**Unformatted text preview:**

Math%126%Final%Study%Guide%Grace%Qiu%%MIDTERM 1 MATERIAL 1. Basic Vectors ! Magnitude of a vector " Vectors have two components: a direction and a magnitude. " If r = <r1, r2, r3>β¦ " Magnitude = | r | = π!!+ π!!+ π!! ! Dot product " a # b = <a1b1 + a2b2 + a3b3> " Result: a scalar.% " When this equals 0, the two vectors are orthogonal to each other! (Angle is pi/2) " When a # b = Β± |a| |b|, parallel. (Angle is 0 or pi) " a # b = |a| | b| cosΞΈ, where ΞΈ is the acute angle between a and b. ! Cross product of two vectors " a Γ b = π π ππ1 π2 π3π1 π2 π3 = (a2b3 β a3b2)i β (a1b3 β a3b1)j + (a1b2-a2b1)k " Result: a vector. " When this equals the zero vector, the two vectors are parallel to each other " | a Γ b | = |a| |b| sinΞΈ = area of parallelogram ! Angle between two vectors " a # b = |a| |b| cosΞΈ, where ΞΈ is the acute angle between a and b. " | a Γ b | = |a| |b| sinΞΈ, where ΞΈ is between a and b and 0 β€ ΞΈ β€ Ο ! Vector given by line segment AB A(x1, y1) and B(x2, y2) " a = <x2-x1, y2-y1> ! Components " projab = (!!!!!!!)π $ Parallel to a, but points in opposite direction " compab = π!!!!|π| = β |projab| $ Parallel to a, but points in same direction Two vectors are also parallel to each other if they are scalar multiples of each other.Math%126%Final%Study%Guide%Grace%Qiu%%2. Basic equations ! Describing a Line " You need a point and a direction vector. " Line equations are NOT UNIQUE, scalar multiples exist. " Parametric equation: x = 1 + 4t, y = 2 β 5t, z = 3 + 6t " Vector: r(t)=<1 + 4t, 2 β 5t, 3 + 6t> " Symmetric: !!!!= !!!!!!= !!!!! **Solve each parametric equation for t and set equal! ! Describing a Plane " You need a point and a normal vector to the plane. " Point P(xo, yo, zo) and vector <a, b, c>β¦ " a(x-xo) + b(y-yo) + c(z-zo) = 0 " Plane equations ARE UNIQUE. 3. Finding equations ! Equation of a plane passing through three given points " Create two lines from the three given points. " Find the cross product of these two lines; this will give you a vector orthogonal to these. " The result is the normal vector to the plane. You can then use this and any of the given points to construct an equation of a plane. ! Equation of a plane passing through a point, parallel to a given plane " The normal vector of the given plane is the normal vector of the unknown plane. " Construct equation using the point and this vector. ! Equation of a plane containing a line and a given point " Make one vector, the coefficients of the line equation. " To get the second vector, set the variable in the line equation to 0. This will give you a point. " Use this point and the given point to make an equation for a vector. " Do the cross product of this new vector and the first vector you made. This will give you the normal vector to the plane. " Use the given point and this normal vector to construct an equation.Math%126%Final%Study%Guide%Grace%Qiu%%4. Distance ! Distance between two points " |P1, P2| = π₯2 β π₯1!+ π¦2 β π¦1!+ π§2 β π§1! " Do three points make a triangle, or a line? $ Use the three points and find the longest distance (A to B, A to C, B to C) $ If smaller distances add up to the longest distance, they are in a line. $ If not, they are in a triangle. Equilateral? Isosceles? Look at the distances. $ Right triangle? Only if this is true: a2 + b2 = c2! ! Distance between a point and a plane " |!!!!!!!!!!!!!|!!!!!!!! ! Distance from a point on a line to a plane " v(t) : x = t, y = 1, z = 5 from plane y = 5 " Process: $ Make an arbitrary point P(x, y, z) $ π₯ β π₯!+ π¦ β 5!+ π§ β π§! = π₯ β π₯!+ π¦ β 1!+ π§ β 5! ! Distance between two parallel planes " |!!!!!|!!!!!!!! ! Midpoint of a line " (!!!!!!,!!!! !!,!!!!!!) ! Center and radius of a sphere given by equation " (x-h)2 + (y-k)2 + (z-l)2 = r2 " Center is at (h, k, l) with radius r (Distance from P to the plane y = 5, only thing that matters is Pβs y-coordinate.) %(Distance from P to the line v(t). Since x = t in v(t), that means the value of point P on line v(t) is x.) %Math%126%Final%Study%Guide%Grace%Qiu%%5. Intersection ! Point of intersection of two lines " Make the variables different, such as one with βsβ and one as βtβ " Set parametric equations (or other) equal to each other accordingly " Use substitution to solve for your two unknowns. " Plus the βsβ value back into the equation with sβs, or βtβ back into the equation with tβs to find the specific point. " Check: they should come out to be the same. ! Line of intersection of two planes " n1 Γ n2 = v " The equations of the two planes give you normal vectors orthogonal to each plane. Finding the cross product gives you a vector orthogonal to the normal vector, which means it is parallel to the planes. A line lying on both planes will be parallel to both. " Let a variable = 0 in the plane equations and use substitution to solve for the remaining two variables. This is using the assumption that at some point, the planes will intersect the xy plane, the yz plane, or the xz plane. ! Angle between two intersection planes " ΞΈ = cos-1(!!!!!!!!|!!|) " If ΞΈ comes out to be negative, you chose normal vectors that created an angle bigger than pi/1. If this is the case, simply doβ¦ Ο β cos-1(!!!!!!!!|!!|) or Ο β ΞΈ ! Angle between two intersecting lines " Find the vectors for both. " a # b = |a| | b| cosΞΈ, where ΞΈ is the acute angle between a and b. ! Find the equation of a plane containing a line of intersection of two other planes and a given point. " Get the two normal vectors of the two planes. " Do their cross product to get the vector for the line of intersection. " Find a point on the line of intersection by setting a variable to 0 in the plane equations, and then solve this system of equations. This results in a point. " Using this found point and the given point, create a vector. " Do the cross product of this new vector and the vector of the line of intersection. This gives the normal vector to the unknown plane. " Use this normal vector β¦

View Full Document