With each point P, corresponds a vector OP.
P is called the image point of OP.
The vector OP is noted P for short.
The vector P can be expressed as x.i + y.j + z.k
(x,y,z) are the coordinates of P.
We write co(P) = (x,y,z) or P(x,y,z) for short.
The vector AB = AO + OB => AB = OB - OA = B - A
It is not difficult to see that
co(A + B) = co(A) + co(B)
co(AB) = co(B - A) = co(B) - co(A)
co(r.A) = r.co(A) (with r a real number)
The center of [AB] is point M <=> AM = MB <=> M - A = B - M <=> 2.M = A + B <=> co(M) = (co(A) + co(B))/2Example:
Let M = the center of [AB]. CZ = 2.ZM <=> Z - C = 2.(M - Z) <=> 3.Z = 2.M + C <=> 3.Z = A + B + C <=> Z = (A + B + C)/3 <=> co(Z) = ( co(A) + co(B) + co(C) )/3Example:
Z - D = 3.(A' - Z) <=> 4.Z = 3.A' + D = A + B + C + D <=> Z = (A + B + C + D)/4 <=> co(Z) = (co(A) + co(B) + co(C) + co(D) )/4We have the same result starting from the centroid of another triangle.
point P is on BC <=> there is a real number r such that BP = r.BC <=> there is a real number r such that P - B = r.(C - B) <=> there is a real number r such that P = B + r.(C - B)The last expression is the vectorial equation of the line. The number r is a parameter.
Now, we take a right-handed rectangular coordinate system.
point P is on BC <=> there is a real number r such that P = B + r (C - B) <=> there is a real number r such that co(P) = co(B) + r.(co(C) - co(B)) <=> there is a real number r such that / x = b + r.(c - b ) (1) | y = b' + r.(c' - b') (2) \ z = b" + r.(c" - b") (3)These equations are called parametric equations of the line BC.
point P is on BC <=> x - b x - b y - b' = ------- (c' - b') and z - b" = ------(c" - b") c - b c - bThese equations are called cartesian equations of the line BC.
x - b y - b' z - b" ------ = -------- = -------- c - b c'- b' c"- b"If a the direction number is zero, the corresponding numerator is zero.
The line BC is defined by the vectors B(b,b',b") and C(c,c',c"). The parametric equations of the line BC with support vector B(b,b',b") and direction vector BC are / x = b + r.(c - b ) | y = b' + r.(c' - b') \ z = b" + r.(c" - b")If all the direction numbers (c - b );(c' - b'); (c" - b") are different from zero, previous equations are equivalent to: x - b y - b' z - b" ------ = -------- = -------- c - b c'- b' c"- b"These equations are the cartesian equations of the line BC. |
/ x = 1 + r.(-1) | y = 2 + r.3 \ z = 3 + r.5The cartesian equations of the line BC are
x - 1 y - 2 z - 3 ------ = -------- = -------- -1 3 5These cartesian equations can be written as a system of two equations.
3(x-1) = -(y-2) 5(y-2) = 3(z-3) <=> / 3x + y - 5 = 0 \ 5y - 3z - 1 = 0The point D(-1,8,13) is a point of BC because there is an r such that
/ -1 = 1 + r.(-1) | 8 = 2 + r.3 \ 13 = 3 + r.5or because (-1,8,13) is a solution of the system
/ 3x + y - 5 = 0 \ 5y - 3z - 1 = 0
/ x = 1 + r.0 | y = 2 + r.3 \ z = 3 + r.5The cartesian equations of the line BC are
y - 2 z - 3 -------- = -------- and x - 1 = 0 3 5 <=> x - 1 = 0 5 y - 3 z -1 = 0The point D(1,8,13) is a point of BC because there is an r such that
/ 1 = 1 + r.0 | 8 = 2 + r.3 \ 13 = 3 + r.5or because (1,8,13) is a solution of the system
x - 1 = 0 5 y - 3 z -1 = 0
/ x = 4 + r | y = 2 + r.3 \ z = 7 + r.2Find m and n such that point P(m+2, -4, n) is on AB.
