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1 2 EP-Program - Strisuksa School - Roi-et Math : Vectors Dr.Wattana Toutip - Department of Mathematics Khon Kaen University 200 :Wattana Toutip 2. Vectors A quantity with both direction and magnitude is a vector. A quantity with magnitude only is a scalar. Fig 2. A vector with magnitude is a unit vector. x A two dimensional vector can be written as or as xy y, or as xi yi unit vectors in the x and y directions respectively. The magnitude or modulus of xy, is x, y x 2 y 2., where i and j are x Three dimensional vectors can be written as y or as x, y, z or as xi yj zk, where z i, j, k are unit vectors in the x, y, z directions respectively. The magnitude of x, y, x is x, y, z x 2 y 2 z Vector geometry If A and B are points, the vector from A to B is written AB. The vector from the origin to a point is the position vector of that point.

2 If the position vectors of A and B are a and b, then AB b a. Vector can be added by a vector triangle or a vector parallelogram. Note that if OABC is a parallelogram, with the position vectors of A, B, C relative to O being abc,, respectively, then b a c. Fig 2.2 One vector is parallel to another if is a scalar multiple of it i.e.: a is parallel to b if there is a scalar so that a b. If wa xb ya zb, where a and b are not parallel to each other, then w y and x z. If A has position vector a, then a general point on the line through A parallel to b has position vector a tb, where t is a scalar. 2.. Examples. OABC is a parallelogram, X and Y lie on OC and AC respectively, with 2 OX OC, AY AC. Show that X, Y, B lie on a straight line. 4 Fig 2. Solution Let the position vector relative to O of A and C be a and c. The position vector of B is a c. The position vector of X is 2 c. The position vector of Y is a c a a c YB a c a c a c

3 2 XB a c c a c. YB XB. It following that YB is parallel to XB. 4 X, Y, B lie on a straight line 2. OABC is a parallelogram, with the position vectors of A and C relative to O being a and c. X is the midpoint of OA, and OB and XC meet at Y. Find the position vector of Y. Solution Fig 2.4 As above, the position vector of B is a c.the position vector of X is 2 a. A general point on OB is t a c. A general point on XC is c s a c 2. Y is on both these lines. ta tc c sa sc. 2 This gives two equations: t s and t s. 2 2 Solve to find that s, t. Y has position vector a c 2..2 Exercises. Which of the following quantities are vectors and which are scalars? (a) Velocity (b) Mass (c) Magnetic field (d) Temperature 2. Find the magnitudes of the following vector.: (a),4 (b) 2,, 7

4 (c) i j (d) 2i j 2.5k. Find the vectors AB for the following points A and B : (a) A,7, B 4,9 (b) A2,7,4, B,2, (c) A x, y, B p, q (d) A2 x, y, z, Bx, y,4z 4. OAB is a triangle, and X and Y are the midpoint of OA and OB respectively. Let OA aand OB b. Find XY in terms of a and b. Show that XY is parallel to AB. 5. OAB is triangle with OA a and OB b. X and Y are such that OX a and OY b. If XY is parallel to AB show that. 6. OABC is a parallelogram, with OA a and OC c. Find OB in terms of a and c. What can you deduce? 7. A, B, C, D are four points in space,with position vectors a, b, c, d respectively. I, J, K, L are the midpoints of AB, BC, CD, DA respectively. Find the position vectors of I, J, K, L in terms of a, b, c, d. Find the vectors IJ, JK, LK, IL. What can you say about the figure IJKL? 8. A, B, C have position vectors abc,, respectively. Find the position vector of M, the midpoint of BC. G lies on AM with AG 2GM. Show that the position vector of G is a b c. Show that G lies also on the lines from B to the midpoint of AC and from C to the midpoint of AB. 9. The position vectors of A and B relative to O are a and b. X is the midpoint of OA and Y lies on OB with OY OB. XY meets AB at Z. Find the position vector of Z. 0. a and b are two non-parallel vectors. The position vectors of XY, and Z are a 2, b a b and a b.find the value of if XYand, Z lie on a straight 2 4 line.. ABCDEF is a regular hexagon, with centre at O.OA a and OB b. Express, in terms of a and b, FA, DE, CD, CF.

