Overview Yesterday we introduced equations to describe lines and - PowerPoint PPT Presentation

Overview Yesterday we introduced equations to describe lines and planes in R 3 : r = r 0 + t v The vector equation for a line describes arbitrary points r in terms of a specific point r 0 and the direction vector v . n · ( r − r 0 ) = 0 The vector equation for a plane describes arbitrary points r in terms of a specific point r 0 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 by the distance between them? Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 2 / 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 by the distance between them? The distance between any 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 . Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 2 / 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 by the distance between them? The distance between any 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 . Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 2 / 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 by the distance between them? The distance between any 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 key 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 = ( x 1 , y 1 , z 1 ) to the plane Ax + By + Cz + D = 0. P 1 z s n b P 0 r x y Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 3 / 17

Distance from a point to a plane We find a formula for the distance s from a point P 1 = ( x 1 , y 1 , z 1 ) to the plane Ax + By + Cz + D = 0. P 1 z s n b P 0 r x y Let P 0 = ( x 0 , y 0 , z 0 ) be any point in the given plane and let b be the � vector corresponding to P 0 P 1 . Then b = � x 1 − x 0 , y 1 − y 0 , z 1 − z 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

The distance s from P 1 to the plane is equal to the absolute value of the scalar projection of b = � x 1 − x 0 , y 1 − y 0 , z 1 − z 0 � onto the normal vector n = � A , B , C � . = | comp n b | s | n · b | = || n || | A ( x 1 − x 0 ) + B ( y 1 − y 0 ) + C ( z 1 − z 0 ) | = √ A 2 + B 2 + C 2 | Ax 1 + By 1 + Cz 1 − ( Ax 0 + By 0 + Cz 0 ) | √ = A 2 + B 2 + C 2 Since P 0 is on the plane, its coordinates satisfy the equation of the plane and so we have Ax 0 + By 0 + Cz 0 + D = 0. Thus the formula for s can be written s = | Ax 1 + By 1 + Cz 1 + D | √ A 2 + B 2 + C 2 Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 4 / 17

Example 1 We find the distance from the point (1 , 2 , 0) to the plane 3 x − 4 y − 5 z − 2 = 0. From the result above, the distance s is given by s = | Ax 0 + By 0 + Cz 0 + D | √ A 2 + B 2 + C 2 where ( x 0 , y 0 , z 0 ) = (1 , 2 , 0), A = 3 , B = − 4 , C = − 5 and D = − 2 . This gives | 3 · 1 + ( − 4) · 2 + ( − 5) · 0 − 2 | s = � 3 2 + ( − 4) 2 + ( − 5) 2 √ 7 7 2 = 7 2 = √ 50 = √ 10 . 5 Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 5 / 17

Distance from a point to a line Question Given a point P 0 = ( x 0 , y 0 , z 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 = ( x 0 , y 0 , z 0 ) and let L be the line through P 1 and parallel to the nonzero vector v . Let r 0 and r 1 be the position vectors of P 0 and P 1 respectively. P 2 on L is the point closest to P 0 if and only if the vector − − − → P 2 P 0 is perpendicular to L . vvvv z P 2 v P 1 θ ℒ r0-r1 s r1 P 0 r0 x y The distance from P 0 to L is given by s = ||− P 2 P 0 || = ||− − − → − − → P 1 P 0 || sin θ = || r 0 − r 1 || sin θ where θ is the angle between r 0 − r 1 and v Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 7 / 17

vvvv z v P 2 P 1 θ ℒ r0-r1 s r1 P 0 r0 x y s = || r 0 − r 1 || sin θ Since || ( r 0 − r 1 ) × v || = || r 0 − r 1 || || v || sin θ we get the formula s = || r 0 − r 1 || sin θ = || ( r 0 − r 1 ) × v || || v || Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 8 / 17

Example 2 Find the distance from the point (1 , 1 , − 1) to the line of intersection of the planes x + y + z = 1 , 2 x − y − 5 z = 1 . The direction of the line is given by v = n 1 × n 2 where n 1 = i + j + k , and n 2 = 2 i − j − 5 k . v = n 1 × n 2 = − 4 i + 7 j − 3 k . z P 1=(1,-1/4,1/4) P 2 v r0-r1 s P 0=(1,1,-1) x y In the diagram, P 1 is an arbitrary point on the line. To find such a point, put x = 1 in the first equation. This gives y = − z which can be used in the second equation to find z = 1 / 4, and hence y = − 1 / 4. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 9 / 17

z P 1=(1,-1/4,1/4) P 2 v Here − − − → P 1 P 0 = r 0 − r 1 = 5 4 j − 5 r0-r1 4 k . So s || ( r 0 − r 1 ) × v || P 0=(1,1,-1) x s = || v || y || ( 5 4 j − 5 4 k ) × ( − 4 i + 7 j − 3 k ) || = � ( − 4) 2 + 7 2 + ( − 3) 2 || 5 i + 5 j + 5 k || = √ 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 respectively. Then parametric equation for these lines are r = r 1 + t v 1 L 1 L 2 ˜ r = r 2 + s v 2 Note that r 2 − r 1 = − − − → P 1 P 2 . We want to compute the smallest distance d (simply 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 simply the distance from the point P 2 to the line L 1 and is given by 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 respectively) 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 . Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 13 / 17

Case 2: L 1 and L 2 are skew lines. 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 . Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 13 / 17

Case 2: L 1 and L 2 are skew lines. 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 || Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 13 / 17

Case 2: L 1 and L 2 are skew lines. 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 || Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 13 / 17