P(m+2, -4, n) is on AB <=> There is an r such that / m+2 = 4 + r | -4 = 2 + 3 r \ n = 7 + 2 r <=> m, n en r are the solutions of / m - r = 2 | r = - 2 \ n - 2r = 7 <=> r = -2 , m = 0 , n = 3
Example 1:
Take the lines a and b
/ x = 1 + r.(-1) | y = 2 + r.3 \ z = 3 + r.5 and / x = 2 + r | y = 1 - r \ z = 3 + rr is the name of a parameter on both lines. But these parameters are independent. Therefore we give the second parameter a different name. For line b, we write
/ x = 2 + r' | y = 1 - r' \ z = 3 + r'There is an intersection point if and only if the following system has a solution for r and r'.
1 - r = 2 + r' 2 + 3r = 1- r' 3+ 5r = 3+ r'After some calculation, we see that this system has no solutions. So the lines a and b do not intersect.
Example 2:
If two lines are given by their cartesian equations and if we want to calculate the intersection point, then we have to solve a system of 4 equations in x, y and z. This can be a rather difficult system. If we first calculate the parametric equations of the lines, this work can be circumvented.
A line r is given by its cartesian equations [ x+y+z=6 , 2x+2y+z=11 ].
A line s is given by its cartesian equations [ x+y-4z=1 , x-z=2 ].
We'll find the intersection point in two ways.
The possible intersection is the solution of the system :
/ x + y + z = 6 | 2x + 2y + z = 11 | x + y - 4z = 1 \ x - z = 2We apply the theory of the systems. The matrix of the coefficients is
[1 1 1] [2 2 1] [1 1 -4] [1 0 -1]The rank of this matrix is 3. We choose the first three equations as main equations. The last equation is the side equation. There is an intersection point if and only if the system has a solution. The condition is : the characteristic determinant of the side equation is zero.
|1 1 1 6| |2 2 1 11| |1 1 -4 1| = 0 |1 0 -1 2|The condition is fulfilled! We omit the last equation. We solve the system and we find the intersection point S(3,2,1).
To find parametric equations we calculate two simple points of each line.
For r: A(0,5,1) and B(5,0,1). We use A as support vector. (5,-5,0) is a direction vector, but then (1,-1,0) is a direction vector too. The parametric equations are :
x = r y = 5 - r z = 1For s: C(2,-1,0) and D (0 -7 -2). We use C as support vector. (2,6,2) is a direction vector, but then (1,3,1) is a direction vector too. The parametric equations are :
x = 2 + r' y = -1 + 3 r' z = r'So we calculate r en r' from
r = 2 + r' 5 - r = -1 + 3r' 1 = r'We find r' =1 and r = 3. The intersection point is S(3,2,1).
Example 3
Given :
The line AB with equations [2x + 2m y + z = 3 , x + y + z = 1]
The line CD with equations [3x - 2 y + z = m , x - y + z = 2]
Find m such that the lines intersect.
The two lines intersect <=> The following system has a solution for x, y and z x + y + z = 1 x - y + z = 2 3x - 2 y + z = m 2x + 2m y + z = 3The rank of the matrix of the coefficients is 3. The system has a solution for x, y and z if and only if the characteristic determinant is zero.
| 1 1 1 1 | | 1 -1 1 2 | | 3 -2 1 m | = 0 | 2 2 m 1 3 | <=> 2m + 11 = 0 <=> m = -11/2
The directions are the same <=> there is a number r such that (w,w',w") = r.(v,v',v") <=> the dimension of span{(v,v',v"), (w,w',w")} is 1 <=> the dimension of the row space of [v v' v"] [w w' w"] is 1. <=> rank of the previous matrix is 1Conclusions:
Two directions (v,v',v") and (w,w',w") are the same <=> The rank of [v v' v"] [w w' w"] is 1.Two directions (v,v',v") and (w,w',w") are different <=> The rank of [v v' v"] [w w' w"] is 2.
point P is on plane ABC <=> There are real numbers r and s such that AP = r.AB + s.AC <=> There are real numbers r and s such that P - A = r(B - A) + s.(C - A) <=> There are real numbers r and s such that P = A + r(B - A) + s.(C - A)The last expression is the vectorial equation of the plane. The numbers r and s are parameters.
point P is on plane ABC <=> There are real numbers r and s such that co(P) = co(A) + r(co(B) - co(A)) + s.(co(C) - co(A)) <=> There are real numbers r and s such that / x = a + r.(b - a ) + s.(c - a ) (1) | y = a' + r.(b' - a') + s.(c' - a') (2) \ z = a" + r.(b" - a") + s.(c" - a") (3)These equations are called parametric equations of the plane ABC.
