 (given one side and two angles) A 104.5o 3 46.60 B C In fact, we know the third angle as well: C = 180o-A-B = 180o-104.5o- 46.6o, thus C = 28.9o Now we can use the sine rule twice 3 BC = BC = 4 sin 46.6 sin 104.5 3 AB = AB = 2 sin 46.6 sin 28.9 11 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 6 (given two sides and a non-included angle) A 2 3 46.60 B C We use the sine rule 3 2 = sinC = 0.484 sin 46.6 sin C Hence, C = 28.9o (by GDC) Then A = 180o - 46.6o - 28.9o, that is A = 104.5o The side BC can be found either by sine or cosine rule! It is BC=4 Notice: In fact, we obtain two values for C. C = 28.9o (by GDC) or C΄ = 180o - 28.9o = 151.1o (since supplementary angles have equal sines). But C΄ = 151.1o is rejected since B + C΄ = 46.6o + 151.1o > 180o But this is not always the case! THE AMBIGUOUS CASE If we are given two sides and a non-included angle (as above) we may have as a solution Two triangles One triangle No triangle at all 12 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis This is because the sine rule provides two values for an unknown angle. For example if we find sinC = 0.5 then C = 30o (this is sin-1C) or C΄ = 180o - 30o = 150o and these two values may result to different solutions. In example 6 we found only one solution. But in the following example we will find two solutions. EXAMPLE 7 (given two sides and a non-included angle) A 5 4 300 B C We use the sine rule: 4 5 = sinC = 0.625 sin 30 sin C Hence, C = 38.7o (by GDC) or C΄ = 180o - 38.7o = 141.3o CASE (1): If C = 38.7o then A = 180o - 30o - 38.7o, thus A = 111.3o and then BC 2 5 2 4 2 2(5 )(4)cos111.3 BC = 7.45 CASE (2): If C΄ = 141.3o then A΄ = 180o - 30o - 141.3o, thus A΄ = 8.7o and then BC2 5 2 4 2 2(5 )(4)cos8.7 BC΄ =1.21 13 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis We may, sometimes, obtain no solution at all. EXAMPLE 8 (given 2 sides and a non-included angle) A 5 1 300 B a C We use the sine rule: 1 5 = sinC = 2.5 sin 30 sin C which is impossible! Hence, there is no such a triangle! JUSTIFICATION OF THE AMBIGUOUS CASE In example 7, we were given B=30o, AB=5, AC=4 and we found two solutions for C, and thus two possible triangles: ABC and ABC΄. Indeed, the two triangles satisfying theses conditions are shown below A 5 4 4 300 B C΄ C Notice that AC=4 can be placed in two different positions. For the two possible values of angle C it holds C+C΄=180o (can you explain why?) 14 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis THE AREA OF A TRIANGLE A c b 1 Area = bc sinA 2 B a C Notice that two sides and an included angle are involved in the formula! We can derive two similar versions for this formula: 1 1 Area = ab sinC Area = a c sinB 2 2 EXAMPLE 9 Look at again the triangle in example 1: A 104.5o 2 3 46.60 28.9o B 4 C 1 Area = 2 3 sin104.5 o 2.90 2 The other two versions give the same result: 1 Area = 2 4 sin46.6o 2.90 2 1 Area = 3 4 sin28.9o 2.90 2 (you may notice little deviations on the result due to rounding!) 15 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.3 APPLICATIONS IN 3D GEOMETRY – NAVIGATION ANGLE OF ELEVATION – ANGLE OF DEPRESSION Suppose that an object is above the horizontal level of an observer. The angle of elevation θ to the object is shown below: Object θ Observer horizontal If the object is below the level of the observer the angle of depression θ to the object is shown below: Observer horizontal θ Object We very often see these notions in 3D shapes. For example, The angle of elevation from A to G is the angle BAˆ G. ˆ F (explain why!) The angle of elevation from A to F is the angle CA The angle of depression from H to B is the angle GHˆ B. ˆ C (explain why!) The angle of depression from H to C is the angle FH 16 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 1 An observer is situated at point A. (a) Find the distance AG and the angle of elevation of point G. (b) Find the distance AF and the angle of elevation of point F. Solution (a) We consider the triangle AGB. By Pythagoras’ theorem, AG 2 4 2 3 2 AG = 5 The angle of elevation is BAˆ G. Hence, tanBA ˆ G 3 BA ˆ G =36.9o 4 (b) For point F we consider the vertical height FC and thus the triangle AFC. We firstly need the side AC. By Pythagoras theorem in ABC AC 2 4 2 5 2 AC = 41 Now, by Pythagoras’ theorem in AFC, 2 AF 2 41 3 2 AF = 50 The angle of elevation