Complex Numbers

Chapter 3

Analytical Solid Geometry

3.3     Parallel, Skew and Perpendicular Lines

Parallel lines:

fig 3.10

If P is (x1, y1, z1) and Q is (x2, y2, z2) with directed values, (PQ) = (l, m, n) = (x2 - x1, y2 - y1, z2 - z1) then segment PQ and segment OA where A = (l, m, n) are parallel and PQ = OA as shown in the given figure.

So line PQ and line OA are parallel if they have same directed values. But if directed values of a line are given by (l, m, n), then (kl, km, kn) are also directed values of the line. Therefore line PQ and line OA are parallel if their directed values multiples of each other by a real number k. Any line parallel to the line OA is also parallel to the line PQ, so that

Two lines are parallel if and only if their directed values are multiples of each other by some real number.


For example, if line PQ has a directed values (2, 3, 5) and line RS has a directed values (4, 6, 10) then Pq ∥ RS.

Skew Lines: In space, there are pair of lines that are neither parallel nor intersect. These pairs of lines are called skew lines.

Example 5.

Given P(2, 1, 3), Q(6, -5, 4), R(2, 3, 4) and S(-1, 5, 1), determine whether the lines PQ and RS are parallel or skew or intersect.

Solution
For P(2, 1, 3) and Q(6, -5, 4),
(PQ) = (4, -6, 1)

For R(2, 3, 4)and S(-1,5,1),
(RS) = (-3, 2, -3) are not parallel.
Since 4 ⁄ -3 ≠ - 6 ⁄ 2 ≠ 1 ⁄ -3, directed values of PQ are not multiple of RS.
So the two lines are not parallel.

If a point (x, y, z) is on the line PQ and RS, then
  x = 2 + 4s             x = 2 - 3t
y = 1 - 6s y = 3 + 2t
z = 3 + s z = 4 - 3t
for real numbers s and t.
Then we have the following system of three equations
2 + 4s = 2 - 3t
1 - 6s = 3 + 2t
3 + s = 4 - 3t
Solving the first two of these equations, we have s = - 3 ⁄ 5 and t = 1 ⁄ 5. But
3 + (- 3 ⁄ 5) ≠ 4 - 3( 4 ⁄ 5)
these values of s and t do not satisfy the last equation.
So the system of equations has no solution and hence the given lines do not intersect each other. Therefore the given lines are skew.

Finding the measure of ∠PAQ : Consider P(x1, y1, z1), A(a, b, c) and Q(x2, y2, z2),

(AP) = (l1,   m1,   n1) = (x1 -a,   y1 - b,   z1 - c)
(AQ) = (l2,   m2,   n2) = (x2 - a,   y2 - b,   z2 - c)
(PQ) = (l3,   m3,   n3)
= (x2 - x1,   y2 - y1,   z2 - z1)
= (l2 - l1,   m2 - m1,   n2 - n1)
(AP)2 + (AQ)2 - (PQ)2 = l12 + m12 + n12   +   l22 + m22 + n22   -   (l2 - l1)2 - (m2 - m1)2 - (n2 - n1)2
= 2(l1l2 + m1m2 + n1n2)

By the law of cosines,

cos ∠PAQ = (AP)2 + (AQ)2 - (PQ)2 ⁄ 2 . AP . AQ
= 2 (l1l2 + m1m2 + n1n2) ⁄ 2 . AP . AQ
= l1l2 + m1m2 + n1n2 ⁄ √ l12 + m12 + n12 √ l22 + m22 + n22

If l1l2 + m1m2 + n1n2 = 0, then cos ∠PAQ = 0 and hence ∠PAQ = 90°.

Perpendicular lines: Two lines are perpendicular if and only if they are intersect and
l1l2 + m1m2 + n1n2 = 0
for any directed values (l1, m1, n1) and (l2, m2, n2) of the line.

Exampole 6.
Given P(0, 0, 1), Q(3, 6, 4), R(0, 3, 1) and S(3, 0, 4), show that lines PQ and RS are perpendicular.

