..READ THIS FIRST..Problems
about Lines - Planes - Distances - Angles
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Problems about Lines - Planes - Distances - Angles
Level 1 problems
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Take line AB with A(4,5,6) and B(6,7,8). Give direction numbers of that line. Is C(1,2,3) on that line?
(2,2,2) are direction numbers but also (1,1,1) and (-10,-10,-10).
Point C is on line AB if and only if the direction AB = direction AC.
Direction AC has direction numbers (-3,-3,-3) or (1,1,1).
So, A B and C are on one line. -
Write the parametric equations and cartesian equations of the x-axis.
Take two points O(0,0,0) and E(1,0,0). Direction numbers are (1,0,0).
The parametric equations are/ x = 0 + r.1 / x = r | y = 0 + r.0 <=> | y = 0 \ z = 0 + r.0 \ z = 0The cartesian equations are y=0 z=0. -
Take the triangle ABC with A(2,2,4) B(4,6,0) and C(0,0,2). Calculate the median lines.
Center of [AB] = C'(3,4,2)
Center of [BC] = A'(2,3,1)
Center of [CA] = B'(1,1,3)
Direction numbers of AA' are (0,1,-3)
Direction numbers of BB' are (-3,-5,3)
Direction numbers of CC' are (3,4,0)The parametric equations of AA' are / x = 2 + r.0 | y = 2 + r.1 \ z = 4 + r.(-3) The cartesian equations are y - 2 z - 4 x-2=0; -------- = --------- <=> x = 2 ; 3y + z - 10 = 0 1 -3 The parametric equations of BB' are / x = 4 + r.(-3) | y = 6 + r.(-5) \ z = 0 + r.3 The cartesian equations are x - 4 y - 6 z - 0 ------ = -------- = ------ <=> -5x + 3y + 2=0 ; x + z = 4 -3 -5 3 The parametric equations of CC' are / x = 0 + r.3 | y = 0 + r.4 \ z = 2 + r.0 The cartesian equations are x y ------ = -------- ; z = 2 <=> 4x = 3y ; z = 2 3 4The center of the triangle is (2,8/3,2) and lies on the three median lines. -
Calculate the parametric equations and cartesian equation of the plane formed by the x-axis and the y-axis.
Direction numbers are (1,0,0) and (0,1,0). The parametric equations are/ x = 0 + r.1 + s.0 / x = r | y = 0 + r.0 + s.1 <=> | y = s \ z = 0 + r.0 + s.0 \ z = 0The cartesian equation is| x y z | | 1 0 0 | = 0 <=> z = 0 | 0 1 0 | -
Calculate the cartesian equation of the plane containing the point A(1,2,3) and parallel to the lines b and c b: 4x = 3y ; z = 2 and c: -5x + 3y + 2=0 ; x + z = 4
First, we calculate direction numbers of the lines b and c.
Take two points on b. B(0,0,2) and B'(3,4,2).
Direction numbers of b are (3,4,0).
Take two points on c. C(4,6,0) and C'(1,1,3).
Direction numbers of c are (-3,-5,3).
The cartesian equation of the plane is| x-1 y-2 z-3 | | 3 4 0 | = 0 <=> 4x - 3y - z + 5 = 0 | -3 -5 3 | -
The plane ABC has equation 4x - 3y - z + 5 = 0. Calculate the equation of the plane parallel to ABC and containing point D(2,1,3).
All planes parallel to ABC have equation 4x - 3y - z + t = 0.
D is in that plane if and only if 8 - 3 - 3 + t = 0 <=> t = -2.
The plane has equation 4x - 3y - z - 2 = 0. -
Given : x - 4 y - 6 z - 2 line b: ------ = -------- = ------ -3 -1 3 x - 1 y - 2 z - 3 line c: ------ = -------- = ------ -1 -2 2 Calculate the equation of the plane such that A(1,2,3) is in that plane and that b and c are parallel to that plane.The cartesian equation of the plane is | x-1 y-2 z-3 | | -3 -1 3 | = 0 | -1 -2 2 | -
Given : x - 4 y - 6 z - 2 line b: ------ = -------- = ------ -3 -1 3 x - 1 y - 2 z - 3 line c: ------ = -------- = ------ -1 -2 2 Are these lines orthogonal?
Direction vectors are A(-3,-1,3) and B(-1,-2,2).
A.B = 3 + 2 + 6 = 11 . The dot product is not 0. The lines are not orthogonal. -
Take plane ABC: 3x-2y-4z=3 and plane DEF: x-y-z=3.
Are these planes orthogonal?