5 2.2 Scalar Product Fig 2.5 The scalar, or dot product of two vectors is defined as follows: a. b a b cos, Where is the angle between a and b. In terms of coordinates the scalar product is as follows: a, b. x, y ax by for two-dimensional vectors. ai bj ck xi yj zk ax by cz, for three-dimensional vectors. In particular, ab. 0if and only if a and b are perpendicular. In particular, putting a b, a. a a The component or projection of a in the direction of b is 2.2. Examples 2 ab. b. Find the angle between the vectors, 2, and 2, 4 Solution Apply the formula above., 2, 4, 2,. 2,, cos 8. 8 cos and. 2, A, B, C are the vertices of a triangle. Let the perpendiculars from A to BC and from B to AC meet at D. Show that CD is perpendicular to AB. (Note: D is called the orthocenter of ABC.) Solution Let A, B, C, D have position vectors a, b, c, d respectively. From A to BC and from information given : d a. b c 0 and d b a c. 0

6 Expand these: d. b d. c a. b a. c 0 d. a d. a b. c a. c 0 Subtract these equations: d. b d. a b. c a. c 0 d cb a 0 DC is perpendicular to AB. Fig Exercises. Evaluate the following: (a),2.,4 (b) 2, 4. 7, (c),2,5. 2,, (d) i 2 j. i 5 j 2. Find the angles between the pairs of vectors in Question.. Find the value of x if x, is perpendicular to 4. Find a vector which is perpendicular to 2,7. 5. Find a unit vector which is perpendicular to 7, 24,. 6. Find the values of x and y if xy,, is perpendicular to both, 2, 4 and 4, 2,. 7. Find a vector which is perpendicular to both,0, and 2,, Find a unit vector which is perpendicular to both,,5 and,,. 9. Find the component of 2, 4 in the direction of,. 0. Find the component of 2i j k in the direction of i j.. Verify that the component of xy, in the direction of,0 is x. 2. Verify that component of xi yj zk in the j direction is y.. Let OAB be a triangle, with OA a and OB b. Let c a b be the third side of the triangle. By expanding cc. prove the cosine rule for OAB. 4. OABC is a parallelogram, with OA a and Oc c. By considering OB. CA show that OABC is a rhombus if and only if its diagonals are perpendicular. 5. Let O be the circumcentre of ABC. Let the position vectors of A, B, C relative to O be abc.,, (So that a b c ). Show that ab c is the orthocenter of the triangle.(see (2) of 2.2.).

7 6. Find the point X on the line 2, t 4, such that OX is perpendicular to the line. Hence find the perpendicular distance from the origin to the line. 7. Find the point X on the line of Question 6 such that AX, where A is at 2, 7,is perpendicular to the line. Hence find the perpendicular distance from A to the line. 2. Lines and planes in three dimensions Suppose a line goes through a point vector a and is parallel to the vector parallel to the vector v. Then a general point on the line is : r a v Where is a scalar. If a ai bj ck and v vi uj wk, then the Cartesian equations of the line are : x a y b z c v u w Suppose a plane goes through a point with position vector a and is parallel to vector u and v. Then its equation can be written as : r. n a. n If n li mj nk, then the Cartesian equation of the plane is : lx my nz d Where d a. n is a constant. The shortest,i.e. perpendicular, distance from the point abc,, to the plane lx my nz d is : la mb nc d l m n The angle between two lines is the angle between their direction vectors. This can be found from the scalar product, defined in 2.2. To find the angle between a line and a plane, find the angle between the direction of the line and the perpendicular to the plane, and subtract that angle from 90. The angle between two planes is the angle between their perpendiculars 2.. Examples. Two lines are given by: 2 2 and r ti j 4k i j k r i j k i j Find the value of t so that the lines intersect. Find the angle between the lines. Solution Equate the values of r. i 2 j k 2i j ti j 4l i j k. This gives the following three equations:

