Even More Vectors

Scalar Equation of Lines in the Plane The standard form of the equation of a line in the plane, Ax+By+C=0, can be developed using one point on the line and a slope. This form is also known as a scalar equation of the line or as a Cartesian equation of the line. An alternative development of this equation makes use of vector concepts.

Perpendicular vectors A normal (vector) to a line l is a vector n which is perpendicular to the line. u l u

Understanding Develop a scalar equation of the line through the point P0(4, -1) with normal n=(3,5) We first pick any point P(x,y) on the line Because P0 and P are on both on the line, we can determine the vector that points in the direction of the line. This tells us that 3x+5y-7=0 is a scalar equation of the line through the point PO(4,-1) with normal n=(3,5)

Scalar Equations

Understanding Find a scalar equation of the line through the point P0(4, -1) with normal n=(3,5) Because we are given the normal, the scalar equation is in the form: 3x+5y+C=0 To determine C, we need only substitute the coordinates of a point on the line: P0(4, -1) Therefore, 3x+5y-7=0 is a scalar equation of the line.

Understanding Find a normal vector for each line: 3x+2y=7 (x-2, y+4)=r(3,-1) a) Placing 3x+2y=7 in the form Ax+By+C=0: Then : b) (3, -1) is a direction vector: Then : Therefore we can choose n1=1 and n2=3 or

Understanding Rewrite the equation (x,y)=(2,-5)+t(2,-3) as a scalar equation for the line The normal: The equation: or or

Understanding Determine the vector equation of a line that is perpendicular to 2x-7y+5=0 and has the same y-intercept as the line (x,y)=(2,-5)+t(2,-3). Vector Equation: Direction vector: Point: (find t that makes x=0) (x,y)=(0,-2)+t(2,-7)

Intersection of Line One method for determining the intersection, if any, of two lines in the plane involves solving a system to two scalar equations. However, the other forms of equations of lines in the plane and in space suggest alternative techniques for investigating intersections.

Understanding Investigate the intersection of the lines l1 and l2 for: Write in parametric form Solve using any 2 equations and verify with the 3rd Since the direction vectors: (3, 2, 6) for l1 and (1, -5, -1) for l2, the lines are not parallel, thus any point of intersection has the property: Substitute into either l1 on l2

Understanding Investigate the point of intersection of the lines l1 and l2 for: Direction vectors: l1=(1,3,7) Therefore these lines are not parallel l2=(5,-4,-2) Parametric: Therefore Therefore these lines do not intersect. These are skew lines Solving (1) and (2) Sub into (3)

Distance between two points y A yA B yB xA xB O x

Distance between Point and Line The distance from point Q(x1,y1) to line Ax+By+C=0 P0 v Q Q’ = P0 +tv l Recall: Since Po(xo,yo) is on the line it satisfies the equation so Axo+By0+C=0 therefore C=-Axo-Byo Now:

Understanding Find the distance from the point Q(2, -3) to the line 4x+5y-6=0

Try writing all vectors as 3D vectors d(-5, 4, 0) Distance in Space P0 v A Q’ = P0 +tv l Find the distance from the point A(1,2,3) to the line l where l is given by l: (x,y,z)= (0,1,5) + t(3,4,1) Since Po(0, 1, 5) lies on the line. Vector PoA is (1-0, 2-1, 3-5)=(1, 1, -2) Since Then Therefore Try writing all vectors as 3D vectors d(-5, 4, 0) Why did it not work? Try it Can you use this technique for 2D lines as in the previous example?

Understanding P0 p A Q’ l Find the distance from the point A(1,2,3) to the line l where l is given by l: (x,y,z)= (0,1,5) + t(3,4,1) D We need the length of the projection of line segment APO on the direction vector d=(3,4,1). Now apply the Pythagorean theorem (0-1,1-2,5-3)

Distance between two lines in 3D Q1 P1 u d P2 v l2 Q2 The distance is attained between two points Q1 and Q2 so that (Q1 – Q2)  u and (Q1 – Q2)  v

Distance between two lines in 3D Q1 P1 u d v P2 l2 Q2 So we will project the position vector Q1Q2 onto a vector  to both u and v This will give us the shortest distance between both lines

