VECTORS IN THREE DIMENSIONS

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1 Mathematics Revision Guides Vectors in Three Dimensions Page of MK HOME TUITION Mathematics Revision Guides Level: ALevel Year VECTORS IN THREE DIMENSIONS Version: Date:

2 Mathematics Revision Guides Vectors in Three Dimensions Page of VECTORS IN THREE DIMENSIONS The sections on vectors at ASLevel were restricted to solving problems in the twodimensional plane, but we can solve problems in three dimensions as well The two standard unit vectors in the twodimensional plane are i and j, where i is parallel to the ais and j is parallel to the ais To etend the idea to three dimensions, we have a third unit vector, k parallel to the ais In column form, the three vectors are i, j and k Note that the positive ais points forwards towards the ee and out of the paper The two forms are interchangeable; thus vector a can be epressed as i + j k Vector arithmetic in three dimensions follows the same rules as in two: Eample (): Given vectors a and b, find i) a+b, ii) ab, iii) a, iv) ab i) a+b ; ii) a+b 7 ; iii) a ; iv) a b 9

3 Mathematics Revision Guides Vectors in Three Dimensions Page of Recap Doubleletter and arrow notation Vectors can be denoted b a single boldface letter, but another notation is to state their end points and write an arrow above them In the righthand diagram, vector a joins points O and A and vector b joins point O and B Therefore OA a and OB b The direction of the arrow is important here; the vector AO goes in the opposite direction to OA although it has the same magnitude Hence AO OA a Eample (): OABCEFGH is a cuboid Note that the verte O at the origin is on the rear face of the cuboid Epress the vectors p, q, r and s in twoletter, ijk component and column notation i) p OA, and because O is the origin and the position vector of A is j or, p j or in column form ii) q DC OC OD i k, or in column form iii) r OF i + j + k or in column form iv) s GB OB OG (i + j) (i + k) j k, or in column form

4 Mathematics Revision Guides Vectors in Three Dimensions Page of The magnitude of a twodimensional vector b a is b a from Pthagoras In three dimensions, the magnitude of a vector c b a is c b a Eample (): Find the magnitudes of the following vectors: i) ; ii) ; iii) ; iii) i) The magnitude of the vector is 7 units ii) The vector has a magnitude of 9 unit iii) Similarl, vector has a magnitude of 9 ) ( units iv) The i and j components of the vector are both ero, so its magnitude is In Eample (ii) we encountered a unit vector distinct from the standard ones of i, j and k

5 Mathematics Revision Guides Vectors in Three Dimensions Page of Eample (): Find the unit vectors having the same direction as the following: i) ; ii) ; iii) ; iv) i) The magnitude of the vector is 7, so we divide the components b 7 to obtain the unit vector in the same direction; it is ii) The vector has magnitude 9 ) (, so we divide the components b 9 to obtain the corresponding unit vector in the same direction, ie These last two eamples have rational components, but this is rarel the case iii) The magnitude of the vector is ) ( ) (, so we divide the components b to obtain the unit vector in the same direction; it is iv) Since onl the jcomponent of vector is nonero, the unit vector with the same direction is simpl

6 Mathematics Revision Guides Vectors in Three Dimensions Page of Angles between vectors in three dimensions The method is the same as in two dimensions; we calculate their lengths using Pthagoras and then use the cosine rule The onl difference is that we are dealing with twodimensional diagrams to represent three dimensions, so it is best to omit an coordinate aes to avoid confusion To find the angle between a vector and an of the coordinate aes, we simpl use a standard unit vector to represent the ais i for the ais, j for the ais, or k for the ais Eample (): Find the angle between the vector OA i + j + k and the ais The ais is represented here b the vector OX, which is equivalent to the standard unit vector i If the question had asked for the angle between OA and the ais, we would have used vector OY, equivalent to the standard unit vector j The same would appl to the ais we would use a vector equivalent to k