is CAˆ F. Hence, tanCA ˆF 3 ˆ =25.1o CAF 41 17 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 2 P 30 45 horizontal B A An object P is above a hill. Two observers A and B are situated as in the diagram above. The angle of elevation from A is 45o. The angle of elevation from B is 30o. The distance between A and B is 10m. Find the vertical height h of the object P above the ground. Solution Consider the triangle P h 30° 45° B 10 A x K h h tan45 1 h x x x h h 1 tan30 h 3 x 10 x 10 x 10 3 Therefore, 10 h 3 = h 10 h( 3 1) = 10 h 13.7 m 3 1 Notice: Another approach is to work in triangle ABP first, to find h AP=19.318 and then by sin 45 , we find h 13.7 19.318 18 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 2 An observer is situated at point A. (c) Find the distance AG and the angle of elevation of point G. (d) Find the distance AF and the angle of elevation of point F. Solution (a) We consider the triangle AGB. By Pythagoras’ theorem, AG 2 4 2 3 2 AG = 5 The angle of elevation is BAˆ G. Hence, tanBA ˆ G 3 BA ˆ G =36.9o 4 (b) For point F we consider the vertical height FC and thus the triangle AFC. We firstly need the side AC. By Pythagoras theorem in ABC AC 2 4 2 5 2 AC = 41 Now, by Pythagoras’ theorem in AFC, 2 AF 2 41 3 2 AF = 50 The angle of elevation is CAˆ F. Hence, tanCA ˆF 3 ˆ =25.1o CAF 41 19 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis NAVIGATION - BEARING When we navigate on a map we should have in mind the four main directions North, East, South, West as well as the four intermediate directions as shown below. The angle between any consecutive directions is 45o. Thus, for example, if two persons walk towards the North and East directions respectively the angle between their directions is 90o. if two persons walk towards the North and Southeast directions respectively the angle between their directions is 135o. Another keyword in navigation is the bearing. Suppose that a moving body goes from point A to point B. The bearing of the course AB is the clockwise angle between the North direction and AB. The following diagram will clarify this notion. 20 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis According to the diagram: North North B 50o A the bearing of the course AB is 50o the bearing of the course BA is 230o (explain why!) EXAMPLE 3 A car travels: from point A to point B in bearing 50o, then from point B to point C with bearing 1500, then goes back to point A with bearing 270o. The distance AC is 10km. Draw a diagram to show the details find the distances AB and AC. Solution B 1500 500 300 500 300 A 10 km C According to the diagram  = 40o, B̂ = 80o, Ĉ = 60o Then, by using the sine rule 10 AB BC sin80 sin60 sin40 we find AB = 8.79km and BC = 6.53km 21 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.4 THE TRIGONOMETRIC CIRCLE – ARCS AND SECTORS The values of the angles can be represented well on the following trigonometric circle: 90o 120o 60o 135o 45o 150o 30o + 0o 180o 360o 210o 330o 225o 315o 240o 300o 270o In fact, each value on the circle indicates the angle between the corresponding radius and the positive x-axis radius (red arrow). The angle formed after a complete circle is 360°. The angle formed after half a circle is 180°. However, after completing a full circle (1st period) we can continue counting: 361°, 362°, 263° and so on The next full circle (2nd period) finishes at 2×360° = 720°. Similarly, we can move clockwise, considering negative angles: -1°, -2°, -3° and so on For example, 270° can also be seen as -90°. Therefore, an angle may have any value from - ∞ to +∞. 22 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis DEGREES AND RADIANS Consider the following circle of radius r =1 (unit circle). B θ A O The circumference of the circle is 2πr = 2π. Start from point A and move anticlockwise. What is the length of the arc AB? If θ = 0ο then AB=0 If θ= 360ο then AB=2π (full circle) If θ= 180ο then AB=π (semicircle) If θ= 90ο then AB=π/2 (quarter of a circle) Thus, an alternative way to measure the angle θ=AÔB is to measure the corresponding arc AB. The new unit of measurement is called radian. DEGREES (deg) RADIANS (rad) 90o π/2 180o 0o π 0 O 360o O 2π 270o 3π/2 23 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis Let us see the basic angles, in degrees and radians, on the trigonometric