Understanding For the lines l1 and l2 given by the following equations l1: x-1 = y = z l2: x/2 = 1-y = z Show that l1 and l2 are skew lines Find the distance between l1 and l2 These are not parallel, so the lines are not parallel Check the direction vectors: Determine a point of intersection No solution Since the lines do not intersect, these are skew lines

Understanding For the lines l1 and l2 given by the following equations l1: x-1 = y = z l2: x/2 = 1-y = z Show that l1 and l2 are skew lines Find the distance between l1 and l2 Point has to satisfy equation Check the direction vectors: Select a point on each line Determine the normal vector Determine distance Find A1A2

Equations of Planes In order to determine a plane it is enough to specify either of the following sets of information: Three non-collinear points on the plane One point (fix location)on the plane and two-non-collinear direction vectors (defines plane’s slant) 

Understanding Develop a vector equation for the plane through the point P0(-3,5,0) with direction vectors d1=(1,2, -1) and d2=(3, -1,4) Pick a point P on the plane: Now apply: Position vector form Parametric form

Understanding Find a vector and parametric equations of the plane through points A(1,7,2), B(4, 0, -1), and C(1, 2, 3) One direction vector of the plane is: Another direction vector is: Using A as the point: Parametric:

Understanding Find three points on each plane. The plane with vector equation : The plane with parametric equation: Each point on the plane corresponds to a pair of values for each of the parameters a) (0,4,2)+1(5,-2,3)+1(1,0,1)=(6,2,6) (0,4,2)+1(5,-2,3)+0(1,0,1)=(5,2,5) (0,4,2) -1(5,-2,3)+3(1,0,1)=(-2,6,2) b) 4=1+4(0)+3(1) 1=-3(0)+1(1) 1=1-2(0) 8=1+4(1)+3(1) -2=-3(1)+1(1) -1=1-2(1) -3=1+4(-1)+3(0) 3=-3(-1)+1(0) 3=1-2(-1)

Normal Vector to a Plane It becomes useful to be able to distinguish between different equations of the same plane and equations of different planes. By recognizing that each point on the plane must correspond to a value of the parameter, it is possible to compare such equations to decide whether they describe to same plane. First we determine whether or not the planes are parallel. A normal (vector) to a plane is a vector n that is perpendicular to every vector in the plane This is not as difficult as might first appear to check that a given vector is normal to a specific plane. In fact, if a vector n is perpendicular to any two independent vectors in a plane then it is normal to the plane

Understanding Determine whether or not both equations describe the same plane To decide where planes π1 and π2 are parallel, we consider n1, a normal to π1, and n2, a normal to π2. We obtain these normals by finding the cross product of the direction vectors of each plane Because these normals are collinear, the planes may be identical or parallel. All we need to do is check that a point on one plane is on the other plane. Let r=s=0, then the point (1,6,0) is on π1, We need to check if this point is on π2

Understanding The point (1,6,0) is on π1,Is this point on π2 ? If so Substituting p=0 into (3), we obtain q=-1 Now substitute p=0, q=-1 into (1) to check: Therefore the point (1, 6, 0) is on both planes, thus the two plane are identical.

Scalar Equations of Planes We have already discussed the scalar equation of the line Ax+By+C=0, We will continue this process and define Ax+By+Cz+D=0 as the scalar or Cartesian equation of the plane. Let’s develop a scalar equation of the plane through point P0(4,-1,3) with normal n=(3,5,2) We pick any point P(x,y,z) on the plane. Since both P0 and P are on the plane, the vector P0P is a vector in the direction of the plane. So P0P is perpendicular to n. Thus Note: the coefficients of x, y, z are the terms of the normal

Understanding The point P0(4, -1,3) satisfies the equation Therefore: 3x+5y+2z-13=0 is a scalar equation of the plane

Understanding Rewrite the equation (x,y,z)=(2,-5,1)+s(2,-3,0)+n(1,1,-1) as a scalar equation for the plane. First we find a normal by the cross product of two non-collinear direction vectors Therefore a scalar equation of the plane is of the form 3x+2y+5z+D=0 The point (2, -5, 1) satisfies this equation Therefore: 3x+2y+5z-1=0 is a scalar equation of the plane or

