Algebra
Linear Algebra
© The scientific sentence. 2010

Linear Algebra
Line and plane
Exercise 1
The parametric equation of the line is:
(x + 1)/2 = (y + 1)/3 = (z  1)/2 (L)
The Cartesian equation of the plane is:
3 x  y + 2 z  5 = 0 (P)
We want to find the components of the point of intersection of
the line and the plane, that is the point where the lane crosses
the plane.
Two methods:
1) Using the parameter of the parametric equations:
(x + 1)/2 = (y + 1)/3 = (z  1/)/ 2 = t
The parametric equations take the form:
x = 2 t  1
y = 3 t  1
z =  2 t + 1
Substituting these values in the equation (P) of the
plane, we obtain:
3(2 t  1)  (3 t  1) + 2 (  2 t + 1) = 5
That is: t =  5
With this value, the parametric equations gives:
x = 2 (5)  1 =  11
y = 3 (5)  1 =  16
z =  2 (5) + 1 = 11
x =  11
y =  16
z = 11
That are the components of the point intersection
of the line (L) and the plane (P).
2) Using Cramer's rules and determinants:
To do that, we will first find the system of equations related to
the two equations (L) and (P).
Let's transform the (L) equations:
(x + 1)/2 = (y + 1)/3
3 x + 3 = 2 y + 2
3 x  2 y =  1
(y + 1)/3 = (z  1)/2
 2 y  2 = 3 z  3
 2 y  3 z =  1
2 y + 3 z = 1
With the equation (P), we obtain the system:
3 x  2 y + 0 z =  1
0 x + 2 y + 3 z = 1
3 x  y + 2 z = 5
We use Cramer's rule to solve this system:
The determinant Δ is
3  2 0 =  1
0 + 2 + 3 = 1
3  1 + 2 = 5
= 3[(2)(2)  (1)(3)] +  2[(3)(3)  (0)(2)] + 0[(0)(1)  (2)(3)] =
3[7]  2[9] + 0 = 3
The determinant Δ_{x} is
1  2 0
1 + 2 + 3
5  1 + 2
= 1[(2)(2)  (1)(3)] +  2[(3)(5)  (1)(2)] + 0 =
1[7]  2[15  2] =  33
The determinant Δ_{y} is
3  1 0
0 + 1 + 3
3 5 + 2
= 3[(1)(2)  (3)(5)] +  1[(3)(3)  (0)(2)] + 0 =
3[ 13]  1[9] =  48
The determinant Δ_{z} is
3  2 1
0 + 2 + 1
3  1 + 5
= 3[(2)(5)  (1)(1)] +  2[(3)(1)  (0)(5)] + 1[(0)(1)  (2)(3)] =
3[(11] +  2[(3] + 1[ 6] = 33
Therefore:
x = Δ_{y}/Δ = 33/3 = 11
x = Δ_{y}/Δ = 36/3 =  16
x = Δ_{y}/Δ = 33/3 = + 11
x =  11
y =  16
z = 11
That are the components of the point intersection
of the line (L) and the plane (P).
It is the same result found with the previous method.
check your results with: this program
Exercise 2
The plane (P) has the equation:
2 x  3 y + 5 z = 15 (P)
The parametric equation of line (L) are:
x = 1 + 2t
y = 1 + 3t
z = 2 + t
That is
(x  1)/2 = (y + 1)/3 = (z  2)/1 (L)
We want to show that the plane (P) contains the
line (L) by two methods:
a) Searching for an intersection possible point;
Let' s transform this double equation to
two Cartesian equations, with the one
of the plane (P) will give the system, as
previously:
(x  1)/2 = (y + 1)/3
3 x  3 = 2 y + 2
3 x  2 y = 5
(x  1)/2 = (z  2)/1
x  1 = 2 z  4
x  2 z =  3
Therefore: (L) and (P) give the system:
3 x  2 y = 5
x  2 z =  3
2 x  3 y + 5 z = 15
The related determinant is:
3  2 0
1 0  2
2  3 + 5
= 3[(0)(5)  (3)(2)] + 2[(2)(2)  (1)(5)] + 0[(1)(3)  (2)(0)] =
3[ 6)] +  2[ 9)] =  18 + 18 = 0
Therefore, we have no solutions for the system. That is, there
is no intersection point of the line and the plane, hence
they are parallel.
b) Showing that two points from the line are also on the
plane
The first point is (+1, 1, +2) .
This point satisfy the equation (L)
(1  1)/2 = (1 + 1)/3 = (+2  2)/1 (L)
This point satisfy the equation (P)
2 (1) 3 (1) + 5 (+2) = 15 (P)
The second point is (+3, +2, +3) .