circle. π/2 2π/3 90o π/3 6 120o 60o 3π/4 π/4 135o 45o 5π/6 π/6 150o 30o + π 0o 0 180o 360o 2π 210o 330o 7π/6 11π/6 225o 315o 5π/4 7π/4 240o 300o 4π/3 270o 5π/3 3π/2 We can also move in the opposite direction (clockwise) and consider negative angles: -π -180o 0o 0 - -150o -30o -5π/6 -π/6 -135o -45o -3π/4 -π/4 -120o -60o -2π/3 -90o -π/3 -π/2 24 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis NOTICE: The ratio between degrees and radians is given by degrees 180 0 radians π EXAMPLE 1 Let θ1 =300, θ2 =800, θ3 =270. Transform in radians. deg We use the ratio : rad 30 o 180 0 30π π For θ1: 180x = 30π x = = rad x π 180 6 80 o 180 0 80π 4π For θ2: 180x = 80π x = = rad x π 180 9 27 o 180 0 27π For θ3: 180x = 27π x = =0.471rad x π 180 EXAMPLE 2 π 4π Let θ1 = rad, θ2 = rad, θ3 =2 rad. Transform in degrees. 3 9 deg We use the ratio : rad x 180 0 180π For θ1: = πx = x = 60ο π/3 π 3 x 180 0 4 180π For θ2: = πx = x = 80ο 4π /9 π 9 x 180 0 360 For θ3: = πx = 360 x = =114.6ο 2 π π NOTICE (not necessary to remember though!!!) 180 π 1 rad = = 57.3o 1o = = 0.0174 rad π 180 25 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis THE ANGLE VALUES OF A POINT ON THE UNIT CIRCLE Consider the point on the unit circle corresponding to 30ο. Let’s start from 0ο and move anticlockwise. We pass through 30ο and after completing a full circle, we pass through the same point at 30ο+360ο =390ο and then again at 30ο+360ο×2 =750ο and so on. In other words, we add (or subtract) multiples of 360ο: In this way, the same point has infinitely many angle values: 30ο+360οk where kZ Working in radians, we add multiples of 2π, so that the point has the infinitely many angle values: π +2kπ where kZ 6 Thus, for k = … -1, 0, 1, 2, … we obtain the values …, -330ο, 30ο, 390ο, 750ο, … [in degrees] 11π π 13π 25π …, , , , , … [in radians] 6 6 6 6 Therefore, any point θ on the circle has infinitely many values: 26 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis ARCS AND SECTORS Consider a circle of radius r. Let θ be the angle shown below measured in radians! B r θ O A The length of the arc AB is given by L=rθ 1 2 The area of the sector OAB is given by A= r θ 2 In particular, if we consider θ=2π (complete circle), we obtain L = rθ = 2πr (the circumference of the circle) 1 2 A = r θ = πr2 (the area in the circle) 2 EXAMPLE 3 Consider the following sector of a circle with r=5m and θ=0.6rad: 5 0.6 Then Length of arc: L = rθ = 5(0.6) = 3 1 2 1 2 Area of sector: A = r θ = 5 (0.6) = 7.5 m2 2 2 Perimeter of sector: L+r+r = 3+5+5 = 13m 27 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 4 B 3 30o A O Let r =3cm and θ=30ο. Find a) the length of the arc AB b) the area of the sector OAB c) the distance AB d) the area of the triangle OAB Solution π First of all we have to transform θ in radians: θ = 6 π π a) L = rθ = 3 = = 1.57 cm 6 2 1 2 1 2π 3π b) Asector = r θ= 3 = = 2.36 cm2 2 2 6 4 c) For AB we use COSINE RULE: π AB2 = 32+32-2.3.3cos AB = 2.41154 = 1.55 cm 6 1. π d) Atriangle = 3 3sin = 2.25 cm2 2 6 Notice in the example above length of arc AB > length of side AB : 1.57 > 1.55 area of sector OAB > area of triangle OAB : 2.36 > 2.25 as expected! Furthermore, the area of the segment between side AB and arc AB is the difference Asector-Atriangle =2.36-2.25=0.11. In general, 1 2 Asegment = r (θ-sinθ) 2 (can you explain why?). 28 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.5 SIN, COS, ΤΑΝ ON THE UNIT CIRCLE sinθ AND cosθ Consider again the unit circle (radius r =1) on the Cartesian plane. 1 y P(x,y) θ -1 O x 1 -1 Let P(x,y) be a point on the circle, OP = r = 1 θ = angle between OP and x-axis Then opposite y adjacent x sinθ = = =y and cosθ = = =x hypotenuse 1 hypotenuse 1 Thus, if we think the angle θ as a point on the circle: sinθ θ sinθ = y coordinate of θ cosθ = x coordinate of θ cosθ 29 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis This description helps us to define sinθ and cosθ not only for angles within 00θ 900, but for any value of θ on the circumference. 900< θ 90o we consider 1800-θ. EXAMPLE 3 Find the angle between the two intersecting lines (see Exercise 2(c)) 1 3 1 2 r1 = 2 +λ 4 and r2 = 4 +μ 2 3 5 4 3 3 2 We find the angle between the direction vectors b1 = 4 , b 2 = 2 . 