The parametric equations of the plane through the points A(a,a',a") ; B(b,b',b") ; C(c,c',c")
are
/ x = a + r.(b - a ) + s.(c - a ) | y = a' + r.(b' - a') + s.(c' - a') \ z = a" + r.(b" - a") + s.(c" - a") |
To obtain the cartesian equation of the plane, we eliminate r and s from the parametric equations.
To eliminate r and s from previous system we write
point P is on plane ABC <=> There are real numbers r and s such that r.(b - a ) + s.(c - a ) = x - a r.(b' - a') + s.(c' - a') = y - a' r.(b" - a") + s.(c" - a") = z - a" <=> The following system has a solution for r and s r.(b - a ) + s.(c - a ) = x - a r.(b' - a') + s.(c' - a') = y - a' r.(b" - a") + s.(c" - a") = z - a"Since the direction vectors AB and AC give a different direction, the rank of the matrix of coefficients is 2.
point P is on plane ABC <=> | (b - a ) (c - a ) (x - a )| | (b' - a') (c' - a') (y - a')| = 0 | (b" - a") (c" - a") (z - a")| and with properties of determinants we have <=> | (x - a ) (y - a') (z - a")| | (b - a ) (b' - a') (b"- a")| = 0 | (c - a ) (c' - a') (c"- a")|This is the cartesian equation of the plane ABC.
u.x + v.y + w.z + t = 0 with u,v and w not all zero.Each plane has an equation of this form.
The cartesian equation of a plane through the points A(a,a',a") ; B(b,b',b") ; C(c,c',c")
is
| (x - a ) (y - a') (z - a")| | (b - a ) (b' - a') (b"- a")| = 0 | (c - a ) (c' - a') (c"- a")| The equation can be written as u.x + v.y + w.z + t = 0 with u,v en w not all zero. |
/ x = 1 + r.1 + s.2 | y = 0 + r.2 + s.1 \ z = 1 + r.(-1)+ s.3The cartesian equation of the plane ABC is
|x-1 y z-1 | | 1 2 -1 | = 0 <=> 7x - 5y - 3z - 4 = 0 | 2 1 3 |
/ 7x - 5y - 3z - 4 = 0 \ x - 2y + 3z - 2 = 0We already know that the cartesian representation of a line consists of two equations. Well, this is precisely what we have in previous system. So, the above system is the expression of the intersection line.
If two planes are parallel there is no line corresponding with the system of the equations of the planes. The following systems do not represent a line
/ 7x - 5y - 3z - 4 = 0 \ 7x - 5y - 3z - 5 = 0 The planes have no common point. / x + 3y + 2z + 3= 0 \ 2x + 6y + 4z + 6 =0 The planes coincide.From parametric equations of two planes to the cartesian equations of a line.
Two planes are given by their parametric equations
x = r + s y = 3s z = 2r and x = 1 + r + s y = 2 + r z = -3 + sFind the cartesian equations of the intersection line. |
Solution :
First, we calculate the cartesian equations of the two planes
| x y z | | 1 0 2 | = 0 <=> 6x - 2y - 3z = 0 | 1 3 0 | |x-1 y-2 z+3 | | 1 1 0 | = 0 <=> x - y - z - 2 = 0 | 1 0 1 | The cartesian equations of the intersection line are / 6x - 2y - 3z = 0 \ x - y - z - 2 = 0
P(a,b,c) is on the intersection of 7x - 5y - 3z - 4 = 0 and x - 2y + 3z - 2 = 0 => 7a - 5b - 3c - 4 = 0 and a - 2b + 3c - 2 = 0 => k.(7a - 5b - 3c - 4) + l.(a - 2b + 3c - 2) = 0 => P(a,b,c) is a point of the plane k.(7x - 5y - 3z - 4) + l.(x - 2y + 3z - 2) = 0So, each point of the intersection of 7x - 5y - 3z - 4 = 0 and x - 2y + 3z - 2 = 0 is a point of the plane k.(7x - 5y - 3z - 4) + l.(x - 2y + 3z - 2) = 0.