7 Mathematics Revision Guides Vectors in Three Dimensions Page 7 of Eample (): Find the angle between the vectors OR i + j + k and OS i + j + k

8 Mathematics Revision Guides Vectors in Three Dimensions Page of Eample (7): Find the angle between the vectors OR i j + k and OS i + j + k We could also deduce that the angle SOR is a right angle because the square of the length of RS is the sum of the squares of the lengths of OR and OS

9 Mathematics Revision Guides Vectors in Three Dimensions Page 9 of Eample (): Find the angle between the vectors OR i + j + k and OS i j k The angle between the vectors is obtuse here, given its negative cosine

10 Mathematics Revision Guides Vectors in Three Dimensions Page of Eample (9): In a triangle ABC, point A has position vector i) find the coordinates of B and C, ii) show that triangle ABC is isosceles iii) show that cos ABC 9 iv) find the area of the triangle, AB 7 and BC 9, 7 7 i) To find the coordinates of B, we add the position vector of A, ie + 7 Hence A (,, ) and B (7, 7, ) 7 The coordinates of C are found in the same wa: 7 + 9, so C (,, ) ii) The length of AB, ie AB, 7 In the same wa, BC, ( ) ( 9) ( ) Hence the lengths AB and BC are equal But AC ( ) ( ) ( _( )) The two sides AB and BC are equal in length, but the length of AC is different Hence triangle ABC is isosceles iii) B the cosine rule, cos ABC ( AB) ( BC) ( AC) ( AB)( BC) 9 9 iv) ABC cos 9 7, so the area of triangle ABC ½ (AB)(BC) sin ABC sin 7 square units

11 Mathematics Revision Guides Vectors in Three Dimensions Page of Eample (): The vertices of a quadrilateral OABC have the following position vectors: O ; A i + j + k; B i + j + k; C i j + k i) Show that OABC is a rhombus ii) Show that OABC is not a square i) Because the position vector of O is the ero vector, we can immediatel sa that OA i + j + k and OC i j + k, where OA and OC are adjacent sides of the quadrilateral The length OA ; length OC, ( ) Hence the adjacent sides OA and OC are equal To prove that OABC is a rhombus, we could either calculate the lengths of OB and BC and show that all four sides are equal, or we could show that OA and CB are equal and parallel, as are OC and AB Since AB OB OA, its vector is (i + j + k) (i + j + k) i j + k Hence AB OC, so its length is also ( ) j Also, CB OB OC, so its vector is (i + j + k) (i j + k) i + j + k Thus CB OA, with length All the sides of OABC are equal, and both pairs of opposite sides are parallel, so OABC is a rhombus ii) The diagonals of a square are equal in length, but those of a rhombus are not The diagonal OB has length To find the length of the other diagonal AC, we reckon AC OC OA and its vector is (i j + k) (i + j + k) i j + k The length is ( ) The diagonals OB and DC are unequal in length, so OABC is not a square

12 Mathematics Revision Guides Vectors in Three Dimensions Page of Eample (): The vertices of a quadrilateral ABCD have the following position vectors: A ; B ; C 7 ; D Show that ABCD is a parallelogram, but not a rectangle i) A sketch would show that AB and AD are two adjacent sides, and that AC is a diagonal AB OB OA ; AC OC OA 7 ; AD OD OA We then work out BC and DC : BC OC OB 7, so sides BC and AD are equal and parallel DC OC OD 7, so sides DC and AB are equal and parallel ABCD is therefore at least a parallelogram, so we use Pthagoras to check if the triangle ABC is rightangled, ie AC AB + BC Now AC + +, AB () + + 9, and BC For ABCD to be a rectangle, angle ABC must be a right angle, and therefore AC AB + BC Now, AB + BC, but AC, so triangle ABC cannot be rightangled, and hence ABCD is not a rectangle

