Equations Involving Lines and Planes Standard equations for lines in space

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1 Equations Involving Lines and Planes In this section we will collect various important formulas regarding equations of lines and planes in three dimensional space Reminder regarding notation: any quantity written with an arrow over the top such as v or in angle brackets such as v 1, v 2, v 3 indicates a vector (we may also refer to this vector using the notation v 1 i + v 2 j + v 3 k) Any quantity written in plain notation like v or in parenthesis such as (v 1, v 2, v 3 ) indicates a scalar or a point (unless otherwise noted) A quantity with a bar over the top, like v, indicates a line segment Standard equations for lines in space We can completely describe a line by specifying (1) a point through which the line passes, and (2) the line s direction In two dimensions, the line s direction is given by its slope In three dimensions, we will specify the line s direction using a vector parallel to the line Suppose that the line L passes through the point r 0 = (x 0, y 0, z 0 ) parallel to vector v = a, b, c (with terminal point v = (a, b, c)) We can move v so that it lies along the line by adding r 0 to vs initial and terminal points; the vector with initial point r 0 and terminal point r 0 + v = (a + x 0, b + y 0, c + z 0 ) lies on L: By stretching or shrinking v and adding this vector to r 0, we can reach any point on L Since stretching and shrinking a vector corresponds to scaling the vector (multiplying the vector by a scalar), any point on the line L may be written as r 0 + tv for some scalar t These observations give us a way to write an equation for the line: 1

2 Vector equation for a line: If the line L passes through the point r 0 = (x 0, y 0, z 0 ) (with position vector r 0 = x 0, y 0, z 0 ) parallel to the vector v = a, b, c, then the line has vector equation r(t) = r 0 + t v, where < t < The numbers a, b, and c are called the direction numbers of the line L Notice that r(t) is a vector; so the vector equation for L actually gives us position vectors whose terminal points lie on L We may also describe the line using the components x, y, and z: Parametric equations for a line: If the line L passes through the point r 0 = (x 0, y 0, z 0 ) (with position vector r 0 = x 0, y 0, z 0 ) parallel to the vector v = a, b, c, then the line may be described by the equations where < t < x = x 0 + ta, y = y 0 + tb, z = z 0 + tc, Another method for specifying the line L is by using symmetric equations If L has parametric equations x = x 0 + ta, y = y 0 + tb, and z = z 0 + tc, then by solving for t in each equation, we have t = x x 0 a, t = y y 0 b, and t = z z 0, c assuming that each of a, b, or c is nonzero Since t is the same in each equation, we have the following description of L: Symmetric equations for a line: If the line L passes through the point r 0 = (x 0, y 0, z 0 ) (with position vector r 0 = x 0, y 0, z 0 ) parallel to the vector v = a, b, c, then the line may be described by the equations x x 0 a = y y 0 b = z z 0 c Example: Find the vector equation and parametric equations for the line passing through the points p 1 = (1, 3, 1) and p 2 = (7, 3, 6) Then paramaterize the line segment p 1 p 2 The two points define a vector p 2 p 1 = 7 1, 3 3, = 6, 6, 5 which is parallel to the line passing through p 1 and p 2 So a vector equation for the line is r(t) = 1, 3, 1 + t 6, 6, 5, < t < Note that we could have written the equation for the line in many different ways; 1, 3, 1 + s 6, 6, 5, < s < 7, 3, 6 + p 6, 6, 5, < p < 2

3 and 7, 3, 6 + q 6, 6, 5, < q < are all equivalent vector equations for this line Parametric equations for the line are given by x = 1 + 6t, y = 3 6t, z = 1 5t, < t < Again, there are many different ways to write equivalent conditions for the line To parameterize the segment from p 1 to p 2, we may use the same parametric equations for the line itself, but we must put bounds on t so that we only describe points on the segment from p 1 to p 2 We need to determine the smallest allowable value for t (which will give us the point p 1 ), and the largest value for t, which will give us p 2 Using the x coordinate of the point p 1, if 1 + 6t = 1, then t = 0 It is easy to see that plugging 0 into each coordinate returns the point p 1 So the smallest value for t that we can allow is t = 0 Using the x coordinate of p 2, suppose that 1+6t = 7 Then t = 1 (and again we see that plugging 1 into each equation will indeed return the point p 2 ) So the equations for the line segment p 1 p 2 are x = 1 + 6t, y = 3 6t, z = 1 5t, 0 < t < 1 Distance from a point to a line To measure the distance from a point p 0 to a line L, we look for the point l on L closest to p 0, then determine the distance from l to p 0 In particular, the segment joining l and p 0 should be perpendicular to L In practice, given any point p on L, the process outlined above can be streamlined using some geometry We want to find the length of the side opposite θ in the right triangle below: 3