Solution
Given P(0, 0, 1) and Q(3, 6, 4), we have (PQ) = (3, 6, 3)
Given P(0, 3, 1) and Q(3, 0, 4), we have (RS) = (3, -3, 3)
And we have 3(3) + 6(-3) + 3(3) = 0
If a point (x, y, z) is on the line PQ and RS then
  x = 0 + 3s             x = 0 + 3t
y = 0 + 6s y = 3 - 3t
z = 1 + 3s z = 1 + 3t
for real numbers s and t. Then we have the following system of three equations
3s = 3t
6s = 3 - 3t
1 + 3s = 1 + 3t
Solving the first two of these equations, we have s = 1 ⁄ 3 and t = 1 ⁄ 3. The two lines intersect each other and the point of intersection is (1, 2, 2). Hence, the two lines are perpendicular to each other.

Example 7.
Find the equation of the line passing through the point (-4, 7, -3) and perpendicular to the line
(x, y, z) = (3 + 2k,   -1 + 3k,   1 - k)
Find also the point of intersection of two lines.

Solution
Directed value of the given lines are (2, 3, -1).
Directed value of the required lines are
(-4 - (3 + 2k),   7 - (-1 + 3k),   -3 - (1 - k) )
    = ( -7 - 2k,   8 - 3k,   -4 + k)
For some real number k.
If the two lines are perpendicular, then
2 (-7 - 2k)   +   3 (8 - 3k)   +   (-1) (-4 + k) = 0
              -14 - 4k   +   24 - 9k   +   4 - k = 0
                                                14 k = 14
                                                        k = 1

So directed values of the required line are (-9, 5, -3) and the equation of the line is
(x, y, z) = (-4 - 9t,   7 + 5t,   -3 - 3t)
The point of intersection is
(x, y, z) = (5, 2, 0)

Exercise 3.3
  1. Find cos ∠PAQ for the followings.
    2. (a)  P(1, 2, -1),   A(-2, 1, 5),   Q(2, -1, 0)
    (b)   P(0, 2, -3),   A(2, -1, 5),   Q(-2, 3, -1)

  2. Determine whether the line PQ and RS are parallel or skew or intersect. If PQ and RS intersect, are they perpendicular?
    3. (a)   P(1, 2, 3),   Q(4, 5, 6),   R(-2, 3, 5),   S(4, 9, 11)
    (b)   P(3, -1, -3),   Q(2, -3, 1),   R(3, -2, 5),   S(-1, -2, 1)
    (c)   P(4, -2, 5),   Q(-2, 6, 1),   R(-1, 1, 4),   S(3, 3, 2)
    (d)   P(-3, -1, 6),   Q(-1, 3, 0),   R(0, 6, 7),   S(-4, -4, -1)

  3. Find the equation of the line passing through the point (8, -1, -10) and perpendicular to the line
    (x, y, z) = (1 + 2k,   2 - k,   3 - 7k)
    Find also the point of intersection of two lines.

Complex Numbers

Exercise 3.3
1.   Find cos ∠PAQ for the followings.
(a)  P(1, 2, -1),   A(-2, 1, 5),   Q(2, -1, 0)
Solution
x1 = 1,   y1 = 2,   z1 = -1 and a = -2,   b = 1,   c = 5

AP = (l1, m1, n1)
= (x1 - a,   y1 - b,   z1 - c)
= (1 - (-2),   2 - 1,   -1 - 5 )
= (3, 1, -6)

x2 = 2,   y2 = -1,   z2 = 0
AQ = (l2, m2, n2)
= (x2 - a,   y2 - b,   z2 - c)
= (2 - (-2),   -1 - 1,   0 - 5 )
= (4, -2, -5)

PQ = (l3, m3, n3)
= (x2 - x1,   y2 - y1,   z2 - z1)
= (2 - 1,   -1 - 2,   0 - (-1) )
= (1, -3, 1)

cos ∠PAQ = l1l2 + m1m2 + n1n2 ⁄ √ l12 + m12 + n12 √ l22 + m22 + n22
= 3 ⋅ 4   +   1 ⋅ (-2)   +   (-6) (-5) ⁄ √ 32 + 12 + (-6)2 √ 42 + (-2)2 + (-5)2
= 12 - 2 + 30 ⁄ √ 9 + 1 + 36 √ 16 + 4 + 25
= 40 ⁄ √ 46 √ 45
= 40 ⁄ √ 2070         (Calculator or logarithms can be used.)
= 0.8792