(3  1)/2 = (2 + 1)/3 = (3  2)/1 (L)
2 (3) 3 (2) + 5 (3) = 15 (P)
That also satisfies the equation (L) and y the equation (P)
Two different points are located both on the line (L) and
the plane (P), so the plane (P) contains the line (L).
Exercise 3
a.
We have two lines (L1) and (L2) defined by their
parametric equations:
For (L1):
x = 4 + 4 t
y = 1 + 2 t
z = 8 + 6 t
That is:
(x  4)/4 = (y  1)/2 = (z  8 )/ 6 (L1)
For (L2)
x = 6 t
y = 1 + 3 t
z = 2 + 9t
That is:
x/6 = (y  1)/3 = (z  2)/9 (L2)
We see that the respective director numbers of the two
lines (L1) and (L2) are
proportional.
Indeed:
4/6 = 2/3 = 6/9 = 3/2
therefore
(L1) and (L2) are parallel.
b.
Two line (L1) and (L2)
(L1) has the following parametric equations:
x = 4 t + 4
y =  2 t
z = 3
(L2) is the intersection of two planes (P1) and (P2)
(P1): x + y  z = 0
(P2): 2 x + 3 y + z = 5
From (P1) y =  x + z
substituting this value in (P2) gives
2 x  3 x + 3 z + z = 5
 x + 4 z = 5
→ (x + 5)/4 = z
From (P1) y =  x + z = 5  4 z + z = 5  3 z
→ (y  5)/3 = z
The parametric equation of the line (L2) is
then:
(x + 5)/4 = (y  5)/3 = z/1 = s
x = 4 s  5
y =  3 s + 5
z = s
The two line (L1) and (l2) are not parallel because
their direction vectors are not proportional.
If they intersect , we will have
x = 4 t + 4 = 4 s  5 or 4t = 4s  9 (1)
y =  2 t =  3 s + 5 (2)
z = 3 = s
(1) → 4t = 4x3  9 = 3 → t = 3/4
(2) →  2 t =  3x3 + 5 → t = 2
The values of t and s are not the same, so the two lines
(L1) and (l2) do not intersect
Since they are not parallel and they do not intersect,
then they must be skew.
c.
The parametric equations of the two lines
(L1) and (L2) are:
(x + 9)/3 = (y  4)/ 2 = (z  5)/1 = (L1)
(x)/2 = (y  6)/4 = (z  1)/3 = (L2)
(L1) and (l2) are not parallel because their
direction vectors (denominators) are not proportional.
(x + 9)/3 = (y  4)/ 2 → 2 x + 3 y =  6 (1)
(x + 9)/3 = (z  5)/1 → x  3 z =  24 (2)
(x)/2 = (y  6)/4 → 2 x + y = 6 (3)
(x)/2 = (z  1)/3 → 3 x  2 z =  2 (4)
(2) → x = 3 z  24
(3) → x = (6  y)/2 in (1) gives y =  6 then
(3) gives x = 6
(2) gives z = 10
x = 6 and z = 10 is verified by (4)
Therefore the intersection point of the two line (L1) and
(L2) is the point (6, 6, 10)
Exercise 4
We want to determine the relative position of two planes:
parallel and distinct, coincident or intersecting.
a)
 x  y  √2 z + 2 = 0 (P1)
x + y + √2 z = 3 (P2)
We have
1 /1 = 1/1 =  √2/√2 =  1
That is the ratio of the respective components of the
two related direction vectors are proportional. Therefore
(P1) and (P2) are parallel.
If we multiply by  1 the equation of (P1), we obtain:
x + y + √2 z  2 = 0 (P1) or
x + y + √2 z = 2 (P1)
Compared to the equation (P2), we find 2 = 3,
which is an impossible result. So The planes (P1) and (P2)
are parallel and distinct.
b)
(1/2) x + y + (1/3) z  7 = 0 (P1)
(3/2) x + 3 y + z  21 = 0 (P2)
We have
(1/2)/(3/2) = 1/3 = (1/3)/1 = 1/3
The ratio of the respective components of the
two related direction vectors are proportional. Therefore
(P1) and (P2) are parallel.
We can simplify, by dividing by 3, the equation of
(P2) and obtain:
(1/2) x + y + (1/3) z  7 = 0 (P2)
This exactly the equation of (P1). Therefore the two
planes (P1) and (P2) are coincident.
c)
2 x + 2 y + z  5 = 0 (P1)
3 x  2 y  2 z  7 = 0 (P2)
These two planes are not parallel, they are coincident.
