(iii) converting between scalar product and parametric forms. (ii) vector perpendicular to two given (3D) vectors

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1 Vector Theory (15/3/2014) Contents (1) Equation of a line (i) parametric form (ii) relation to Cartesian form (iii) vector product form (2) Equation of a plane (i) scalar product form (ii) parametric form (iii) converting between scalar product and parametric forms (3) Angle between two direction vectors (4) Perpendicular vectors (i) vector perpendicular to given (2D) vector (ii) vector perpendicular to two given (3D) vectors (5) Intersections (lines are 3D) (i) point of intersection of two lines (ii) point of intersection of a line and a plane (iii) line of intersection of two planes (6) Shortest distances (i) from a point to a plane (ii) between two parallel planes (iii) between parallel lines / from a point to a line (iv) between two skew lines 1

2 (1) Equation of a line (i) parametric form (2D example; but can be extended to 3D) The vector equation of the line written in various forms: through the points A & B can be (a) r = a + d (b) r = a + ( (c) r = a + (a weighted average of ; when ; when shows ) ; when, is the average of ; the diagram (d) ( ) = ( ) + ( ) or ( ) where a = ( ) and d = ( ) is any vector in the direction from A to B (normally d1 & d2 are chosen to be integers with no common factor) 2

3 Note the difference between (a) the vector equation of the line through the points A & B and (b) the vector : The vector has magnitude AB (the distance between A & B) and is in the direction from A to B. Whereas the vector equation of the line through A & B is the position vector of a general point P on the line, with completely different magnitude and direction to that of the vector (ii) relation to Cartesian form ( ) = ( ) + ( ) the straight line through with gradient (iii) vector product form (3D lines only) r = a + d can be written as ( ) (since and are parallel) or eg line through (1, 0, 1) and (0, 1, 0): ( ) ( ) ( ) ( ) 3

4 Thus equation is ( ) ( ) Note: Textbooks tend to write the determinant with the elements transposed (it gives the same result though). (2) Equation of a plane (i) scalar product form Let be the position vector of a point in the plane, and Let ( ) be a general point in the plane. be a vector perpendicular to the plane. As and are perpendicular, (a constant) (Cartesian form) 4

5 Example If ( ) and ( ), then (Another way of thinking of this is that, since is a point on the plane, it is a solution of, so that, or ) (ii) parametric form This is an extension of the parametric form of the vector equation of a line. Let and be non-zero vectors in the plane (that are not parallel to each other). Then 5

6 Note that and are direction vectors, whilst is a position vector. and can of course be determined from 2 points and in the plane, as and (or ) (iii) converting between scalar product and parametric forms (a) to convert from scalar product to parametric form Example Suppose that the equation of the plane is Let and, so that and a general point is ( ) ( ) ( ) ( ) ( ) (b) to convert from parametric to scalar product form Method 1 Example: ( ) ( ) ( ) ( ) Then eliminate s and t to obtain an equation in. 6

7 Method 2 For the above example, create normal vector: ( ) ( ) giving (3) Angle between two direction vectors Example 1 To find the acute angle between the line with equation ( ) ( ) and the plane with equation ( ) The two direction vectors in this case are ( ) and ( ) Then ( ) ( ) (*), so that This gives Whether is acute or obtuse depends on the relative direction of the normal vector to the plane and the direction vector of the line - see the diagram below. 7

8 In this case, the angle we want (between the plane and the line) is in the diagram. Thus Example 2 If we need to find the angle between two planes, then the angle in question will be in the diagram below. This will be acute, so that we expect to be obtuse (as ). If one of the normals to the planes has its direction reversed, then we obtain an acute angle from the scalar product result (*), and this has to be converted to the required angle by subtracting from. 8

9 (4) Perpendicular vectors (i) vector perpendicular to given (2D) vector Example Given direction vector ( ): gradient is ; hence perpendicular gradient and perpendicular direction vector is ( ) or ( ) (ii) vector perpendicular to two given (3D) vectors Let given vectors be and Method 1 Method 2 Let ( ) be required vector. Then eliminate two of from and (*) to give a direction vector in terms of parameter. eg ( ) (note: form of eq'ns (*) ensures that will be multiples of ) which is equivalent to the direction vector ( ) 9

10 (5) Intersections (lines are 3D) (i) point of intersection of two lines Note: Lines may not have a point of intersection, if the equations are not consistent; in which case they are termed 'skew'. Example: intersection of where has equation ( ) ( ) ( ) and has equation ( ) ( ) ( ) Eliminate, to give ( ) =( ) (ii) point of intersection of a line and a plane Example: has equation ( ) ( ) ; plane has equation ( ) Then (( ) ( )) ( ) creates a linear equation in. (It is possible that the line is either parallel to the plane or lies in the plane; in which case the term corresponding to 10

11 ( ) ( ) above will vanish, since the scalar product will be zero; then the remaining numbers will only be consistent if the vector corresponding to ( ) lies in the plane; ie if the scalar product corresponding to ( ) ( ) equals the right-hand side.) (iii) line of intersection of two planes Method 1 Starting with the equations of the planes in the form, let (eg), to obtain ( ) ( ); ie will be expressible as linear functions of. Note that we are effectively choosing a point on the line which has coordinate 0 (and coordinates ). Example 2 covers an unusual case. Example 1 Planes Let, so that and So that the equation of the line of intersection of the planes is: 11