5 3 We have b1. b2 = 3.2+4.2+5.3 = 29 and | b1 |= 50 , | b2 |= 17 so 29 cosθ = = 0.995, and the GDC gives θ=5.73° 50 17 EXAMPLE 4 Show that the angle between the following lines is 90°: 1 3 1 - 4 r1 = 2 +λ 4 and r2 = 4 +μ 3 3 5 4 0 The dot product of the direction vectors is 3(-4)+4.3+5.0 = 0. 101 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis DISTANCES In this paragraph we will study the distance between two points a point and a line two lines Let us present our methodology by considering particular examples: Distance between Points Consider A(1,2,3) and B(5,7,9) B(5,7,9) A(1,2,3) The well-known formula gives 2 2 2 d= (51) (72) (93) = 77 Distance between Point and Line Consider 5 3 point A(1,2,3) and line L: r = 7 +λ 2 9 1 A(1,2,3) b P(5+3λ,7+2λ,9+λ) Key point: Vector AP is perpendicular to line L. We first find the foot P(5+3λ,7+2λ,9+λ) on the line L. 3 4 3λ 3 AP L AP 2 5 2λ 2 1 1 6 λ 102 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis Thus 4 3λ 3 5 2λ 2 =0 6 λ 1 3(4+3λ)+2(5+2λ)+(6+λ)=0 14λ=-28 λ=-2 Hence, the foot of the distance is P(-1,3,7) The distance between the point and the line is d(A,P)= (1 1) 2 (2 3) 2 (3- 7) 2 = 21 Distance between Lines (A) If the lines are parallel: Consider 1 3 5 3 Line L1: r = 2 +μ 2 and line L2: r = 7 +λ 2 3 1 9 1 The lines are clearly parallel (equal direction vectors). A(1,2,3) P(5+3λ,7+2λ,9+λ) Key point: We select a point in line L1 and find the distance from line L2. Here, the distance of point (1,2,3) of Line L1 from line L2 is exactly the case B above. 103 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis (B) If the lines are skew: Consider 1 4 5 3 Line L1: r = 2 +μ 5 and line L2: r = 7 +λ 2 3 0 9 1 It is given that the lines are skew b1 P(1+4μ,2+5μ,3) b2 Q(5+3λ,7+2λ,9+λ) Key point: Vector PQ is perpendicular to both lines L1 and L2. 4 3λ - 4μ We first find foots P and Q. Notice that PQ= 5 2λ - 5μ 6 λ 4 3λ - 4μ 4 a) PQ L1 5 2λ - 5μ 5 0 6 λ 4(4+3λ-4μ)+5(5+2λ-5μ)+0(6+λ)=0 22λ-41μ=-41 4 3λ - 4μ 3 b) PQ L2 5 2λ - 5μ 2 1 6 λ 3(4+3λ-4μ)+2(5+2λ-5μ)+(6+λ)=0 14λ-22μ=-28 41 7 The system gives λ= , and μ= 15 15 13 1 16 23 94 Hence we find P( , ,3) and Q( , , ) 15 3 5 15 15 and hence we can find the distance |PQ|. 104 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.15 KINEMATICS A nice application of the vector equation of line is the following: VELOCITY AND SPEED Suppose that a body is moving along a straight line with a constant velocity and its position at time t is given by r = a +t b Then a is the position of the body at time t=0 b is the velocity vector of the body (usually v ) |b| is the speed of the body (usually | v |) The vectors (and thus the motion) can be in either 2D or 3D space. EXAMPLE 1 Suppose that a body is moving according to the equation 1 3 r = +t 2 4 where time is measured in seconds and distance in meters. The initial position (at t=0) of the body is (1,2). So it is 1 2 2 2 = 5 =2.23m far from the origin. The position at time t=1sec is (4,6) So it is 4 2 6 2 = 52 =7.21m far from the origin. 3 The velocity vector is v = 4 The speed is | v |= 32 4 2 =5 m/sec NOTICE If r = a +λ b is an equation of line, the direction vector b can be substituted by any multiple of b. If r = a +t b is an equation of motion, the velocity vector b CANNOT be substituted by a multiple of b. 105 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis This is because the velocity vector corresponds to one unit of time t. To explain the difference, consider the following situations: Suppose that a body is initially at position A(1,2) and after 1 second at position B(5,8). Then 4 1 4 velocity vector: v =AB= , equation of motion: r = +t 6 2 6 Suppose that a body is initially at position A(1,2) and after 2 4 seconds at position B(5,8). Then the direction vector b =AB= 6 corresponds to 2 seconds, hence 1 2 1 2 Velocity vector: v = b = , equation of motion: r = +t 2 3 2 3 EXAMPLE 2 Suppose that a body is moving on a straight line (in 3D space) in 1 the direction of the vector b = 2 with speed 15 ms-1. Its initial 2 position is A(1,1,1). Find the equation of the motion of the body. Since b = 12 + 2 2 + 2 2 = 3 The unit vector if b is 1 ˆ 1 b = 2 3 2 and since the speed is 15 1 5 v = 15b = 5 2 = 10 ˆ 2 10 Therefore, the equation of the motion is 1 5 r = 1 + t 10 1 10 106 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis In paragraph 3.14 we found that the point of intersection of the lines 1 3 1 2 r1 = 2 +λ 4 and r2 = 4 +μ 2 3 5 4 3 7 is the point (7,10,13) (i.e. with position vector 10 ) 13 Let us see the same question in terms of Kinematics. EXAMPLE 3 Two bodies are moving in 3D space according to the equations 1 3 1 2 r1 = 2 +t 4 and r2 = 4 +t 2 3 5 4 3 respectively. (notice that here we use the same parameter t for time). (a) Do their paths meet? (b) Do the two bodies collide? Solution We have to solve the equation r1 = r2. If we use the equations of r1 , r2 as they are (with t) we will answer only question (b) (the two bodies do not collide). It helps to call the time parameters t1 and t 2 respectively. 