The plane k.(7x - 5y - 3z - 4) + l.(x - 2y + 3z - 2) = 0 is a variable plane through the intersection of 7x - 5y - 3z - 4 = 0 and x - 2y + 3z - 2 = 0If k and l vary, the plane rotates around this fixed intersection line. Then we have a bundle of planes. The parameters k and l are homogeneous parameters.
Let t = l/k.
Then, the equation of the bundle becomes
(7x - 5y - 3z - 4) + t.(x - 2y + 3z - 2) = 0.t is a non-homogeneous parameter.
General formulation
The plane k.(ux + vy + wz + q ) + l.(u'x + v'y + w'z + q') = 0 is a variable plane through the line with equations / ux + vy + wz + q = 0 \ u'x + v'y + w'z + q'= 0 k and l are homogeneous parameters. The variable plane can be written with the use of a non-homogeneous parameter t. (ux + vy + wz + q ) + t(u'x + v'y + w'z + q') = 0 |
Example 1:
Point D(1,2,3) is on a line d with direction (3,2,1). Find the plane through d and through the point F(0,4,0). |
The line d has equations x -1 y - 2 z - 3 ------ = ------- = -------- 3 2 1 <=> x - 3 z + 8 = 0 and y - 2 z + 4 = 0The variable plane through d is (x - 3 z + 8)+ t(y - 2 z + 4) = 0
Example 2:
Given : The lines p , q and l
p [ x + y - z + 1 = 0 ; 2x - y + z + 3 = 0] q [ 2x + y - 1 = 0 ; x + y - z - 1 = 0 ] l [ x + y + z = 0 ; 2x - y + 3z = 0 ]Find the equations of the variable line m intersecting p, q and l. |
Procedure:
Take a variable point L on the line l. The requested line m is the intersection of the planes (L,p) and (L,q).
Working out :
A variable plane through the line p is (x + y - z + 1) + t(2x - y + z + 3) = 0 L is in that plane if and only if (-4r + r - 3r + 1) + t (-8r - r + 3r + 3) = 0 <=> t = (6r-1)/(-6r+3) The plane (L,p) is (-6r+3)(x + y - z + 1) + (6r-1)(2x - y + z + 3) = 0
A variable plane through the line q is (x + y - z - 1) + t ( 2x + y - 1) = 0 L is in that plane if and only if (-4r + r - 3r - 1) + t (-8r + r - 1) = 0 <=> t = (6r+1)/(-7r-1) The plane (L,q) is (-7r-1)(x + y - z - 1) + (6r+1) ( 2x + y - 1) = 0
/ (-6r+3)(x + y - z + 1) + (6r-1)(2x - y + z + 3) = 0 \ (-7r-1)(x + y - z - 1) + (6r+1) ( 2x + y - 1) = 0
Then, A.B =(a.i + a'.j + a".k).(b.i + b'.j + b".k) Using distributivity, this becomes A.B = a.b + a'.b' + a".b" All other term disappear using i.i = j.j = k.k = 1 and i.k = k.j = j.i = 0
Relative to an orthonormal basis in space, we have: The dot product of two vectors A(a,a',a") and B(b,b',b") is a.b + a'.b' + a".b" We write A.B = a.b + a'.b' + a".b" |
Take two vectors A(3,2,5) and B(1,0,4).
A.B = 3 + 0 + 20 = 23.
(The magnitude of vector A)^{2} = A.AWe write this magnitude as ||A||. And if A has coordinates (a,a',a"), A.A = a.a + a'.a' + a".a"
||A|| = sqrt(a^{2} + a'^{2} + a"^{2} )Example:
The magnitude of vector A(2,-1,4) is sqrt(4+1+16) = sqrt(21).