4 The length of the hypotenuse is pp 0, so the length of the sides are pp 0 sin θ and pp 0 cos θ In particular, the opposite side has length pp 0 sin θ Recall that the length of the cross product of pp 0 with v is pp 0 v = pp 0 v sin θ Rewriting the above equation as pp 0 v = pp 0 sin θ v gives us a method for finding the length pp 0 sin θ: we can instead calculate pp 0 v v Theorem 001 If the point p lies on the line L, p 0 is any point not on L, and v is a vector parallel to L, then the distance from p 0 to L is given by pp 0 v v Example: Determine the distance from the point p 0 = ( 2, 4, 1) to the line L with vector equation r(t) = 3, 0, 0 + t 4, 1, 1 In order to use the formula pp 0 v, v 4

5 we must know a point p on the line L and a vector v parallel to L Fortunately, it is easy to find such a v Since L has vector equation the vector r(t) = 3, 0, 0 + t 4, 1, 1, v = 4, 1, 1 is parallel to L We can find a point on L by plugging a value for t into t = 0 Then r(0) = 3, 0, , 1, 1 = 3, 0, 0 is a vector whose terminal point lies on L So the point p = (3, 0, 0) is on L The vector pp 0 is given by pp 0 = 5, 4, 1 So to determine the distance from p 0 to L, we must calculate pp 0 v 5, 4, 1 4, 1, 1 = v 4, 1, 1 5, 4, 1 4, 1, 1 = = 5 i + j + 21 k 18 r(t); for ease of computation, let s let = = units Equation for a plane In order to completely describe a plane P, we must specify (1) a point p 0 that lies on the plane, as well as (2) the tilt of the plane We talk about the plane s tilt or direction using any vector n normal (ie orthogonal) to the plane Notice that, if p 0 is a point on the plane P with normal vector n, then 5

6 every vector that lies on the plane P is orthogonal to n, and the point p lies on the plane P if and only if the vector pp 0 is orthogonal to n Since vectors are orthogonal if and only if their dot product is 0, we may reformulate the latter idea: point p lies on the plane P if and only if pp 0 n = 0 Let us make the arguments more precise Suppose that a normal vector of the plane P is given by n = a i + b j + c k, and that the point r 0 = (x 0, y 0, z 0 ) lies on P With r = (x, y, z), the dot product rr 0 n is given by rr 0 n = x x 0, y y 0, z z 0 a, b, c = a(x x 0 ) + b(y y 0 ) + c(z z 0 ) Then the point r lies on P if and only if rr 0 n = a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0 So P is the set of all points (x, y, z) so that rr 0 n = n 1 (x x 0 ) + n 2 (y y 0 ) + n 3 (z z 0 ) = 0 Theorem 002 If P is a plane passing through the point r 0 = (x 0, y 0, z 0 ) with normal vector n = a i + b j + c k, then P is the set of all points r = (x, y, z) so that (vector equation) n rr 0 = 0 (scalar equation) a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0 (modified scalar equation) ax + by + cz = ax 0 + by 0 + cz 0 Examples Find the equation for the plane P passing through the point p = (2, 1, 0) and parallel to the plane P 0 defined by the equation 3x 7y + 4z = 12 Since the plane we want is parallel to the plane 3x 7y + 4z = 12, the two planes have the same normal vector Now P 0 has normal vector 3, 7, 4, so using the form a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0, P is defined by 3(x x 0 ) 7(y y 0 ) + 4(z z 0 ) = 0, 6