1 3 1 2 3t1 2t 2 0 r1 = r2 2 + t 1 4 = 4 + t 2 2 4t1 2t 2 2 3 5 4 3 5t 3t 1 1 2 The first two equations give t 1 =2, t 2 =3. These values satisfy the third equation 5 t 1 -3 t 2 =1. Hence 7 (a) The two paths intersect at 10 (use t 1 in equation r1 ) 13 (b) The two bodies do not collide since t1 ≠ t 2. 107 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.16 CROSS PRODUCT (for HL) This definition applies only for 3D vectors. THE GEOMETRIC DEFINITION (the “ugly” one) Let u and v be two vectors and θ be the angle between those two vectors ( where 0 θ π). The cross product (or vector product) of u and v is defined to be a vector given by u v = (| u || v |sinθ) n where n is the unit vector which is perpendicular to both u and v and follows the “screw rule”†: v u n u v n That is, u v is a new vector perpendicular to both u and v (and so to the plane determined by u and v ) with magnitude | u || v |sinθ and direction n. u u v v v u u v Notice that the commutative law does not hold. However, u v = -v u †If we place a screw at the common starting point of u and v and rotate it form u to v , then the screw will move in the direction of n. 108 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis THE ALGEBRAIC DEFINITION (the “pretty” one) a1 a2 Let u = b1 and v = b 2 be two vectors. The cross product (or c 1 c2 vector product) of u and v is given by b 1 c 2 b 2 c1 u v = c1 a 2 c 2 a 1 a b a b 1 2 2 1 Well, it doesn’t look as pretty as the title promised! But the is a kind of symmetry in it! For the first row of the result, you forget the first rows of u and v and you move along the arrow below a1 a 2 b1 c 2 b 2 c1 b1 b 2 = c 1 c2 Then you carry on in a similar way for the 2nd and the 3rd row. Mind though the order of the operations for the three rows: NOTICE For those who know determinants, the definition can be given in the form i j k u v = a1 b1 c1 a2 b2 c2 expanded in terms of the first row vectors, i.e. b c1 a1 c1 a1 b1 u v = 1 i j k b2 c2 a2 c2 a2 b2 109 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 1 1 4 Let u = 2 and v = 5 . Then 3 6 . a) find u v , b) find u v and v u (by using the “pretty” definition) c) verify that u v is perpendicular to both u and v a) u. v = 4+10+18=32 1 4 12 - 15 - 3 b) u v = 2 5 = 12 - 6 = 6 3 6 5 - 8 - 3 4 1 15 - 12 3 v u = 5 2 = 6- 12 = - 6 . That is v u = - u v 6 3 8- 5 3 c) u v u and u v v -3 1 -3 4 since 6 2 =-3+12-9 = 0, 6 5 =-12 +30-18 = 0 -3 3 -3 6 Notice that the “ugly” definition cannot be applied directly as we need the unit vector n. Let us choose below two more convenient vectors u and v in order to compare the two definitions. EXAMPLE 2 3 1 Let u = 2 and v = 4 . Then 0 0 a) find u v by using the “pretty” definition b) find the angle θ between u and v c) find the unit vector n. d) find u v by using the “ugly” definition e) verify that u v is perpendicular to both u and v 110 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3 1 0 a) u v = 2 4 = 0 0 0 10 uv 11 b) cosθ= = 0.74, hence θ = 42.27ο | u || v | 13 17 c) both vectors u and v are on the plane Oxy so the unit vector n is parallel to axis Oz (if we draw u and v we will realize n is in the positive direction so 0 n =0 1 0 d) u v = (| u || v |sinθ) n = ( 13 17 sin42.3 ) n = 10 n = 0 ο 10 e) clearly u v is parallel to n and thus perpendicular to both u and v. THE MAGNITUDE | u v | Notice that the ugly definition u v = (| u || v |sinθ) n implies | u v | = | u || v |sinθ since n is a unit vector. But, if we consider the triangle determined by u and v v θ u 1 we know that its area is given by | u || v |sinθ. 2 Therefore, the area of this triangle is given by 1 Area of triangle = |u v | 2 111 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis