Vectors+Dot Product + Cross Product (1) Findavector v with representation given by the directed line segment AB. Draw AB and the equivalent position vector where: A = (0,6, 1), B = (3,4,4) (2) Sketch the vector 3 43 + 3kand find its magnitude. (3) Give a unit vector (a) in the opposite directionas v/ = 21 + 37 k (b) perpendicular to (4) Determine whether the vectors are orthogonal, parallel or neither. @ (b) , (c) i+k, 212j (d) u={c,cc),v=/c0,c) (5) Refer to the figure. A=(0,0,1) D=(0,1,0) B=(1,0,0) C=(1,1,0) Caculate the projection of AC onto AB (4) (a) Find a vector normal to the plane z = 2 + 3x - y. (b) Find a vector parallel to the plane. (s) Find a parametric equation for the line through point (2, 1, 4) and perpendicu lar to the plane 3x + y + z = 3. (6) Find an equation of the plane passing through (0, 1, 1), (1, 2, 3) and (3, -1, -2). (7) How do you check to see if the line r(t) = +t is perpendicular to the plane 4(x - 2) + 2(y + 1) + 2z = 13? Explain and then investigate. (8) Find an equation of the plane that contains the lines r(t) = and r s) =Question 4 (a) Find a vector normal to the plane z = 2 + 3z y. 1. Identify the plane equation in standard form: The given equation z = 2 + 32 y can be rewritten in standard form as 3 y 2z = 2. Extract the normal vector: The normal vector to the plane is given by the coefficients of x, 3, and z in the equation Az By+Cz=D.Here A=3B=-1and = -1 Thereforg, the normal vector is (3, 1, 1). (b) Find a vector parallel to the plane. . Understand what a parallel vector is: A vector parallel to the plane is any vector that lies in the plane. These vectors are orthogonal to the normal vector. Choose vectors orthogonal to (3, 1, 1: To find vectors parzllel to the plane, we need vectors that are orthogonal to the normal vector {3, -1, -1. Examples of such vectors could be: . (1,0,1) = (0,1,-1) Question 5 Find a parametric equation for the line through point (2, 1, 4) and perpendicular to the plane dr+y+=2z=23 1. Identify the normal vector of the plane: The normal vector to the plane 3z + y + =z = 3is (3,1, 1. 2. Use the point (2, 1, 4) and the direction vector: The direction vector of the line is the same as the normal vector of the plane, which is -::3._ 1. 1::-. Therefore, the parametric eguations of the line are: r=2+423t y=1+1t =4+t Question 6 Find an equation of the plane passing through (0,1,1), (1,2, 3}, and (3, ~1, 2). 1. Find two vectors in the plane: n=(1-02-1,3-1) h=(3-0,-1-1,-2-1) 3 2. Find the normal vector using the cross product: =1 =1 =1112) = {3 -2,-3, ={-1,9,-5 3. Use the point (0, 1, 1) and the normal vector: The equation of the plane can be written as: Question 7 How do you check to see if the line 7(t) = (1,0,1) + #{2,1, 1} is perpendicular to the plane Az 2)+2(y+1)+22=13 1. Find the direction vector of the line: The direction vector of the lineis (2,1, 1} 2. Find the normal vector of the plane: The normal vector of the plane 4(x 2) + 2(y + 1) + 2z = 1315 (4,2, 2}, 3. Check the dot product: To check if the line is perpendicular to the plane, the dot product of the direction vector of the line and the normal vector of the plane should be zero: (=2,1,1) - (4,2,2) = (-2 x4) + (1 x2)+ (1 x2)=-8+2+2=4 Since the dot product is not zero, the line is not perpendicular to the plane. Question 8 Find an equation of the plane that contains the lines 7(#) = (,2f, 3f) and F(s) = (1 8, 3,2+ st 1. Find direction vectors of the lines: 2. Check if lines are skew or intersect: If the lines intarsect, they will li2 in the same plane. If they are skew, they do not lie in the same plane. 3. Cross product of direction vectors: The cross product gives a normal vectar to the plane containing both lines: ii=d xdy=(1,2,3) x (~1,0,5) = (2s, -3,2) 4. Use a point from one of the lines: Using the point (1, 3, 2) from 7(s), the equation of the plane can be written as: 25(x - 1) -3y 3)+2(z2)=0 287 - 25 -3y +9+2:-4=0 2sr 3y+2:4+5-2s=10 Lines in 3D Space A line in three-dimensional space can be represented in several ways, with the most common being the parametric form and the vector form. Parametric Form of a Line A line passing through a point (To, 10, 20) and having a direction vector (a, b, c) can be represented parametrically as: r = co + at y = yo + bt 2 = 20+ ct where t is a real number parameter. Vector Form of a Line The vector form of the same line can be written as: T (t ) = To+td where r(t ) is the position vector of any point on the line, To is the position vector of the given point (To, Bo, 20), and d = (a, b, c) is the direction vector. Planes in 3D Space A plane in three-dimensional space can be described by a point on the plane and a normal vector to the plane. Standard Form of a Plane The standard form of a plane equation is: Ar + By + Cz = Dwhere A, B, and (' are the components of the normal vector to the plane, and D) is a constant. Key Concepts Used 1. Normal Vector to a Plane: The normal vector to a plane is a vector that is perpendicular to every vector lying in the plane. For the plane equation Ax + By + C'z = D), the normal vector is (A, B, ). 2. Parallel and Perpendicular Vectors: s Two vectors are parallel if they are scalar multiples of each other. s Two vectors are perpendicular if their dot product is zero, 3. Parametric Equation of a Line: Given a point (xq, yn, 20) and a direction vector (a, b, ), the parametric equations of the line are: r =1y + at 2=z +ct 4. Cross Product: The cross product of two vectors @ and T results in a vector that is perpendicular to both 4 and . This is used to find a normal vector to a plane containing two given lines. 5. Equation of a Plane from Three Points: To find the equation of a plane passing through three points, we can: s Find two vectors in the plane using the given points. s Find the cross product of these vectors to get the normal vector, s Use one of the points and the normal vectar to write the eguation of the plane