8 2 t ;2 ; 4. From the last two equations and 2.Substitute these values into the first equation to find t. t 4 The angle between the lines is the angle between the two direction vectors. Find this by the scalar product. 5 cos. The angle is 9 2. Find the equation of the plane which contains the two lines of example. Solution Let a perpendicular to the plane be li mj nk. This vector is at right angles to the direction vectors of the lines. 2lm 0 and l m n. These equations are satisfied by l, m 2, n.hence a perpendicular vector is i 2 j k. The equation of the plane is x 2y z i 2 j k. i 2 j k The equation is x 2y z Exercises. Find vector equations for the following lines: (a) Through,, 2, parallel to 2i j 7k (b) Through, 0,, parallel to 4i 2j (c) Through,, and 2,5, 2 (d) Through,2, and, 2, 2. Find the Cartesian equations of the lines in Question.. Find the values of t so that the following pairs of lines intersect: (a) r,,2 ti j and r, 4, 2i j k x y 4 z 5 x y z (b) and t Find the angles between the pair of lines in Question. 5. Find in the form r a u v the equations of the following planes: (a) Through, 2,, parallel to i j k and i k. (b) Through,2,, 2,, and,,. 6. Find in the form r. n k the equation of the following planes: (a) Through, 2,,perpendicular to 2i j k (b) Through 2,, 4, perpendicular to the z -axis. (c) Through 2,, 4,parallel to i j and to k. (c) Through 2,,2, 2,,0 and 0, 4,.

9 7. Find the angles between the following: (a) The line 0,,2 i j and the plane r i j k (b) The line x 4 y z 2.. and the plane x y z 0 (c) The line 0,, 2i j k and the plane 8. Find the angles between the following pairs of planes : (a) x y 2z and 2x y z (b) r. i 2 j k 0 and r i j k. 2 (c) r,,0 i k j and 2,,7,, i j vk. r i j k 9. Find the perpendicular distance between the following: (a) The point,, and the plane x 2y z 4 (b) The point 2,0, and the plane 2x y 4z 0. Find the point where the line r 5, 7, i j 2k meets the plane x 2y z 4. Find the angle between the line and the plane.. Find the point where the line r, 0, i 2 j meets the plane 2x y z. Find the angle between the line and the plane. 2.4 Examination questions. Referred to the point O as origin, the position vectors of the points A and B are i j and 8i 4j respectively. The point P lies on the line OB, between O and B, such that 4OP OB. The point Q lies on the line AB, between A and B, such that 5AQ 2AB. Find, in term of i and j, the vectors OP, AB, AQ, OQ and AP. The lines AP and OQ intersect at the point R. Given that OR : OQ and AR : AP, express the position vector of R in terms of (a),i and j (b),i and j. Hence find the values of and. Prove that the length of OR is The position vectors, relative to an origin O, of three points A, B, C are 2i 2 j,5i j and i 9 j respectively. (i) Given that OB moa noc, where m and n are scalar constant, find the value of m and of n. (ii) Evaluate AB. BC and state the deduction which can be made about ABC. (iii)evaluate AB. BC and hence find BAC.. The position vectors of the points AB, with respect to an origin O are a and b respectively, where ABand, O are not collinear. Point L, M, N have position

10 vectors pa, qb and 2b a respectively, where p and q are non- constants and LM is not parallel to AB. Show that N lies on the line AB. Prove that L, M, N are collinear if 2p q p. If p and q, find LM LN. 2 If, further, a 2i j 6k and b 4i j 4k,find the position vector of the foot of the perpendicular from N to OA. 4. A Cartesian frame of reference, having origin O, is defined by the by mutually perpendicular unit vectors i, j and k. (i) Find the vector equation of the line which passes through the two points, having position relative to O, 2i j 2k and i j in the form r a tb. (ii) A second line has vector equation r i 4 j k s 2i j k 2 2. Find the point of intersection of the two lines. 5. Relative to an origin O,points A and B have position vectors 2i 9 j 6k and 6i j 6k respectively, i, j and k being orthogonal unit vectors. C is the point such that OC 2QA and D is the midpoint of AB. Find (i) The position vectors of C and D (ii) A vector equation of the line CD (iii)the angle AOB correct to the nearest degree. 6. Calculate unit vectors, or otherwise,calculate the equation of one of the bisectors of the angles between the lines By considering the sum of these two vectors, or otherwise, calculate the equation of one of the bisectors of the angles between the lines x s y 4 and x t y 4 Deduce the equation of one of the bisectors of the angles between the lines x 4 s y 4 and x 5 4 t, giving your answer in a similar form. y 9 7. The lines L and M have the equations r 2 s and 4 5 respectively. The plane has the equation 2 r r 4 t 2 6 2