Understanding Rewrite the equation (x,y,z)=(2,-5,1)+s(2,-3,0)+n(1,1,-1) as a scalar equation for the plane. Another solution is found by finding the dot product of both sides with the normal to the plane Will always be zero because (3,2,5) is normal to the plane

Understanding Determine the scalar equation of the plane passing through the point (2,-5,1) and parallel to the lines (x,y,z)=(1,6,-1)+s(2,-3,0) and (x,y,z)=(4,5,3)+n(3,2,-1) . If we place in Scalar form, we need only find a vector perpendicular to (2,-3,0) and (3, 2,-1). Now for D:

Understanding Find the distance from the point Q(3, -1, 1) to the plane 4x-8y-z=-41 Find normal: Find Point: Then: n P0 Q’  Q

Intersection of a Line and a Plane Find the point of intersection of the line and the plane Write the symmetric equations for l as parametric equations If a point (x,y,z) lies on the plane, it must satisfy the equation of the plane Hence

Distance from a Point to a Plane The distance from the point Q(x1, y1, z1) to the plane Ax+By+Cz+D=0 is:

Understanding Find the distance from the point Q(1, 3, -2) to the plane 4x-y-z+6=0

Understanding Find the point of intersection between the line and the plane If (x,y,z) is the point of intersection

Understanding Find the point of intersection of the line and the plane Write the symmetric equations for l as parametric equations If a point (x,y,z) lies on the plane, it must satisfy the equation of the plane There is no solution, no point of intersection

Understanding Find the point of intersection of the line and the plane Write the symmetric equations for l as parametric equations If a point (x,y,z) lies on the plane, it must satisfy the equation of the plane There is infinite solutions, the line is on the plane

Understanding Three vectors are coplanar if: Given: Determine z, that makes them coplanar Therefore: Now:

Intersection of Two Planes If two distinct planes meet, their intersection is a line. Because the line of intersection of the planes lies in each plane, the direction vector d of the line is perpendicular to both the normals of each plane What we need to find now is a point that is in both planes (not easy) and combine this with the direction vector. Or simpler, use elimination to solve for their intersection. Only if the planes are parallel can you determine the distance between them.

Understanding Find the parametric form of the line of intersection of the planes: Eliminate x Eliminate y Let z=t A parametric form of the line of intersection

Understanding Find the parametric form of the line of intersection of the planes (using direction vectors) Find direction vector Pick an axis to zero, say z-axis Solve these two equations for x, and y We can zero any axis, x or y or z Therefore:

Understanding Find the parametric form of the line of intersection of the planes: Scalar equation: Eliminate x Eliminate y Let z=t

Understanding Find the parametric form of the line of intersection of the planes: Let’s find the direction vector of the line Therefore these planes are either parallel or the same. Since plane 1 is not a multiple of plane 2, they are not the same and thus are parallel and do not intersect.

Understanding Find the vector equation of the lines that passes through the point (2, -1,7) and is parallel to the line of intersection of the planes: Let’s find the direction vector of the line The vector equation is:

Intersection of Three Planes The system has a unique solution, in which case the three planes intersect at only one point Case 1: Case 2: The system has an infinite number of solutions described by one parameter, in which the three planes intersect in a line The system has an infinite number of solutions described by two parameter, in which the three planes are coincident (same plane) and the solution consists of the coordinates of all points in the plane. Case 3: Case 4: The system has no solution, that is it is inconsistent. This will happen if at least two of the planes are parallel and distinct. It will also happen if the three lines of intersection of pairs of planes are parallel.

Case 1 Determine the intersection of the three planes: Therefore the planes intersect at (x,y,z)=(1, -2, -4)

Case 2 Determine the intersection of the three planes: Gauss-Jordan elimination This reduces to: x+4z=1 y-2z=-3 Or (x,y,z)=(1-4t, -3+2t,t) The line passes through (1, -3, 0) with direction (-4, 2, 1)

Case 3 Determine the intersection of the three planes: Gauss-Jordan elimination This reduces to: x+y+2z=-2 The planes are the same

Case 4 Determine the intersection of the three planes: Gauss-Jordan elimination The last line states : 0=-13 Which has no solution.

Case 4 Determine the intersection of the three planes: Gauss-Jordan elimination The last line states : 0=7 Which has no solution.