Vectors Section 1: Lines and planes Notes and Examples These notes contain subsections on Reminder: notation and definitions Equation of a line The intersection of two lines Finding the equation of a plane The equation of a plane through three points Finding the intersection of a line and a plane Finding the distance of a point from a plane The angle between a line and a plane The angle between two planes Reminder: notation and definitions In typescript algebraic letters representing vectors are written in bold. When writing by hand you cannot easily write in bold, so the convention is that you underline letters representing vectors. It is very important that you underline vectors to avoid confusing them with scalar quantities. e.g. typescript a = hand-written a. For a vector between two points, say L and M, write the letters representing the points next to one another with an arrow above. e.g. vector between point L and point M = LM It is usual to denote a point by a capital letter and the position vector of that point by the same letter in lower-case. e.g. we could write: l = position vector of point L m = position vector of point M So LM m l The modulus of a vector is the magnitude of the vector, or the length of the line representing the vector. The modulus of a vector a is written as a and can be found using Pythagoras theorem. e.g. for the vector a a1i a2j a3k, a a a a. 2 2 2 1 2 3 The scalar product of two vectors a and b is written as a.b and defined by ab. a b cos where is the angle between the vectors a and b. 1/9
The equation of a line A line is defined by a vector in the direction of the line, and a point on the line. If a point on the line has position vector a, and the direction vector of the line is d, then the vector equation of the line is r a d or a1 d1 r a d 2 2 a3 d3 where is a parameter. Different values of correspond to different points on the line. This vector equation can be written in parametric form: x a d, y a d, z a d. 1 1 2 2 3 3 By rearranging each of these parametric equation to give expressions for, and equating these three expressions, you obtain the equivalent Cartesian equation of the line: x a 1 2 3 y a z a. d d d 1 2 3 The intersection of two lines In two dimensions, two non-parallel lines always intersect. In three dimensions, however, two lines may not intersect even if they are not parallel. Lines which do not intersect are called skew lines. You can find out if two lines intersect simply by attempting to find a point of intersection, as shown in the example below. Example 1 4 5 5 2 Lines l 1 and l 2 have vector equations r 3 2 and r 4 1 3 1 respectively. Do these lines intersect? The lines are clearly not parallel as their directions are not parallel (if they were then 5 2 their direction vectors 2 and 1 would be multiples of one-another). 3 As they are not parallel, we must establish whether or not they are skew. 2/9
If the lines meet then their parametric equations can be solved simultaneously i.e. there will exist values of and which will satisfy each of these equations simultaneously: 4 5 5 2 (A) 3 2 4 (B) 3 1 (C) 1 1 (C) ; substitute in (B). 3 3 Note that different symbols must be used to denote the parameter in the two lines. This must be done in order to avoid confusion. If the same symbol were used it would imply that the parameters in each line always have equal values, which is certainly not true. 1 1 17 13 Substituting and into (A). 3 3 3 3 This is clearly a contradiction so the equations cannot be solved simultaneously and the lines must be skew. In the example above, if the lines did meet, the point of intersection could be found by substituting one or other of the parameters into the appropriate parametric equation. Finding the equation of a plane The equation of a plane is defined in terms of a vector n normal to the plane. The vector equation of a plane is rn. d where d is a constant, determined by the position of the plane in space. If you know the normal vector and the coordinates of a point on the plane, you can easily find the value of d by substitution. The equivalent Cartesian equation of the line is n x n y n z d. 1 2 3 Example 2 (i) Write down the equation of the plane through the point (4, 5, -2) given that the 2 vector 1 is perpendicular to the plane. 3 (ii) Verify that the point (2, 4, -1) also lies on the plane. (i) The equation of the plane is n1 x n2 y n3 z d where d an 3/9
2 4 n 1 and a 5 3 2 4 2 an. 5 1 4 2 5 ( 1) ( 2) 3 8 5 6 3 2 3 All planes with these coefficients of x, y Now d an d 3 and z will be parallel... Substituting into n1 x n2 y n3 z d gives: 2x y 3z 3 2 and the vector 1 will be 3 the value of d locates the plane. perpendicular to each plane (ii) You need to verify that 2x y 3z 3 when x 2, y 4and z 1 So: 22 4 3 ( 1) 3 4 4 33 as required. For further practice in examples like the one above, use the interactive questions Vector plane equation using the normal. The equation of a plane through three points You need three points to define a plane. A plane through the points A, B and C can be defined by the vector equation: You can get to any point on a plane by moving to a point on the plane... r OA λab μac moving λ steps in the direction AB moving μ steps in the direction AC Example 3 Find the equation of the plane through A(-2, 3, 1), B(2,, -4) and C(-1, 3, 2). 2 OA 3 1 4/9