In other words, the magnitude of the cross product u v gives directly the area of the parallelogram determined by u and v v u Area of parallelogram = | u v | EXAMPLE 3 1 4 - 3 For u = 2 and v = 5 , we have seen that u v = 6 3 6 - 3 Therefore, the area of the parallelogram determined by u and v is given by Area = | u v | = 9 36 9 = 7.35 1 Also, the area of the corresponding triangle is (7.35)=3.67 2 EXAMPLE 4 Find the area of the triangle determined by the three points A(1,1,1), B(1,3,1) and (-3,3,4) B A C It suffices to find the area of the triangle determined by any two vectors; let’s choose the vectors AB and AC. 0 - 4 0 - 4 6 AB = 2 , AC = 2 and so AB AC= 2 2 = 0 0 3 0 3 8 Hence, 1 1 Area of triangle = |AB AC| = 36 0 64 = 5 2 2 112 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.17 PLANES (for HL) VECTOR EQUATION a1 Given: Point A(a1,a2,a3) (the position vector is a = a 2 ) a 3 b c1 1 Two vectors b = b 2 , c = c2 (which are non-parallel) c b3 3 There is a unique plane passing through A, parallel to both b and c b A c x The position vector r = y of any point P(x,y,z) of this plane is z given by r = a +λ b +μ c or a1 b1 c1 r = a 2 +λ b 2 +μ c2 a c 3 b3 3 where λ,μ are two parameters. 113 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis SHORT EXPLANATION If P(x,y,z) is any point on the plane then AP lies in fact on the plane determined by b and c. P(x,y,z) c A b Hence AP = λ b +μ c (for some λ,μ). Then, the position vector of P is given by r =OP=OA+AP= a +λ b +μ c PARAMETRIC EQUATIONS x a1 b1 c1 x=a1+λb1+μc1 y = a 2 +λ b 2 +μ c2 gives y=a2+λb2+μc2 z a c 3 b3 3 z=a3+λb3+μc3 CARTESIAN EQUATION If we eliminate λ and μ we will obtain an equation of the form Ax+By+Cz=D Remark: Although the method of eliminating λ and μ is not necessary (a much easier method will be given in a while!) we will demonstrate the procedure by using the example below, just to persuade ourselves. The steps are as follows Eliminate λ from the first two equations; Eliminate λ from the last two equations; Eliminate μ from the two resulting equations. 114 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 1 Let A(1,2,3) be the given point 4 7 b = 5 and c = 8 be the parallel vectors 6 8 Then the plane passing through A, parallel to b and c is Vector equation: 1 4 7 r = 2 +λ 5 +μ 8 3 6 8 Parametric equations: x=1+4λ+7μ (1) y=2+5λ+8μ (2) z=3+6λ+8μ (3) We eliminate λ from equations (1) and (2) 5 (1)-4 (2): 5x-4y=-3+3μ (4) We eliminate λ from equations (2) and (3) 6 (2)-5 (3): 6y-5z=-3+8μ (5) Next, we eliminate μ from (4) and (5) 8 (4)-3 (5): 40x-32y-18y-15z = -24+9 40x-50y+15z=-15 We simplify the equation by dividing by -5 and we obtain Cartesian equation: -8x+10y-3z = 3 As we said, a much quicker process will give the same result! 115 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis VECTOR EQUATION IN NORMAL FORM a1 Given: Point A(a1,a2,a3) (the position vector is a = a 2 ) a 3 Α Normal vector n = Β C There is a unique plane passing through A, perpendicular to n. n P A The equation of the plane is r.n =a.n Indeed, if P(x,y,z) is a random point of the plane then AP n But AP =OP-OA=r - a , and so AP. n =0 ( r - a ). n =0 r. n - a. n =0 r. n = a. n NOTICE The equation r. n = a. n derives immediately the Cartesian form Ax+By+Cz=D x Α Indeed, r n = y Β = Ax+By+Cz z C while a n is a constant scalar, say D In fact, given point A and n, we directly find the Cartesian form: write down the LHS using n Ax+By+Cz plug in A to find the RHS D 116 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 2 Find the equation of the plane passing through A(1,2,3) which is 8 perpendicular to n = 10 (normal vector) -3 . . The equation r n = a n implies x 8 1 8 y 10 = 2 10 z -3 3 -3 or directly -8x+10y-3z =-8+20-9 and finally -8x+10y-3z =3 NOTICE In examples 1 and 2 we obtained the same plane: -8x+10y-3z =3 We had: 1 4 7 EXAMPLE 1: Point: a = 2 Parallel vectors: b = 5 and c = 8 3 6 8 1 8 EXAMPLE 2: Point: a = 2 Normal vector: n = 10 3 -3 Indeed, if we consider as n the cross product b c (which is to both b , c and hence perpendicular to the plane) we obtain 4 7 8 n = 5 8 = 10 6 8 -3 Thus, given the vector equation of the plane, the Cartesian equation can be easily derived in this way instead of following the elimination process of λ and μ. 117 