Hence, |AB| = sqrt((b - a)^{2} + (b' - a')^{2} + (b" - a")^{2} )
A.B = ||A||.||B||.cos(t) <=> A.B cos(t) = -------------- ||A||.||B||Example :
Take A(1,2,3) ; B(4,5,6) ; C(3,2,0) Calculate the angle between the lines AB and AC. |
The line AB has a direction numbers (3,3,3) and line AC has
direction numbers (2,0,-3).
Hence
6 + 0 - 9 cos(t) = ----------------- sqrt(27) .sqrt(13) for the sharp angle we find 80.78 degrees.
Example :
v(1,4,-1) en w(5,-1,1) are orthogonal vectors
Example:
Find the m-values such that the following lines are orthogonal
/ x = 2 + (2m-1)r | y = 3 + m r \ z = 2 + r / x = 2 - r | y = 3 + m r \ z = 6 |
The direction vectors are ( 2m-1 , m , 1 ) and ( -1 , m , 0)
The lines are orthogonal if and only if -2m + 1 + m^{2} = 0 <=> m = 1
Example :
The line b contains the point B(1,2,1) and has a direction vector v(3,2,1). The line c contains the point C(-2,2,-1) and has a direction vector w(1,-1,2). Find the common perpendicular to the lines b and c. |
The vector PQ must be perpendicular to v and to w. The conditions are:
PQ . v = 0 <=> ... <=> -14 r + 3 s = 11 PQ . w = 0 <=> ... <=> - 3 r + 6 s = 7From this system we find r = -3/5 and s = 13/15.
x = -4/5 - r y = 4/5 + r z = 2/5 + r
| ux + vy + wz = 0 | ux + vy + wz + t = 0It is easy to see that this system has no solution fot t not zero. The planes are parallel.
| c_{1} c_{2} c_{3} | | x y z | | x' y' z'|These cofactors are
(y z' - y' z) -(x z' - x' z) (x y' - x' y)Example
The cross product of a(1,3,2) and b(-1,2,4) is the vector with coordinates the cofactors of * in the determinant
| * * * | | 1 3 2 | | -1 2 4 |We find (8,-6,5). The vector c(8,-6,5) is orthogonal with a and c.
The line b contains the point B(1,2,1) and has a direction vector v(3,2,1). The line c contains the point C(-2,2,-1) and has a direction vector w(1,-1,2). Find the common perpendicular d to the lines b and c. |
The plane beta through b and with direction (-1,1,1) is
| x - 1 y - 2 z - 1 | | 3 2 1 | = 0 | -1 1 1 | <=> x - 4 y + 5 z + 2 = 0The plane gamma through c and with direction (-1,1,1) is
| x + 2 y - 2 z - 1 | | 1 -1 2 | = 0 | -1 1 1 | <=> x + y = 0The common perpendicular d is the intersection line of these planes. The equations of d are :
/ x - 4 y + 5 z + 2 = 0 \ x + y = 0
P(a,a',a") is in the plain ux + vy + wz = 0 <=> u.a + v.a'+ w.a"= 0 <=> The vectors P(a,a',a") and N(u,v,w) are orthogonalHence, the vector N(u,v,w) is orthogonal to all the vectors P in plane ux + vy + wz = 0. The direction of N is orthogonal to this plane and to all parallel planes. N is called a normal vector to these planes. Conclusion:
The direction of vector N(u,v,w) is orthogonal to the plane ux + vy + wz + t = 0 This vector is a normal vector to that plane. |
Example 1:
The plane 3x + 2y + z -12 = 0 has a normal vectors (3, 2, 1) , (-6, -4, -2) ....
Example 2:
Given: point P(1,2,3) and the vector v(4,5,6).
The plane through point P and with normal vector v is
4 (x-1) + 5 (y-2) + 6 (z-3) = 0 <=> 4 x + 5 y + 6 z - 32 = 0
Example 1:
m are n are different real parameters different from zero. Find the values of m and n such that the following planes are parallel. m x + m y + n z = 1 (1) n x + n y + m z = 1 (2) |
A normal vector to plane (1) is a(m, m, n).