12 ( ) ( ) Example 2 Planes This implies that and, so that the equation of the line of intersection of the planes is: ( ) ( ), where any value can be chosen for p Method 2 The required line will be perpendicular to the normal vectors of both planes. Therefore the vector product of the normal vectors to the two planes has the direction vector of the required line. Using Example 1 above, with planes, ( ) ( ) In order to find the equation of the line, we just need a point on it; ie a point on both planes, so that eg let ; then and the equation of the line is ( ) ( ) 12

13 Note: This can be seen to be equivalent to ( ) ( ) in Example 1, as follows: Let, so that Then ( ) ( ) ( ) (6) Shortest distances (i) shortest distance from a point to a plane (See "Equation of a plane" to convert between the scalar product and parametric forms of the equation of a plane, if necessary.) Method 1 Example 1 Point, P is ( ) ; plane has equation (*) The position vector of the point in the plane at the shortest distance from P is: ( ) ( ) for some (to be determined), as ( ) is the direction vector normal to the plane. Since this point lies in the plane, it satisfies (*); hence (**) giving 13

14 The shortest distance is the distance travelled from P to the plane, along the direction vector ( ); ie ( ) Example 2 In the special case where P is the origin O (and the plane has equation as before), (**) becomes ie, where is the normal to the plane, ( ) As before, the shortest distance from O to the plane is Alternatively, if the equation of the plane is given in 'normalised' form (ie the direction vector has unit magnitude; the word 'normal' being used here in a different sense to that of the normal to a plane); ie right-hand side of the equation., then the distance required is simply the Method 2 Using the above example, we can find the equation of the plane parallel to and passing through ( ). The equation of the parallel plane will be 14

15 ie From the special case of the Origin in Method 1, the distance between the two planes (and hence between the point and the plane) is Note: This method gives rise to the standard formula:, as the shortest distance from the point ( ) to the plane Method 3 Using the same example, where P is ( ) and the plane has equation (*), we first of all find a point Q in the plane (as in the diagram above) and create the vector The required distance will then be the projection of onto (the normal to the plane); namely 15

16 In this case, putting (say) in (*) gives, so that ( ), and ( ) Then ( ) and the shortest distance (ii) distance between two parallel planes Using the method for finding the shortest distance from the origin to a plane (method 1, example 2 of "shortest distance from a point to a plane"), the two planes need first of all to be put into normalised form; the constant term of each equation then gives the distance of the plane from the origin, so that the distance between the planes is then the difference between the constant terms. Example: Find the distance between the planes and As, the normalised equations are and so that the distance between the planes is (iii) distance between parallel lines / shortest distance from a point to a line Assuming that A and B are given points on the two lines, and that is the common direction vector: 16

17 Method 1 Let C be the point on with parameter, so that (*) Then we require ( ) Solving this equation for and substituting for in (*) gives, and the distance between the two lines is then. Example Let lines be ( ) ( ) and ( ) ( ) If ( ), ( ) and ( ) ( ), then ( ) ( ) Hence ( ) and the distance between the lines is 17

18 Method 2 Having obtained the general point, ( ) on in Method 1, we can minimise the distance BC by finding the stationary point of either or : Then and, as before Method 3 As, In the above example, =( ) and ( ) Then (iv) shortest distance between two skew lines Method 1 eg & ; A has position vector etc 18

19 XY is shortest distance, as it is perpendicular to both and unit vector in direction of XY is AE = XY ; (as CE is in the plane of the 'back wall' of the cuboid - because C lies on ) So AE is the projection of onto the direction of XY; ie onto So XY = AE = ( ) (the modulus sign ensuring that the distance is +ve) Note: This method can't be used to find the distance between two parallel lines, as, since 19

20 Example 1a: To find the shortest distance between the lines ( ) ( ) and ( ) ( ) The direction normal to the two lines is ( ) ; and we can take ( ) instead As the unit vector in this direction is ( ) We then require (( ) ( )) ( ) ( ) ( ) so that the required distance is or 9.2 Method 2 Find the vector perpendicular to both and, as in Method 1: Then the equation of the plane with normal, containing line (ie the front face of the cuboid in Method 1) will be Similarly the equation of the plane with normal (ie the back face of the cuboid) will be, containing line 20

21 The distance between these two planes (ie XY) is obtained by first adjusting the equations of the planes, so that they are based on a normal vector of unit magnitude. Thus Then planes"] and [See "Distance between two parallel [Note that this method is algebraically equivalent to method 1.] Example 1b (Lines as in 1a) From Example 1a, ( ) Then the equations of the planes in which the front and back faces of the cuboid in method 1 lie are [ ] and [ ] So the distance between the two planes, and hence between the two lines is Method 3 Referring to the earlier diagram, suppose that X and Y have position vectors & respectively. Then, if is the vector normal to both and, (*) 21

22 (ie Y is reached by travelling first to X and then along XY) and XY will then (*) gives 3 simultaneous equations in ( ) ( ), from which can be found Example 1c: (Lines as in 1a) From Example 1a, ( ) We need to find such that ( ) ( ) ( ) ( ) ( ) So or ( ) ( ) ( ) ( ) ( ) ( ) 22

23 and Method 4 As in method 3, suppose that X and Y have position vectors & respectively. Then and (*) Solving (*) enables to be determined, from which can be found Example 1d: (Lines as in 1a) ( ) ( ) ( ) ( ) ( ) Then ( ) ( ) and ( ) ( ) 23

24 and ie or and or or ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Then ( ) ( ) and 24