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis NOTICE If we know the Cartesian form Ax+By+Cz=D Α n = Β we also know a normal vector of the equation. It is C EXAMPLE 3 Consider the plane 3x-2y+z = 6 a) Find a normal vector n b) Find three points on the plane c) Find two vectors b and c parallel to the plane d) Confirm that n b and n c e) Write down all the forms of equation for this plane Solution 3 a) n = -2 1 b) For y=z=0 it is x=2, thus we obtain the point A(2,0,0). Similarly we obtain the points B(0,-3,0) and C(0,0,6) 0 2 -2 0 2 -2 c) Let b =AB= -3 - 0 = -3 and c =AC= 0 - 0 = 0 0 0 0 6 0 6 . . d) We can easily see that n b =-6+6 = 0 and n c =-6+6 = 0 2 -2 -2 e) Vector form: r = a +λ b +μ c or r = 0 +λ -3 +μ 0 0 0 6 Parametric form: x=2-2λ-2μ, y=-3λ, z=6μ 3 . Normal form: r -2 =6 [since a. n =6] 1 Cartesian form: 3x-2y+z = 6 118 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis EXAMPLE 4 Consider the plane 3 1 5 r = 1 +λ 2 +μ 0 2 3 2 a) Find two parallel vectors b and c b) Find three points on the plane c) Find a normal vector n d) Write down all the forms of equation for this plane Solution 1 5 A(3,1,2) (the obvious one) a) b= 2 , c = 0 b) B(4,3,5) for λ=1,μ=0 3 2 C(8,1,4) for λ=0, μ=1 1 5 4 c) Let n = b c = 2 0 = 13 3 2 -10 3 1 5 d) Vector form: r = a +λ b +μ c or r = 1 +λ 2 +μ 0 2 3 2 Parametric form: x=3+λ+5μ, y=1+2λ, z=2+3λ+2μ Normal form: r.n =a.n Cartesian form: 4x+13y-10z = 5 4 since 13 is a normal vector and a. n =12+13-20 = 5 -10 119 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.18 INTERSECTIONS AMONG LINES AND PLANES (for HL) In this section we will study the relative position between two lines a line and a plane two planes three planes TWO LINES Given: Lines L1: r1 = a1 +λ b1 L2: r2 = a 2 +μ b 2 We have already seen this study in paragraph 3.14. Let us remember all possible cases. Lines Look like Method parallel Check if b1 // b2 Check if b1 // b 2 coincide + a common point Intersect r1 = r2 at some point has a solution r1 = r2 skew has no solution θ = angle between the two lines b1 b 2 θ = angle between b1 and b 2 cosθ= | b1 || b 2 | 120 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis A LINE AND A PLANE Given: Line L: r1 = a +λ b Α Plane Π: Ax+By+Cz=D (so n = Β ) C Line and Plane Look like Method x n b Intersect plug r1 = y into z at some point Ax+By+Cz=D to find λ b Check if b n parallel n or no intersection point Check if b n + a common point Line lies on Plane or intersection points θ = angle between line and plane If φ = angle between b and n b n sinθ= then θ = 900 - φ | b || n | Notice: if the line and the plane are given in other forms, we transform them into the forms L: r1 = a +λ b and Π: Ax+By+Cz=D EXAMPLE 1 1 4 Consider the line L: r1 = 2 +λ 5 and the plane Π: 2x+5y-3z=18 3 6 Find the the angle between L and Π and the point of intersection. For the angle between L and P we have bn 15 sinθ= = =0.277, hence θ=16.10 | b || n | 77 38 121 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis The point of intersection lies on L, so it has the form (x,y,z)=(1+4λ,2+5λ,3+6λ) We plug it into the equation of the plane 2x+5y-3z=18: 2(1+4λ)+5(2+5λ)-3(3+6λ)=18 15λ+3=18 λ =1 Hence, x=5, y=7, z=9 and the intersection point is (x,y,z)=(5,7,9). EXAMPLE 2 1 4 Show that line L: r1 = 2 +λ 5 is parallel to plane Π: 2x+2y-3z=1 3 6 4 2 Method A: If b = 5 and n = 2 , then 6 - 3 b n =0 b n L //Π The point (1,2,3) of the line does not satisfy 2x+2y-3z = 1, hence the line does not lie on the plane. Method B: A point on L has the form (x,y,z)=(1+4λ,2+5λ,3+6λ) We plug it into the equation of the plane 2x+2y-3z=1: 2(1+4λ)+2(2+5λ)-3(3+6λ)=1 0λ=4 The last equation is impossible, thus there is no intersection point. EXAMPLE 3 1 4 Show that line L: r1 = 2 +λ 5 lies on plane Π: 2x+2y-3z = -3 3 6 Method A: Again b n =0 b n L //Π But this time, the point (1,2,3) of the line satisfies the equation 2x+2y-3z = -3, hence the line lies on the plane. Method B: A point on L has the form (x,y,z)=(1+4λ,2+5λ,3+6λ) We plug it into the equation of the plane 2x+2y-3z=-3: 2(1+4λ)+2(2+5λ)-3(3+6λ)=-3 0λ=0 The last equation is true for any λ, so the line lies on the plane. 