13 Mathematics Revision Guides Vectors in Three Dimensions Page of Eample (): The vertices of a quadrilateral ABCD have the following position vectors: A ; B ; C ; D i) Show that ABCD is a rectangle, but not a square ii) Find the area of the rectangle i) A sketch would show that AB and AD are two adjacent sides, and that AC is a diagonal AB OB OA ; AC OC OA ; AD OD OA We then work out BC and DC : BC OC OB, so sides BC and AD are equal and parallel DC OC OD, so sides DC and AB are equal and parallel ABCD is therefore at least a parallelogram, so we appl Pthagoras in reverse to show that the triangle ABD is rightangled, ie AC AB + BC Now AC 7 +, AB 7 9 and BC 9 Hence AC AB + AD, angle BAD 9, and ABCD is a rectangle Because the lengths of AB and AD are different, ABCD cannot be a square ii) The area of the rectangle is 7, or 9, square units

14 Mathematics Revision Guides Vectors in Three Dimensions Page of Eample (): The diagram below shows a quadrilateral ABCD i) Show, using Pthagoras, that angle BAD is a right angle ii) Show that ABCD is a kite iii) Find the area of the kite iv) The point C is moved to C such that ABC D is a square a) Find the position vector of C b) Find the area of the resulting square v) The point A is moved to A such that A BCD is a rhombus a)find the position vector of A (C is the point (, 7) again) b) Find the area of the rhombus i) Using Pthagoras, (AB) () + () + () (AD) () + () + () + () + () 9 (DB) () + () + () + + Since (DB) (AB) + (AD), the triangle BAD is rightangled ii) A kite has two adjacent pairs of sides equal, and from the last part, AB AD 9 We work out the lengths of BC and DC in the same wa: (BC) () + () + () + () + (DC) () + () + () + + BC DC, so both pairs of adjacent sides of ABCD are equal, therefore ABCD is a kite The area of a kite (like that of a rhombus) is half the product of the diagonals, so we need to find the length of AC B Pthagoras, (AC) () + () + () + + 7, so AC 7 or As (BD), BD or The area of the kite is therefore square units

15 Mathematics Revision Guides Vectors in Three Dimensions Page of iv) a) From part i), we know that angle BAD is a right angle, and that lengths AB and AD are equal So for ABC D to be a square, the vectors BC and AD must be equal and parallel, as must AB and DC Let the position vector of C be Now AD OD OA, and ' BC ' OC OB Thus, hence the position vector of C,, for ABC D to be a square Using AB and DC would lead to the same result, since if three sides of a quadrilateral are equal and two of them form a parallel pair, then the fourth side is equal to the other three, as well as parallel to the unmatched side AB OB OA, and ' DC ' OC OD Thus, hence the position vector of C,, for ABC D to be a square b) We have worked out in part ii) that AB, so the area of the square ABC D is 9 square units

16 Mathematics Revision Guides Vectors in Three Dimensions Page of v) Unlike angle BAD, BCD is not a right angle, but the process of finding the direction vector of A is identical The lengths of AB and AD are equal, so for A BCD to be a rhombus, the vectors BC and A D must be equal and parallel, as must A B and DC Again, let the position vector of A be, and choose to work with vectors D A' and BC Now BC OC OB, so D A' OD ' OA, Thus, hence the position vector of A,, for A BCD to be a rhombus The area of a rhombus is half the product of the diagonals, so we need to find the length of A C B Pthagoras, (A C) (()) + () + () + () +, so A C or 9 As (BD) from part i), BD or Hence the area of the rhombus is 7 9 square units

Visit: ImperialStudy.com For More Study Materials Class IX Chapter 12 Heron s Formula Maths

Visit: ImperialStudy.com For More Study Materials Class IX Chapter 12 Heron s Formula Maths Exercise 1.1 1. Find the area of a triangle whose sides are respectively 150 cm, 10 cm and 00 cm. The triangle whose sides are a = 150 cm b = 10 cm c = 00 cm The area of a triangle = s(s a)(s b)(s c) Here