2 2 4 AB 3 3 4 1 5 1 21 AC 3 3 2 1 1 Substituting these into the general vector equation of a plane r OA λab μac : 2 4 1 r 3 λ 3 μ 1 51 The following example shows you how to convert the vector equation to a cartesian equation. Example 4 Find the cartesian equation of the plane 2 4 1 r 3 λ 3 μ 1 51 Step 1: Read across the vector equation: x 2 4 1 r y 3 λ 3 μ z 1 51 So: x 2 4λ μ y 3 3λ z 1 5λ μ Step 2: Now eliminate the parameters λ and μ between the 3 equations. From - 3 y y 3 3λ λ 3 x 2 4λ μ z 1 5λ μ x z 3 9λ So: x z 3 9λ 5/9
Substitute into : 3 y x z 3 9 x z 3 3(3 y) 3 Simplifying: x z 3 9 3y x 3y z 6 Check: Check the points A(-2, 3, 1), B(2,, -4) and C(-1, 3, 2) lie on the plane: x 3y z 6 A : 2 3 3 1 2 9 1 6 B : 2 3 ( 4) 2 4 6 C : 1 3 3 2 1 9 2 6 In practice the Cartesian equation of the plane is usually easier to use. Finding the intersection of a line and a plane You can find the point of intersection of a line and a plane by substituting the vector equation of the line into the vector equation of the plane, and solving to find the value of (unless the line is parallel to the plane, in which case there is no value of, or the line lies in the plane, in which case there are an infinite number of possible values of ). You can then substitute this value of into the equation of the line to find the coordinates of the point of intersection. The following example shows you how to find the intersection of a line and a plane. Example 5 Find the point of intersection of the line 2x 3y z 6 2 1 r 3 λ 2 and the plane 4 3 The general point on the line is: x 2 1 r y 3 λ 2 z 4 3 So reading across: x 2 λ y 3 2λ z 43λ 6/9
Substitute these into the equation of the plane 2x 3y z 6 : 2(2 λ) 3( 3 2 λ) (4 3 λ) 6 Simplifying: 4 2λ 9 6λ 4 3λ 6 17 11λ 6 11λ 11 λ 1 Now substitute λ 1 into the equation of the line to find the position vector of the point of intersection. x 2 1 1 r y 3 1 2 1 z 4 3 1 So the coordinates of the point of intersection are: (1, -1, 1) Check this point lies on the plane 2x 3y z 6 : 21 3 ( 1) 1 2 3 1 6 as required.. Finding the distance of a point from a plane The shortest distance of a point A to a plane is the distance AP where AP is a line perpendicular to the plane and P is a point on the plane. To do this: Step 1: Find the equation of the line through A perpendicular to the plane Step 2: Find the point of intersection, P, of the line and the plane Step 3: Find the distance AP. The following example shows you how to do this. Example 6 Find the distance of the point A(-1, 2, 4) from the plane 3x 2y z 11. Step 1: The direction vector perpendicular to the plane is: So the equation of the line through A(-1, 2, 4) is: 1 3 r 2 λ 2 4 1 Step 2: The general point on the line is: 3 2 1 The coefficients of x, y and z in the equation of the plane. 7/9
x 1 3 r y 2 λ 2 z 4 1 So reading across: x 1 3λ y 22λ z 4 λ Substitute these into the equation of the plane 3x 2y z 11: 3( 1 3 λ) 2(2 2 λ) (4 λ) 11 Simplifying: 3 9λ 4 4λ 4 λ 11 3 14λ 11 14λ 14 λ 1 Now substitute λ 1 into the equation of the line to find the position vector of the point of intersection. x 1 3 2 r y 2 1 2 z 4 1 5 So the coordinates of the point of intersection are: P(2,, 5) Step 3: 2 1 3 AP 2 2 5 4 1 2 2 2 AP 3 ( 2) 1 9 4 1 14 So the distance of the point A to the plane is 14 units. This method of finding the distance of a point from a plane can be generalised to give the formula below: The distance of a point (x 1, y 1, z 1 ) from a plane ax + by + cz + d = is given by the formula ax by cz d 1 1 1 a b c 2 2 2 This formula is given in your formula book, but it is easy to learn. Example 7 shows this formula being applied. Example 7 Find the distance of the point (1, -2, 4) from the plane 2x y 3z 4. 8/9
The equation of the plane can be written as 2x y 3z 4, so a = 2, b = 1, c = -3 and d = -4. 2 2 12 3 4 4 Distance of point from plane 2 2 2 2 1 3 14 14 14 The angle between a line and a plane The angle between a line and a plane is the complement of the angle between the line and the normal to the plane (i.e. the angle between a line and a plane = 9 - the angle between the line and the normal to the plane). You can find the cosine of the angle between the normal and the line by using the scalar product: For the line r a b and the plane rn. d, the acute angle between the direction vector of the line, b, and the normal vector to the plane, n, is given by. cos bn bn and so the acute angle between the line and the plane is given by. sin bn bn The angle between two planes The acute angle between two planes is equal to the acute angle between the normals to the planes. This means that the angle between two planes can be found quite easily using the scalar product. 1. 2 cos nn n n 1 2 9/9