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Chapter 8 Review

Chapter 8 Review. Bryan Olive, Sahej Brar , Manvir Bains. A direction vector is defined to be a nonzero  vector, Vector m  = (a, b) collinear to the given line . To find this direction vector you need 2 points that the line passes through. Using this you find the vector

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Chapter 8 Review

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  1. Chapter 8 Review Bryan Olive, SahejBrar, ManvirBains

  2. A direction vector is defined to be a nonzero vector,A direction vector is defined to be a nonzero vector, Vector m = (a, b) collinear to the given line. To find this direction vector you need 2 points that the line passes through. Using this you find the vector Vector AB= B-A=(x2-x1,y2-y1) 8.1 Vector and Parametric Equations of a Line in R2

  3. Using this and a point on the line you are trying to find, you can create vector equation • Vector Line = Point (x,y) +t(direction vector) • : r = (xo, yo) + t(a, b), t∈R • t is an element of all real numbers and as such lets you create every point using that direction vector • Example Direction Vector m= (2,1) Point on line (3,2) Line = (3,2) +t(2,1) t∈R Vector Equation of A Line in R2

  4. Once you have your vector equation it is very easy to make your parametric equations by seperating the x, and y components. • Example Line(x,y) = (3,2) +t(2,1) Parametric Eqns: X= 3 + 2t Y=2+t • Remember if the direction vectors of two equations are the same or scalar multiples of each other the lines must be parallel • if they are the same line than a point from one line subbed into the parametric eqn of the other line will give the same t values for both x and y Parametric Equation of a line in R2

  5. Cartesian Eqn: Ax+By+c=0 • To find your cartesian equation you first need to find the normal vector, basically the vector perpendicular to your line, and you must have a point on the line • This can be done by first finding the direction vector of your line, and finding the negative reciprocal of it Example M=(3,2) N=(2,-3) Cartesian Equations of Lines

  6. Now that you have your normal vector you can use the cartesian equation and solve for your c Value Ax+By+c=0 Normal vector= (A,B) Example Line(x,y) = (3,2) +t(2,1) M=(2,1)n=(1,-2) 1x-2y+c=0 1(3)-2(2)=-c 1=c Eqn: 1x-2y+1=0 Remember if two lines are perpendicular than their normals are also perpendicular, likewise parallel lines have parallel normals. Based on this you can use dot product or scalar multiplication to find unknowns Cartesian Equations of Lines

  7. The same as equations in R2 except now a z component. • Vector Line = Point (x,y,z) +t(direction vector) • : r = (xo, yo,zo) + t(a, b,c),t∈R • Parametric is also the same, just a z equation also • Example Direction Vector m= (2,1,1) Point on line (3,2,1) Line = (3,2,1) +t(2,1,1) t∈R Parametric : x=3+2t Y=2+t Z=1+t 8.3 Vector, Parametric, and Symmetric Equations of a Line in R3

  8. Example • P(2,3,4) Direction vector (1,2,3) (X-2)/1=(y-3)/2=(z-4)/3 Remember if a comonent of the direction vector is 0 than do the other two components first than add the 0 component using a comma and the component=Point component (ex. If direction z=0 on direction vector ,z=4 Symetric Equations

  9. To see if lines are the same simply check if their direction vectors are scalar multiples, and then take a point from one equation and substitute it into the parametric equation of the other line. If all the t values match the lines must be the same • To find angles between intersecting lines use dot product of the direction vectors of the two lines, and simply solve for the angle (the other angle will be 180degrees subtract this angle) Things to remember!

  10. Linear combinations of the two direction vectors will create/generate  every possible point on aplane Similar to the equation of a line except now you use two separate direction vectors that are not collinear Plane (pi) = Point on plane +t(direction vector 1) + s(direction vector 2) Example Point on plane (2,3,4) Direction vector 1 (1,2,3) Direction vector 2 (4,5,6) Plane (pi) = (2,3,4) +t(1,2,3) + s(4,5,6) t,s are all real numbers 8.4 Vector and Parametric Equations of a Plane

  11. To create the parametric equations we still separate the vector equation into components Example : Plane (pi) = (2,3,4) +t(1,2,3) + s(4,5,6) t,s are all real numbers X=2+t+4s Y=3+2t+5s Z=4+3t+6s You can make these two equations when you have one of four situations: 1)Eqn for line + one point not on that line 2) Two intersecting lines 3)Three Points 4)Two parallel and non-coincident lines

  12. To see if a point falls within a plane then you sub that point into the parametric equation for the plane, using two equations do elimination/substitution, solve for the two variables (s,t) and using those values do a LS=RS check on the third eqn • If the LS=RS that means the point lies on that plane

  13. The cartesian equation of a plane is very similar to that of a line in R2 but now there is a third component • To use the equation you need a normal vector on that plane as well as one point on that plane Eqn: Ax+By+Cz+D=0 Normal vector n= (A,B,C) Example: N=(1,2,3) Point (4,5,6) 1x+2y+3z+D=0 1(4)+2(5)=3(6)= -D -32=D Cartesian Eqn: 1x+2y+3z-32=0 8.5 The Cartesian Equation of a Plane

  14. If two planes are perpendicular that means that their normals are also perpendicular meaning you can use dot product (equal to zero) to solve for unknowns • If two planes are parallel that means that their normals are parallel and so the normals must be scalar multiples of each other Parallel and Perpendicular Planes

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