Overview Yesterda we introduced equations to describe lines and planes in R 3 : r + tv The vector equation for a line describes arbitrar points r in terms of a specific point and the direction vector v. n (r ) 0 The vector equation for a plane describes arbitrar points r in terms of a specific point and the normal vector n. Question How can we find the distance between a point and a plane in R 3? Between two lines in R 3? Between two planes? Between a plane and a line? (From Stewart 10.5) Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 1 / 17 Distances in R 3 The distance between two points is the length of the line segment connecting them. However, there s more than one line segment from a point P to a line L, so what do we mean b the distance between them? The distance between an two subsets A, B of R 3 is the smallest distance between points a and b, where a is in A and b is in B. To determine the distance between a point P and a line L, we need to find the point Q on L which is closest to P, and then measure the length of the line segment PQ. This line segment is orthogonal to L. To determine the distance between a point P and a plane S, we need to find the point Q on S which is closest to P, and then measture the length of the line segment PQ. Again, this line segment is orthogonal to S. In both cases, the ke to computing these distances is drawing a picture and using one of the vector product identitites. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 2 / 17 Distance from a point to a plane We find a formula for the distance s from a point P 1 ( 1, 1, 1 ) to the plane A + B + C + D 0. P 1 s n b r P 0 Let P 0 ( 0, 0, 0 ) be an point in the given plane and let b be the vector corresponding to P 0 P 1. Then b 1 0, 1 0, 1 0. The distance s from P 1 to the plane is equal to the absolute value of the scalar projection of b onto the normal vector n A, B, C. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 3 / 17
s comp n b n b n A( 1 0 ) + B( 1 0 ) + C( 1 0 ) A 1 + B 1 + C 1 (A 0 + B 0 + C 0 ) Since P 0 is on the plane, its coordinates satisf the equation of the plane and so we have A 0 + B 0 + C 0 + D 0. Thus the formula for s can be written s A 1 + B 1 + C 1 + D Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 4 / 17 Eample 1 We find the distance from the point (1, 2, 0) to the plane 3 4 5 2 0. From the result above, the distance s is given b where ( 0, 0, 0 ) (1, 2, 0), s A 0 + B 0 + C 0 + D A 3, B 4, C 5 and D 2. This gives s 3 1 + ( 4) 2 + ( 5) 0 2 3 2 + ( 4) 2 + ( 5) 2 7 7 50 5 2 7 2 10. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 5 / 17 Distance from a point to a line Question Given a point P 0 ( 0, 0, 0 ) and a line L in R 3, what is the distance from P 0 to L? Tools: describe L using vectors u v u v sin θ Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 6 / 17
Distance from a point to a line Let P 0 ( 0, 0, 0 ) and let L be the line through P 1 and parallel to the nonero vector v. Let and r 1 be the position vectors of P 0 and P 1 respectivel. P 2 on L is the point closest to P 0 if and onl if the vector P 2 P 0 is perpendicular to L. vvvv L v P P 2 1 θ -r r 1 s 1 P 0 The distance from P 0 to L is given b s P 2 P 0 P 1 P 0 sin θ r 1 sin θ where θ is the angle between r 1 and v Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 7 / 17 Since we get the formula ( r 1 ) v r 1 v sin θ s r 1 sin θ ( r 1 ) v v Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 8 / 17 Eample 2 Find the distance from the point (1, 1, 1) to the line of intersection of the planes + + 1, 2 5 1. The direction of the line is given b v n 1 n 2 where n 1 i + j + k, and n 2 2i j 5k. v n 1 n 2 4i + 7j 3k. P 1 (1,-1/4,1/4) v P 2 -r 1 s P 0 (1,1,-1) In the diagram, P 1 is an arbitrar point on the line. To find such a point, put 1 in the first equation. This gives which can be used in the second equation to find 1/4, and hence 1/4. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 9 / 17
Here P 1 P 0 r 1 5 4 j 5 4k. So s ( r 1 ) v v ( 5 4 j 5 4k) ( 4i + 7j 3k) ( 4) 2 + 7 2 + ( 3) 2 5i + 5j + 5k 74 75 74. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 10 / 17 Distance between two lines Let L 1 and L 2 be two lines in R 3 such that - L 1 passes through the point P 1 and is parallel to the vector v 1 - L 2 passes through the point P 2 and is parallel to the vector v 2. Let r 1 and r 2 be the position vectors of P 1 and P 2 respectivel. Then parametric equation for these lines are L 1 r r 1 + tv 1 L 2 r r 2 + sv 2 Note that r 2 r 1 P 1 P 2. We want to compute the smallest distance d (simpl called the distance) between the two lines. If the two lines intersect, then d 0. If the two lines do not intersect we can distinguish two cases. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 11 / 17 Case 1: L 1 and L 2 are parallel and do not intersect. In this case the distance d is simpl the distance from the point P 2 to the line L 1 and is given b d P 1 P 2 v 1 (r 2 r 1 ) v 1 v 1 v 1 Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 12 / 17
Case 2: L 1 and L 2 are skew lines. If P 3 and P 4 (with position vectors r 3 and r 4 respectivel) are the points on L 1 and L 2 that are closest to one another, then the vector P 3 P 4 is perpendicular to both lines (i.e. to both v 1 and v 2 ) and therefore parallel to v 1 v 2. The distance d is the length of P 3 P 4. Now P 3 P 4 r 4 r 3 is the vector projection of P 1 P 2 r 2 r 1 along v 1 v 2. Thus the distance d is the absolute value of the scalar projection of r 2 r 1 along v 1 v 2 d r 4 r 3 (r 2 r 1 ) (v 1 v 2 ) v 1 v 2 Observe that if the two lines are parallel then v 1 and v 2 are proportional and thus v 1 v 2 0 (the ero vector) and the above formula does not make sense. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 13 / 17 Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 15 / 17 Eample 3 Find the distance between the skew lines { + 2 3 + 2 3 and { + + 6 2 5 vvvv v P 1 P 3 1 L 1 r 2 -r 1 L 2 P 4 v 2 P 2 v 1 v 2 We can take P 1 (1, 1, 1), a point on the first line, and P 2 (1, 2, 3) a point on the second line. This gives r 2 r 1 j + 2k. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 16 / 17
Now we need to find v 1 and v 2 : and This gives v 1 (i + 2j) (j + 2k) 4i 2j + k, v 2 (i + j + k) (i 2k) 2i + 3j k. v 1 v 2 i + 2j + 8k. The required distance d is the length of the projection of r 2 r 1 in the direction of v 1 v 2, and is given b d (r 2 r 1 ) (v 1 v 2 ) v 1 v 2 (j + 2k) ( i + 2j + 8k) ( 1) 2 + 2 2 + 8 2 18 69. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 17 / 17