A normal vector to plane (2) is b(n, n, m).
The planes are parallel <=> a and b have the same direction <=> the rank of the following matrix is 1 [ m m n ] [ n n m ] <=> / m n - n m = 0 \ m^{2} - n^{2} = 0 <=> m^{2} - n^{2} = 0 and since m and n are different <=> m + n = 0Conclusion : The planes are parallel if and only if m+ n = 0.
Example 2:
Find the m-values such that the following system does not represent a line.
/ x - my = 2-m \ 3x - mz = 6 - 3m |
The system does not represent a line <=> The planes x - my = 2-m and 3x - mz = 6 - 3m are parallel <=> The normal vectors of the planes have the same direction <=> the rank of the following matrix is 1 [ 1 -m 0 ] [ 3 0 -m ] <=> 3m = 0 en m^{2} = 0 <=> m = 0
Property:
If two planes are orthogonal, a normal vector to one plane is a direction vector of
the other plane.
Example:
Find the angle between
the plane x + y + z = 4 and the plane x + 2y +3z = 5 |
n(1,1,1) is normal to the first plane.
m(1,2,3) is normal to the second plane.
cos(m,n) = (m.n)/(||m||.||n||) = 0.925
The sharp angle between m and n is 22.2 degrees.
The angle between the planes is 22.2 degrees.
Example:
Find the angle between the plane x + y +z = 4 and the line [x = 1 + r ; y = 1+ 2r ; z = 1 + 3r]. |
n(1,1,1) is normal to the plane.
m(1,2,3) is a direction vector of the line.
cos(m,n) = (m.n)/(||m||.||n||) = 0.925
The sharp angle between m and n is 22.2 degrees.
The sharp angle between the line and the plane is 67.8 degrees.
3x -2 y + 7z - 4 = 0 2x - 5y + 3z - 5 = 0In fact, this line is the intersection of the planes [3x -2 y + 7z - 4 = 0 ; 2x - 5y + 3z - 5 = 0]
General formulation
The direction vector of the line / u x + v y + w z + t = 0 \ u'x + v'y + w'z + t'= 0 is the direction of the cross product of the vectors (u , v, w) and ((u', v', w') |
The line m has equations [2x + ry + 1 = 0 ; x + 3z + 1 = 0] The line n has equations [x + y + rz + 3 = 0 ; x - 2y + r = 0] Find the r-values such that the lines m and n are orthogonal. |
A direction vector of n is given by the cross product of (1,1,r) and (1,-2,0). We find (2r, r, -3).
The lines m and n are orthogonal if and only if the dot product of their direction vectors is zero. This condition is
6r^{2} + 6r + 3r = 0 <=> 6r^{2} + 9r = 0 <=> 3r(2r+3) = 0 <=> r = 0 of r = -3/2Example 2:
The line m has equations [x - r y = 2 - r ; 3x - r z = 6 - 3r] The line n has equations [x + 2z = 1 ; y - (r-1)z = 0 ] r is different from 0. Find the r-values such that the lines m and n are parallel. |
A direction vector of n is given by the cross product of (1, 0, 2) and (0, 1, 1-r). We find (-2, r-1, 1).
m and n are parallel <=> m and n have the same direction <=> The following matrix has rank = 1 [ r 1 3] [ -2 r-1 1] <=> / r+6 = 0 \ 1-3(r-1) = 0The last system has no solution.
Example 0.5x -0.5y +(1/sqrt(2))z + 5 = 0 is a normal equation of a plane.
r^{2} (u^{2} + v^{2} + w^{2} ) = 1 <=> r =+1/sqrt(u^{2} + v^{2} + w^{2} ) or r =-1/sqrt(u^{2} + v^{2} + w^{2} )Usually we choose the + sign.