7 or 3(x 2) 7(y 1) + 4(z 0) = 0; we may rewrite this equation as 3x 7y + 4z = 1 Find the equation for the plane P passing through the points (4, 0, 3) and (2, 1, 1), and parallel to the line L with parametric equations x = 1 + t, y = 3 3t, z = 12t In order to use the equation a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0, we need to find a vector normal to P The plane must be parallel to the line L, which in turn is parallel to the vector v = 1, 3, 12 Since the cross product of a pair of (nonparallel) vectors is normal to both of the original vectors, the cross product of v with any other (nonparallel) vector on P will be a vector normal to P So we need to find another vector on P Since we know two points on P, this is easy to accomplish: u = 2, 1, 2 is parallel to P Then v and u have cross product v u = 1, 3, 12 2, 1, 2 = 6 i + 22 j + 5 k, and this vector is normal to P So we may write the equation for P as 6(x 4) + 22(y 0) + 5(z 3) = 0, or 6x + 22y + 5z = 39 Lines of Intersection A pair of lines is parallel if they have the same slope A pair of planes is parallel if they have the same tilt In 3 dimensions, we measure the tilt of the plane using vectors normal to the plane; so if planes N and M are parallel, there normal vectors must also be parallel If planes N and M are not parallel, then the planes intersect in a line L that lies on both planes; thus L must be perpendicular to the normal vectors of each plane Example Find an equation for the line of intersection of the planes 3x 4y+12z = 14 and x+2y z = 3 Recall that, if we are to write an equation for a line L in 3 dimensional space, we must find a point p on the line and a vector v parallel to the line The vector v is not difficult to find; L is perpendicular to the normal vectors of each plane, so the cross product of the normal vectors will be a vector parallel to L 7

8 We calculate v using the cross product formula: i j k v = = i(4 24) j( ) + k(6 4) = 20 i 9 j + 2 k Now that we have a vector parallel to the line of intersection of the two planes, we need to find a point on the line of intersection A point on the line of intersection is a point (x 0, y 0, z 0 ) that lies on both planes In other words, this point (x 0, y 0, z 0 ) should satisfy both equations 3x 4y + 12z = 14 and x + 2y z = 3, ie 3x 0 4y z 0 = 14 and x 0 + 2y 0 z 0 = 3 We currently have two equations with three unknowns, but we can reduce the problem to one we can solve (two equations with two unknowns) by making a choice for one variable Since the x value for the line of intersection is clearly not constant, there is some point on the line where x 0 = 0 Let s determine the values for y 0 and z 0 where this occurs If x 0 = 0, we have 4y z 0 = 14 and 2y 0 z 0 = 3 Let s solve this system of equations: Multiply the second line by 2: Now add the two equations to eliminate y 0 : 4y z 0 = 14 2y 0 z 0 = 3 4y z 0 = 14 4y 0 2z 0 = 6 4y z 0 = y 0 2z 0 = 6 10z 0 = 20 So z 0 = 2 Substituting 2 into the second equation and solving for y 0, we have 2y 0 2 = 3 2y 0 = 5 y 0 = 5 2 So the point ( 0, 5 2, 2) is on the line of intersection Thus the parametric equations for L are given by x = 20t, y = 5 9t, z = 2 + 2t 2 Distance from a point to a plane 8

9 To measure the distance from a point p to a plane P, we will compare the point to the plane s normal vector n: if the initial point of n is p 0, then we may measure the distance from the point p to the plane along the normal vector n by projecting the vector p 0 p onto n Recall from 123 that the length of vector projection of p 0 p onto n is given by proj n p 0 p = n p 0p n So the distance from p to the plane with point p 0 and normal vector n is n p 0 p n Example Find the distance from the point p = ( 2, 1, 5) to the plane P with equation 5x 12y + z = 7 P has normal vector n = 5, 12, 1 ; we will also need a point on the plane; p 0 = (0, 0, 7) will work nicely Finally we need to find the length of the projection of the vector p 0 p = 2, 1, 2 onto n So the distance is given by n p 0 p p 0 p = = 5, 12, 1 2, 1, = Angles between planes The easiest way to find the angle of intersection of two planes P 1 and P 2 is by noting that the angle of intersection of their normal vectors n 1 and n 2 is precisely the same as the angle of intersection of the planes 9

10 In particular, the angle θ between vectors n 1 and n 2 can be calculated using the dot product From section 123, we know that angle θ between vectors n 1 and n 2 is ( ) θ = cos 1 n1 n 2 n 1 n 2 Theorem 003 If planes P 1 and P 2 have normal vectors n 1 and n 2 respectively, then the angle θ between P 1 and P 2 is given by ( ) θ = cos 1 n1 n 2 n 1 n 2 10

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