122 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis TWO PLANES Α 1 Given: Planes Π1: A1x+B1y+C1z=D1 so n 1 = Β1 C 1 Α2 Π2: A2x+B2y+C2z=D2 so n 2 = Β2 C 2 Planes Look like Method Find two common points n2 and thus the line n1 intersecting or into a line one common point a and r = a +λ b direction vector b = n 1 n 2 or solve simultaneous equations parallel Check if n 1 // n 2 Check if n 1 // n 2 + coincide The equations are multiple to each other θ = angle between the two planes n1 n 2 θ = angle between n 1 and n 2 cosθ= | n 1 || n 2 | EXAMPLE 4 Consider the planes x+2y+3z=6 4x+5y+6z=15 Find the angle between the two planes and the line of intersection. n1 n 2 32 For the angle: cosθ= = θ=12.93ο | n 1 || n 2 | 14 77 123 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis For the line of intersection Method A: Let us first find two common points For z=0 the equations become x+2y=6 and 4x+5y=15 which give x=0, y=3. Hence, a common point is A(0,3,0). For z=1 the equations become x+2y=3 and 4z+5y=9 which give x=1, y=1. Hence, a common point is B(1,1,1). The two points A,B determine the equation of the intersecting line 0 1 r = 3 +λ - 2 0 1 Method Β: We find only one common point, say A(0,3,0) and as a 1 4 - 3 1 direction vector we consider n 1 n 2 = 2 5 = 6 which is // - 2 3 6 - 3 1 Method C: We solve the system of the two linear equations (GDC or Gauss elimination). The general solution is x λ, y 3 2λ , z λ. These are the parametric equations of the same line. EXAMPLE 5 Consider the planes x+2y+3z=10 2x+4y+6z=30 1 2 Their normal vectors 2 and 4 are clearly parallel. 3 6 Moreover, the two planes do not have a common point (since one equation is not a multiple of the other). EXAMPLE 6 Consider the planes x+2y+3z=10 and 2x+4y+6z=20 The two planes coincide (one is a multiple of the other) 124 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis THREE PLANES Given: Planes A1x+B1y+C1z=D1 A2x+B2y+C2z=D2 A3x+B3y+C3z=D3 The problem reduces to the solution of a 3x3 system of simultaneous equations (see paragraph 1.9). Then SYSTEM CONCLUSION The three planes have one common Unique solution (x,y,z) point (x,y,z) No common point: No solution The planes form a triangular prism or 2 of the planes are parallel Planes intersect into a line Infinitely many solutions or at least 2 planes coincide EXAMPLE 7 Consider the planes 2x +3y +3z = 3 x + y -2z = 4 5x +7y +4z = 10 We may see (either by Gauss elimination or by a GDC) that this system has infinitely many solutions: x = 14 + 16λ y = -5 + 7λ z = λ R (free variable) 14 16 The solution represents the line r = - 5 +λ 7 . 0 1 125 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis 3.19 DISTANCES (for HL) We have already studied the distance between two points a point and a line two lines In this section we will also study the distance between a point and a plane a line and a plane two planes Let us remember the distance between a point and a line: 5 3 point A(1,2,3) and line L: r = 7 +λ 2 9 1 A(1,2,3) b P(5+3λ,7+2λ,9+λ) The advantage here is that the foot P(x,y,z) has the form P(5+3λ,7+2λ,9+λ) thus, it is enough to find the parameter λ. 3 4 3λ 3 AP L AP 2 5 2λ 2 1 1 6 λ 3(4+3λ)+2(5+2λ)+(6+λ)=0 14λ=-28 λ=-2 Hence, the foot of the distance is P(-1,3,7) The distance between the point and the line is d(A,P)= 21 126 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis For the distance between a Point and a Plane, again the first task is to find the corresponding foot P(x,y,z) on the plane. Distance between Point and Plane Consider point A(3,4,6) and Plane Π: 2x+3y+5z=10 A(3,4,6) n P Key point: Line AP is parallel to the normal vector n. The equation of line AP is: 3 2 r = 4 +λ 3 6 5 Thus the foot has coordinates P(3+2λ,4+3λ,6+5λ). But it also lies on Π, so that 2(3+2λ)+3(4+3λ)+5(6+5λ)=10 38λ=-38 λ=-1 Hence the foot is P(1,1,1). The distance is |AP|, that is d(A,P)= (3- 1) 2 (4 - 1) 2 (6 - 1) 2 = 38 Distance between Line and Plane This case occurs only if the line is parallel to the plane. Consider 3 3 Line L: r = 4 +λ - 2 and Plane Π: 2x+3y+5z=10 6 0 It is given that the line is parallel to the plane. A(3,4,6) n P We just find the distance of point A of line L from plane Π 127 TOPIC 3: GEOMETRY AND TRIGONOMETRY Christos Nikolaidis Distance between Planes This case occurs only if the planes are parallel. Consider Plane Π1: 2x+3y+5z=10 and Plane Π2: 2x+3y+5z=48 Clearly planes Π1 and Π2 are parallel. A We just find a point of plane Π1: for x=y=0, we obtain z=2, and thus A(0,0,2) Then we find the distance be