Take the plane x + 2y + 2z - 1 = 0 . 1 ---(x + 2y + 2z - 1) = 0 is a normal equation of that plane. 3
| l.a + m.a' + n.a" + t |Example 1:
1 |---(1.1 + 2.2 + 2.3 - 1) | = 10/3 3Example 2:
The vertices of the triangle base of a pyramid are A(0,0,0) ; B(1,2,3) ; C(3,4,4). The fourth vertex is T(2,5,8). Find the volume of the pyramid TABC. |
We first calculate the area of triangle ABC with Heron's formula c = |AB| = sqrt(14) b = |AC| = sqrt(41) a = |BC| = 3 The semiperimeter of the triangle is s = 6.57 With Heron's formula, the area of the triangle ABC is sqrt(s(s-a)(s-b)(s-c))= 3.354 The height of the pyramid is the distance from T to the plane ABC. The plane ABC has equation 4x - 5y + 2z = 0 The distance from T to the plane ABC is | 4 . 2 - 5 . 5 + 2 . 8| -------------------------- = 1/sqrt(45) sqrt(45) The volume of the pyramid is (1/3) (Area of triangle ABC). height = 0.1666
Example:
We start with two planes 2 x - y + 2 z + 5 = 0 and x - 2 y - 2 z - 3 = 0.
P(x,y,z) is equidistant from the given planes <=> 2 x - y + 2 z + 5 x - 2 y - 2 z - 3 | ----------------- | = | ------------------- | sqrt(4 + 1 + 4) sqrt(1 + 4 + 4) <=> | 2 x - y + 2 z + 5 | = | x - 2 y - 2 z - 3 | <=> 2 x - y + 2 z + 5 = ± (x - 2 y - 2 z - 3) <=> x + y + 4 z + 8 = 0 or 3 x - 3 y + 2 = 0
The line d has equations
x + y + z -3 = 0 2x - y + z -1 = 0Find the planes through the line d such that the distance from P(2,1,1) to these planes is equal to one. |
All the planes through the line d form a bundle. A variable plane of this bundle is
(x + y + z -3) + t(2x - y + z -1) = 0 <=> (1 + 2t)x + (1 - t)y + (1 + t)z -3 -t = 0t is parameter. We look for the values of t such that the distance from P to this plane is one. The condition is :
(1 + 2t).2 + (1 - t) + (1 + t) -3 - t --------------------------------------------- = 1 sqrt( (1 + 2t)^{2} + (1 - t)^{2} + (1 + t)^{2} )We find the values : t_{1} = 0.55 and t_{2} = -1.22
The required planes are:
(x + y + z -3) + 0.55 (2x - y + z -1) = 0 and (x + y + z -3) - 1.22 (2x - y + z -1) = 0
Find the distance from point P(1,1,1) to the line with parametric equations
x = 2 + r y = 3 + 2r z = 1 + r |
2 + 6 + 1 - 4 5 | ---------------- | = ------- = |DS| sqrt(6) sqrt(6)
Example:
To find the direction of line a we take two simple points on line a.
A_{1}(1,0,1) and A_{2}(0, -1,-1). The direction of line a is (1,1,2).
The plane through line b and parallel to line a has the equation
| x y+2 z | | 1 1 1 | = 0 | 1 1 2 | <=> x - y - 2 = 0Now, we calculate the distance from A_{1} to this plane.
1 - 0 - 2 1 | -------------- | = ------ sqrt(1 + 1 + 0) sqrt(2)1/sqrt(2) is the distance between the lines a and b.
So, V = P + P'
Example:
Relative to a given orthonormal basis V = V(3,5,2).
The plane has as equation x + y -2 z = 0.
We calculate the orthogonal projection of V on that plane.
N(1,1,-2) is a normal vector of the plane. The projection P' of V on N is 4 N /6 = 2 N /3 P'(2/3, 2/3, -4/3) The projection P of V on the plane is V(3,5,2) - P'((2/3, 2/3, -4/3) = P(7/3, 13/3, 10/3)The orthogonal projection of vectors on a plane through point O is a linear transformation of the three dimensional space. With this projection corresponds a matrix A_{o} and with this matrix we can find the projection of any vector in a simpel way.
Finding the matrix A_{o} is not obvious. For this purpose we rely on the knowledge about changing the basis of the vector space and its consequences on the matrix of the projection.