Normal vectors to the planes are A(3,-2,-4) and B(1,-1,-1) A.B = 3 + 2 + 4 = 9 .The dot product is not 0. The planes are not orthogonal.
Level 2 problems
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Are the lines b and c intersecting? parallel? / x = 4 + r.(-3) b: | y = 6 + r.(-5) \ z = 0 + r.3 / x = 3 + r.3 c: | y = 1 + r.1 \ z = 1 + r.3
Since (-3,-5,3) is not a multiple of (3,1,3), b is not parallel to c. If there is an intersection point, this point does not necessarily corresponds with the same parameter value in b and in c. Therefore we call the parameter inthe equations of line c r' instead of r. If there is an intersection point p(x,y,z), there is a r and r' such that4 + r.(-3) = 3 + r'.3 6 + r.(-5) = 1 + r'.1 0 + r.3 = 1 + r'.3 <=> 3r + 3r' = 1 5r + r' = 5 -3r + 3r' = -1This system has no solution. So, the lines are not parallel and not intersecting. -
Are the lines b and c intersecting? parallel? line b: 2x + 3y + z = 5 ; x + y + z = 3 line c: x + 2y - z = 2 ; x - z = 0
If there is an intersection point, the coordinates are the solutions of the system 2x + 3y + z = 5 x + 2y - z = 2 x + y + z = 3 x - z = 0 This system has just one solution x = 1; y = 1; z = 1. The intersection point is p(1,1,1) -
Calculate the orthogonal projection A' of point A(1,2,3) on the plane 3x-y+4z = 0.
The normal direction to the plane is (3,-1,4). The line from A orthogonal to the plane is
/ x = 1 + r.3 | y = 2 + r.(-1) \ z = 3 + r.4The variable point P(1 + r.3, 2 + r.(-1),3 + r.4) of that line is in the plane if and only if3(1 + r.3)-(2 + r.(-1))+4(3 + r.4) = 0 <=> r = -1/2The point A' is (-1/2, 5/2, 1). -
Calculate the sharp angle between the lines / x = 1 + r | y = 2 - r \ z = 1 + r and / x = 1 + r.3 | y = 2 \ z = 3 + r.4
The direction vectors of the lines are A(1,-1,1) and B(3,0,4).
A.B = 7 = sqrt(3).5.cos(t) => cos(t) = 7/(sqrt(3).5) = 0.808
t = 36 degrees -
Calculate the sharp angle between the planes
2x + y + 4z = 2 and x + y - 4 = 0
The normal vectors are N(2,1,4) and N'(1,1,0).
N.N' = 3 = sqrt(21).sqrt(2).cos(t) => t = 62 degrees -
Given: a(2,1,0) ; b(1,0,1) ; c(3,0,1) d(0,0,2)
Point d is on a line L orthogonal to the plane abc.
Calculate the equations of L, the intersection point s with the plane and the distance from d to the plane abc.
The plane abc has equation|x-2 y-1 z | |-1 -1 1 | = 0 <=> y + z - 1 = 0 |1 -1 1 | The direction of the line L is (0,1,1) The parametric equations of the line L are x = 0 + 0 y = 0 + r z = 2 + r A variable point p on L is p(0, r, 2 + r) Point p is in the plane if and only if r + 2 + r - 1 = 0 <=> r = -1/2 So, the intersection point s is (0, -1/2, 3/2). The distance from d to the plane abc is ______________________ | 2 2 2 \| 0 +(1/2) + (1/2) = 0.707
Level 3 problems
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Take a plane x + y - z = 1 and point A(1,2,-3). A line l has equations / x = 1 + r.3 | y = 2 + r.(-1) \ z = 3 + r.4 Calculate the coordinates of a point B of line l, such that AB is parallel to the plane.Consider B(1+3r,2-r,3+4r) as a variable point of line l. The direction numbers of AB are (3r,-r,6+4r). The normal direction to the plane is (1,1,-1). AB is parallel to the plane <=> AB is normal to the direction (1,1,-1) <=> 3r -r-6-4r = 0 <=> r = -3 So, B is point (-8,5,-9) -
Take a point A(1,2,0). A line l has equations / x = 1 + r | y = 2 - r \ z = 1 + rCalculate the coordinates of the points B of line l, such that |AB| is sqrt(6).
Consider B(1+r,2-r,1+r) as a variable point of line l.
The coordinates of vector AB are (r,-r,1+r).
Now, ||AB|| must be sqrt(6).
<=> 3r.r +2r +1 = 6 <=> r = 1 or r = -5/3
The points are (2,1,2) and (-2/3, 11/3, -2/3)
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