Problems about Lines - Planes - Distances - Angles

• READ THIS FIRST
  
• Problems about Lines - Planes - Distances - Angles
  
  • Level 1 problems
    
  • Level 2 problems
    
  • Level 3 problems
    

READ THIS FIRST

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Lines - Planes - Distances - Angles

Level 1 problems

• 
  

  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 = 0
  

  The 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          4
  

  The 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 = 0
  

  The 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

• 
  

  The points P(2,-2,1) and Q(1,2,-2) belong to a sphere with center O(0,0,0). Calculate the angle between the two tangent planes in P and Q to the sphere.

  
  
  OP is a line perpendicular to the tangent plane in P and OQ is a line perpendicular to the tangent plane in Q.

  The angle between two planes is the angle between the two perpendiculars and it is the angle t between the two vectors P en Q.

  Now, P.Q = ||P||.||Q||.cos(t) <=> -4 = 3 . 3. cos(t) <=> cos(t) = -4/9

  The sharp angle between the two planes is 63.6 degrees.

• 
  

  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 in the 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 that

   
          4  + 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' = -1
  

  This 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 = 6 ; 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 = 6
          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.4
  

  The variable point P(1 + r.3, 2 + r.(-1),3 + r.4) of that line is in the plane if and only if

   
          3(1  + r.3)-(2  + r.(-1))+4(3  + r.4) = 0
  <=>
          r = -1/2
  

  The 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
  

• 
  

  Find the equation of the plane which is perpendicular bisector of the segment [AB] with A(1,2,3) and B(5,6,7).

  
  

  The center of [AB] is M(3,4,5). The direction of AB is (4,4,4) or also (1,1,1). So, v(1,1,1) is a normal vector to the requested plane.
  The requested plane has an equation (x-3)+(y-4)+(z-5) = 0

Level 3 problems

• 
  

  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 + r Calculate the coordinates of the points B of line l, such that |AB| = 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)
  

• 
  

  Two planes have respectively an equation 2 x - 2 y - z + 5 = 0 and x + 5 y - z - 8 = 0 Point A = A(3,5,7) Find the equation of a plane through A and perpendicular to the two planes.

  
  

  The normal vectors of the given planes are direction vectors of the requested plane. Point A is a support vector of the requested plane.
  The requested plane is

   
         | x - 3     y - 5    z - 7 |
         |   2         -2      -1   |  = 0
         |   1         5       -1   |
  

• 
  

  Given: The plane alpha with equation 2x + 3y - z -7 = 0 The line d with equations [ 3x + y - z = 0 ; x - y - z + 2 = 0 ] Find the plane gamma through d ans perpendicular to alpha.

  
  

  The point A = A(0,1,1) on the line d can act as support point for gamma.
  The normal vector (2,3,-1) of alpha can serve as a direction vector for gamma.
  The vector (1, -1, 2) is a direction vector of d. This vector is a direction vector of gamma.

  The requested plane is

   
    |  x   y - 1   z - 1 |
    |  2     3      -1   | = 0
    |  1    -1       2   |
  

• 
  

  A plane alpha has an equation x + y + z = 3. P(1,1,1) is in alpha. Point P is on the line d and d has a direction vector (1,2,3). Find the line c, in alpha, such that c and d are orthogonal lines and P is on c.

  
  

  All lines through P and orthogonal with d are in the plane beta through P and perpendicular to d. The given vector (1,2,3) is a normal vector of beta. The plane beta has an equation

   
      1.(x - 1) + 2.(y-1) + 3.(z-1) = 0
  <=>
      x + 2 y + 3 z  - 6 = 0
  

  The intersection line of alpha and beta is the requested line c. The line c has equations

   
    x +   y +   z = 3
    x + 2 y + 3 z = 6
  

• 
  

  A pyramid has base ABC A(1, 2, -3) ; B(0, -1, 5) ; C(-3, 0, 9) The vertex T is a variable point of the line x = 1 + r y = 2 + r z = -1 - 2 r 1) Find T such that the height of the pyramid is 4. 2) Find the volume of the pyramid.

  
  

  1. First, we calculate the equation of the plane ABC. We use support vector A and direction vectors BA and BC.

      
       |  x  y+1  z-5 |
       |  1   3   -8  | = 0  <=> 2x + 2y + z -3 = 0
       | -3   1    4  |
     

     T has coordinates (1 + r, 2 + r, -1 - 2 r).
     The height of the pyramid must be 4.

      
          2(1 + r) + 2(2 + r) + (-1  - 2 r) - 3
       |  -------------------------------------  | = 4
              sqrt( 4 + 4 + 1)
     
     <=> ...
     
     <=> r = 5 of r = -7
     

     So, T(6, 7, -11) or T(-6, -5, 13)

  2. We calculate the area of the triangle ABC.
     The lengths of the sides of the triangle ABC are :

      
        |AB| = sqrt(74) = c
        |BC| = sqrt(26) = a
        |AC| = sqrt(164) = b
     

     Now, we can calculate the area with Heron's formula.
     

      
       s = (a+b+c)/2
                   _________________
       The area = V s(s-a)(s-b)(s-c)
     

     We find that the area is 15.
     

  3. The volume of the pyramid is

      
       I = (1/3) . 15 . 4 = 20
     

• 
  

  M is the centroid of the triangle DEF with D(6,0,0), E(0,6,0) and F(0,0,6). C is a circle, in plane DEF, with radius 2 en center M. Find point L on C such that L is as close as possible to the plane z = 0.

  
  

  M = M(2,2,2).
  Point L is on the median FM.
  Th direction of FM is (2,2,-4) or also (1,1,-2).
  We take a variable point L on FM: L(2+ r, 2+ r, 2 - 2r)
  

   
      L is on the circle
  <=>
        | M L|2 = 4
  <=>
       r2 + r2 + 4r2 = 4
  <=>
      r2 = 2/3
  

  Point L is ( 2 + sqrt(2/3) , 2 + sqrt(2/3) , 2 - 2 sqrt(2/3) )

• 
  

  Given: a fixed point P and two variable lines a and b. Find the line c through point P such that line c intersects both lines. Find the value of the parameter so that there are infinitely many solutions P(3,1,6) line a / x + (m-1)y -2 = 0 \ y + z - 3 = 0 line b / (2m-3)x - 1 = 0 \ y - z + 1 = 0

  
  

  The line c is the intersection of plane (P,a) and plane (P,b).

  1. Plane (P,a)

     The direction of a is (m-1, -1, 1); it is a first direction of the plane (P,a).
     A simple point of a is A(2,0,3).
     A second direction of the plane (P,a) is the direction A,P and this is (1,1,3).
     The plane (P,a) has as equation

      
       | x -3   y - 1  z - 6 |
       | m - 1    -1     1   | = 0 <=> -4x + (4-3m)y + mz -3m+8 = 0
       |   1       1     3   |
     

  2. Plane (P,b)

     In the same way as for plane (P,a) we find

      
       (4m-6)x + (3m-5)y - (3m-5)z + 3m-7 = 0
     

  3. The requested line c through P is the intersection of both planes.

     So we have immediately the equations for c

      
        /   -4x + (4-3m)y + mz -3m+8 = 0
        \   (4m-6)x + (3m-5)y - (3m-5)z + 3m-7 = 0
     

  4. There are infinitely many solutions if and only if P, a and b are all in one plane.

     The planes (P,a) and (P,b) have point P in common.
     The planes coincide if and only if the simple point A(2,0,3) of line a is in the plane (P,b).
     

     One finds that the condition is fulfilled if and only if m = 2.