Indeed:
The vector n1 (2, 2, 1) is normal to the plane (P1)
The vector n2 (3,  2,  2) is normal to the plane (P2)
But the dot product n1.n2 = 0 , that is n1 is perpendicular to n2.
n1 ⊥ n2 and n1 ⊥ (P1) and n2 ⊥ (P2) → (P1)⊥ (P2)
Two points of intersection are evident:
x = 0 gives:
2 y + z  5 = 0 (1)
 2 y  2 z  7 = 0 (2)
That gives the components of the first point A1:
x = 0, y = 17/2, z =  12
y = 0 gives:
2 x + z  5 = 0 (3)
3 x  2 z  7 = 0 (4)
That gives the components of the second point A2:
x = 17/7 , y = 0 , z = 1/7
Therefore
The director vector of the line intersection (L)
of the two planes (P1) and (P2) is:
A1A2 (17/7  0, 0  17/2, 1/7 + 12) = (17/7,  17/2, 84/7)
The parametric equation of the line (L) is:
(x  0)/(17/7) = (y  17/2)/(17/2) = (z + 12)/(84/7)
d)
3 x  2 y = 0 (P1)
z = 4 (P2)
These two planes are not parallel, they are coincident.
Indeed:
The vector n1 (3,  2, 0) is normal to the plane (P1)
The vector n2 (0, 0, 1) is normal to the plane (P2)
But the dot product n1.n2 = 0 , that is n1 is
perpendicular to n2.
n1 ⊥ n2 and n1 ⊥ (P1) and n2 ⊥ (P2)
→ (P1)⊥ (P2)
We can use another method:
If n1 ⊥ n2, then their cross product is
a vector on the line intersection (L), and it is its director
vector, so
n = n1 x n1 =
i j k
3  2 0
0 0 1
= (2,  3, 0)
An evident point is A (2, 3, 4). It satisfies (P1) and (p2).
Hence
The parametric equations of the line (L) are:
(x  2)/(2) = (y  3)/(3) = t
or
x  2 =  2t
y  z =  3t
z = 4
Exercise 5
The shortest distance between:
a) The point A(1, 2, 3) and the plane 2 x  2 y + z = 4
D = [(2)(1) + (2)(2) + (1)(3)  4]/[(2)^{2} + (2)^{2}
+ (1)^{2}]^{1/2} = 5/3
b)The two following parallel planes:
2 x  3 y + 4 z  7 = 0 (P1)
4 x  6 y + 8 z  3 = 0 (P2)
We determine a point on the plane (P1) and solve
as previously.
So
x = 0 and y = 1 gives:
 3 + 4 z  7 = 0 (P1)
Then z = 5/2, and the point on the plane (P1) is
A(0, 1, 5/2)
Therefore, the distance is between A and (P2), that
is the distance between (P1)and(P2) is:
D = [(4)(0) + (6)(1) + (8)(5/2)  3]/[(4)^{2} + (6)^{2}
+ (8)^{2}]^{1/2} = 11/√116.
c) The point P(1, 0, 2) and the line
that crosses the point Po(1,  1, 1) and
parallel to the vector d(1,  2,  2)
Let M(x, y, z) a point on the line. As Po is
also on the line, we have PoM = (x  1, y + 1, z  1).
This vector is parallel to the vector d.
The vector PPo is equal to
PPo = (1  1,  1  0, 1  2) = (0, 1 , 1)
The distance is given by
D = (vector PPo) x (vector d)/d
We first calculate the cross product (vector PoP) x (vector d):
(vector PoP) x (vector d) =
i j k
0  1  1
1  2  2
(2  2) i + ( 1  0) j + (0 + 1) k =
0 i + ( 1) j + (1) k
Therefore
D = [o^{2} + (1)^{2} + 1^{2}]^{1/2}/
[1^{2} + (2)^{2} + (2)^{2}]^{1/2}
= √2/3.
d)
The lines:
(x  6)/1 = (y  5)/1 = (z  5)/3 (L1)
(x  3)/1 = (y + 1)/1 = (z  2)/ 3 (L2)
The two line (L1) and (L2) are
parallel.
First, we find a point on the line (L1) such as
P1 (6, 5, 5), as search for the distance as previously.
The point on the line (L2) is P2(3,  1, 2).
The vector b (1, 1, 3) is a vector parallel to (L2).
Then
P2P1 = ( 3,  6,  3)
We have the distance D = (vector P2P1) x (vector b)/b
(vector P2P1) x (vector b) =
i j k
3  6  3
1 1  3
(18 + 3) i + ( 3  9) j + ( 3  6) k =
21 i + ( 12) j + ( 9) k
Therefore
D = [21^{2} + (12)^{2} + ( 9)^{2}]^{1/2}/
[1^{2} + (1)^{2} + (23)^{2}]^{1/2}
= √530/√530 = 1.