11 (i) Verify that the point A with coordinates, 4,4 lies on L and on M but not on. (ii) Find the position vectors of the point of intersection B of L and. (iii)show that M and have no common point. 2 (iv) Find the cosine of the angle between the vectors and 0. 5 (v) Hence find, to the nearest degree, the angle between L and. 8. With respect to the origin O the points A, BC have position vectors 5, 4 4, 5 2 a i j k a i j k a i j k respectively, where a is a nonzero constant. Find (a) A vector equation for the line BC, (b) a vector equation for the plane OAB, (c) the cosine of the acute angle between the lines OA and OB. Obtain, in the form r. n p, a vector equation for, the plane which passes through A and is perpendicular to BC. Find Cartesian equation for (d) the plane, (e) the line BC. 9. The line L passes through A0,, and is parallel to 0, and the line M passes through B 2,, and is parallel to. 2 Show that L and M do not intersect. M ' is the line through A parallel to M. The lines M ' and L lie the plane. (i) Find the direction of the normal to. (ii) Find the angle between AB and. (iii)write the equation to in standard Cartesian form. (iv) Use the formula for the distance of a point from a plane to show that the shortest distance from B to is AB. Explain how you could have deduced this result from answer your answer to (ii).

12 Common errors. Points and Vectors (a) The vector from the origin to from say,7 to position. (b) The vector from 2,5 to,4 is,4. But this vector also goes 4,. Vectors have magnitude and direction, but not,0 is,5 and, 5. (c) In general, if A and B have position vectors a and b, then AB is b a. It is not b, or a b, or a b. 2. Modulus (a) The modulus of 2 2,4 is 4 5, not 7. (b) Be careful of negative signs. Any number squared is positive,so the modulus of, 4,4. is the same as the modulus of. Scalar Product (a) ab. is scalar, not a vector: 2,., 4. It should be (b) Do not be confused if there are many possible answers to a problem. There are infinitely many vectors perpendicular to, 2,,for example. Just give the simplest one you can find. 4. Lines and Planes (a) Do not confuse the position vector of a point on a line and the direction vector of that line. (b) In there dimension, two Cartesian equations are needed to define a line. If you have only one equation then that will describe a plane. (c) Do not confuse vectors perpendicular to a plane with vector parallel to the plane.

13 Solution (to exercise) ac, 2. (a) 5 (b) 54 (c) 0 (d) (a), 2 (b), 5, (c) p x, q y (d) x,2y, z 4. 2 b a 6. a c, a c 2 a b, b c, d a, c a, d b, c a, d b Parallelogram 8. 2 b c 9. 2a b b, a b, b,2a 2b (a) (b) 8 (c)9 (d) 7 2. (a)0. (b) 55. (c) 47.9 (d) , , x.8, y 5. 7., 4, ,, , 0.96, , 0.6,8.. (a),,2 2,,7 (b),0, 4,2,0 (c),,, 2, (d),2 0, 4,2

14 x y z 2 x y z 2. (a) (b) x y z (c) (d) x y 2 z (a) t (b) t 2 4. (a) 40.5 (b)98 5. (a), 2,,, (, 0,) (b),2,,,2, 0,,0 6. (a) r. 2,, (b) r. 0,0, 4 (c) r.,,0 0 (d) 7. (a) 25 (b)4 (c)7 8. (a) 60 (b) 50 (c) (a) 0.86 (b).67 0.,,.79.,,.47 2 =========================================================== References: Solomon, R.C. (997), A Level: Mathematics (4 th Edition), Great Britain, Hillman Printers(Frome) Ltd. r.,, 5

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