Vectors. Section 3: Using the vector product

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1 Vectors Section 3: Using the vector product Notes and Examples These notes contain subsections on Using the vector product in finding the equation of a plane The intersection of two planes The distance of a point from a line The distance between two skew lines Using the vector product in finding the equation of a plane If you want to find the equation of a plane in the scalar product or Cartesian form, you need to find a vector perpendicular to the plane Often the information you will have will be about points or lines that lie in the plane The vector product is very useful in situations like these If you know two vectors which lie in the plane, then the vector product of these vectors is a vector perpendicular to the plane So if you are given three points that lie within a plane, you can find two vectors in the plane by subtraction, and then use these to find a vector perpendicular to the plane This method is shown in Example 441 on page 284 in the textbook The intersection of two planes The line of intersection of two planes can be found in several different ways One approach uses the vector product The line of intersection of two planes obviously lies in both planes, so it is perpendicular to the normals of both planes Therefore, finding the vector product of the two normals gives the direction vector for the line of intersection Once you know this, you can use any point in both planes to find the equation of the line Example 1 Find the vector equation of the line of intersection of the planes: x 2y z 2 and 3x y 2z 1 Solution The normal vectors to the plane are and /5

2 The direction vector of the line of intersection Choosing x = 1 to find a point on both planes gives 2yz 1 y 2z 2 Solving simultaneously gives y = 0 and z = The equation of the line is r You can check that this equation is correct by using it to give the coordinates of any two points on the line and showing that these points each lie on both planes eg 1gives the point 4, 5, 8 and 0 gives the point 1, 0, 1 Check for yourself that these points satisfy both plane equations Why do you need two points to be sure your equation is correct? The distance of a point from a line The formula for the distance of a point P with position vector p from a line with equation r = a + d is ( p a) d ˆ or ( p a) d d The explanation of this formula is given on pages of the textbook If you find this difficult to follow at first, don t worry When you have done the exercise, using the formula above, go back and re-read the text: it is sometimes easier to understand something after you have used it in practice This formula is not given in your formula book, so you need to learn it Example 2 shows how this formula is applied Example 2 Find the distance of the point (2, 1, -2) from the line x 1 y 1 z /5

3 Solution 1 2 The equation of the line can be written as r p a ˆ 1 1 d1 d ˆ p a d ˆ p a d Note that the shortest distance of a point from a line is the length of the perpendicular line from the point to the line M r = a + d P The distance between two skew lines The shortest distance between two skew lines r = a + d and r = b + e is given by ( a b)( de) de The explanation for this formula is given on pages of the textbook As before, if you find this difficult at first, come back to it when you have worked through the exercise You need to learn this formula, as it is not given in your formula book 3/5

4 Example 3 shows how this formula is applied Example 3 Find the shortest distance between the two skew lines x 1 y 1 z 2 and x 3 y 2 z Solution a 1 and b 2 so ab d 1 and e 2 so de a b de ( 2 1 9) a bde de If you want to find the points on the two lines closest to each other (the ends of the shortest line connecting the two skew lines), there are several possible approaches, based on the fact that the shortest line connecting two skew lines is perpendicular to both lines Example 4 shows two possible approaches, using the lines from Example 3 Example 4 x 1 y 1 z 2 Find the points A on the line L, given by, and B on the line M, given by x y z, such that the distance AB is as short as possible Solution 1 A general point P on line L is (1 + 2p, -1 + p, -2 + p) A general point Q on line M is (3 + q, -2 2q, 7 + 3q) 3 q 1 2 p 2 q 2 p PQ 2 2q 1 p 1 2q p 7 3q 2 p 9 3q p 4/5

5 2 q2 p 2 PQ is perpendicular to L 1 2 q p q p 1 4 2q 4 p 1 2q p 9 3q p 0 3q 6 p12 0 q 2p 4 2 q2 p 1 PQ is perpendicular to M 1 2 q p q p 3 2 q 2 p 2 4q 2 p 27 9q 3p 0 14q 3p 31 0 Substituting q = 2p 4 into 14q 3p + 31 = 0: 14(2 p 4) 3p p 56 3p p 25 p1, q2 The point A is (3, 0, -1) and the point B is (1, 2, 1) Solution The direction of the vector AB is de Point A lies on line L so has coordinates (1 + 2, -1 +, -2 + ) The equation of the line AB can be written as r 1 t Line AB meets line M at the point B, so t 3 2 t 2 1 t t t t t 9 Solving these three simultaneous equations in, t and gives 1, 2, t 2 So A = (3, 0, -1) and B = (1, 2, 1) Note that the distance AB is given by , the same shortest distance as found in Example 3, as you would expect! 5/5