| 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? |
| Write the parametric equations and cartesian equations of the x-axis. |
/ 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. |
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. |
/ 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 |
| 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). |
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.
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
|
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 = 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.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. |
|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
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| 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)