We start by choosing 3 new base vectors. As base vectors we choose
two independent vectors in the given plane and as third vector
we choose a normal vector to that plane. For example:
( A(1,-1,0) ; B(2,0,1) ; N(1,1,-2) )
Relative to this new basis all vectors have new coordinates. From the theory of the vector spaces we know that these new coordinates are connected to the old coordinates by a matrix C. The columns of C are the coordinates of the new basis vectors relative to the original (old) basis. In our example is
[ 1 2 1] C = [-1 0 1] [ 0 1 -2]Relative to this new basis the linear transformation has a simple matrix. The columns of this matrix are the new coordinates of the projected new basis vectors.
new basisvector new coordinates of the projected basis vector ---------------- ---------------------------------------------- A (1,0,0) B (0,1,0) N (0,0,0) The matrix of the projection relative to the new basis is [1 0 0] A_{n} = [0 1 0] [0 0 0] Between the old and the new matrix of the projection we have the connection A_{n} = C^{-1} A_{o} C <=> A_{o} = C A_{n} C^{-1} With this we calculate the matrix A_{o} of the projection relative to the natural basis. We find: [ 5 -1 2] A_{o} = (1/6) [-1 5 2] [ 2 2 2]Once this result is found, it is very easy to find the reflection of any vector.
As an example, we retake the vector V(3,5,2) from above. The image is :
[ 5 -1 2] [3] [ 7/3] (1/6) [-1 5 2] [5] = [13/3] [ 2 2 2] [2] [10/3]So,in a very simple way, we find the same result as above.
To find the reflection V ' it is sufficient to find P (see previous paragraph).
Example:
Relative to a given orthonormal basis V = V(3,5,2).
The plane has as equation x + y -2 z = 0.
We calculate the reflection V ' of V in the given plane.
In the previous paragraph, we have found that te projection of V on the plane is equal to P(7/3, 13/3, 10/3).
V' = 2 P - V 2 (7/3, 13/3, 10/3) - (3, 5, 2) = (5/3, 11/3, 14/3) V' = V'(5/3, 11/3, 14/3)The orthogonal reflection of vectors in a plane through O is a linear transformation in a three dimensional space.
With this reflection corresponds a matrix A_{o} and with this matrix we can calculate the reflection of any vector in a simple way.
Finding the matrix A_{o} is not obvious. For this purpose we rely on the knowledge about changing the basis of the vector space and its consequences on the matrix of the projection.
We start by choosing 3 new base vectors. As base vectors we choose
two independent vectors in the given plane and as third vector
we choose a normal vector to that plane. For example:
( A(1,-1,0) ; B(2,0,1) ; N(1,1,-2) )
Relative to this new basis all vectors have new coordinates. From the theory of the vector spaces we know that these new coordinates are connected to the old coordinates by a matrix C. The columns of C are the coordinates of the new basis vectors relative to the original (old) basis. In our example is
[ 1 2 1] C = [-1 0 1] [ 0 1 -2]Relative to this new basis the linear transformation has a simple matrix. The columns of this matrix are the new coordinates of the mirrored new basis vectors.
new basis vector new coordinates of the mirrored basis vector ------------------- -------------------------------------------- A (1,0, 0) B (0,1, 0) N (0,0,-1) The matrix of the reflection relative to the new basis is [1 0 0] A_{n} = [0 1 0] [0 0 -1] Between the old and the new matrix of the reflection we have the connection A_{n} = C^{-1} A_{o} C <=> A_{o} = C A_{n} C^{-1} With this we calculate the matrix A_{o} of the reflection with respect to the natural basis. We find: [ 2 -1 2 ] A_{o} = (1/3) [ -1 2 2 ] [ 2 2 -1 ]Once this result is found, it is very easy to find the reflection of any vector.
As an example, we retake the vector V(3,5,2) from above. The image is :
[ 2 -1 2 ] [3] [ 5/3] (1/3) [ -1 2 2 ] [5] = [11/3] [ 2 2 -1 ] [2] [14/3]So,in a very simple way, we find the same result as above.