(b)   P(0, 2, -3),   A(2, -1, 5),   Q(-2, 3, -1)

Solution
x1 = 0,   y1 = 2,   z1 = -3 and a = 2,   b = -1,   c = 5
AP = (l1, m1, n1)
= (x1 - a,   y1 - b,   z1 - c)
= (0 - 2,   2 - (-1),   -3 - 5 )
= (-2, 3, -8)

x2 = -2,   y2 = 3,   z2 = -1
AQ = (l2, m2, n2)
= (x2 - a,   y2 - b,   z2 - c)
= (-2 - 2,   3 - (-1),   -1 - 5 )
= (-4, 4, -6)

PQ = (l3, m3, n3)
= (x2 - x1,   y2 - y1,   z2 - z1)
= (-2 - 0,   3 - 2,   -1 - (-3) )
= (-2, 1, 2)

cos ∠PAQ = l1l2 + m1m2 + n1n2 ⁄ √ l12 + m12 + n12 √ l22 + m22 + n22
= -2 ⋅ (-4)   +   3 ⋅ 4   +   (-8) (-6) ⁄ √ (-2)2 + 32 + (-8)2 √ (-4)2 + 42 + (-6)2
= 8 + 12 + 48 ⁄ √ 4 + 9 + 64 √ 16 + 16 + 36
= 68 ⁄ √ 77 √ 68
= 68 ⁄ √ 5236         (Calculator or logarithms can be used.)
= 0.9397



2.   Determine whether the line PQ and RS are parallel or skew or intersect. If PQ and RS intersect, are they perpendicular?

(a)   P(1, 2, 3),   Q(4, 5, 6),   R(-2, 3, 5),   S(4, 9, 11)
Solution
For P(1, 2, 3) and Q(4, 5, 6)
PQ = (3, 3, 3)

For R(-2, 3, 5) and S(4, 9, 11)
RS = (6, 6, 6)

Since 3⁄6 = 3⁄6 = 3⁄6, directed values of PQ are multiple of RS, the two lines are parallel.

∴ The two lines are not intersect.
∴ The two lines are not skew.

(b)   P(3, -1, -3),   Q(2, -3, 1),   R(3, -2, 5),   S(-1, -2, 1)
Solution
For P(3, -1, -3) and Q(2, -3, 1)
PQ = (-1, -2, 4)

For R(3, -2, 5) and S(-1, -2, 1)
RS = (-4, 0, -4)

Since -1⁄-4 ≠ -2⁄0 ≠ 4⁄-4, directed values of PQ are not multiple of RS.
So the two lines are not parallel.
If point (x, y, z) is on the line PQ and RS, then
x = 3 - s           x = 3 - 4t
y = -1 - 2s           y = -2
z = -3 + 4s           z = 5 - 4t
for real number s and t.

3 - s = 3 - 4t
-s = - 4t
s = 4t       eq_______ (1)

-1 - 2s = -2
-2s = -1
s = 1 ⁄ 2

Substituting s = 1 ⁄ 2 in eq_______(1),
1 ⁄ 2 = 4t
t = 1 ⁄ 8

-3 + 4s = 5 - 4t       eq_______(3)
Substituting s = 1 ⁄ 2 and t = 1⁄8 in eq_______(3),
-3 + 4 (1⁄2) = 5 - 4 (1⁄8)
-3 + 2 = 5 - 1 ⁄ 2
-1 = 9 ⁄ 2       (impossible)
-1 ≠ 9 ⁄ 2

The value of s and t do not satisfy the equation. So the system of equation has no solution.
∴ The two lines are not intersect.
∴ The two lines are skew.

(c)   P(4, -2, 5),   Q(-2, 6, 1),   R(-1, 1, 4),   S(3, 3, 2)
Solution
For P(4, -2, 5) and Q(-2, 6, 1)
PQ = (-2 - 4,   6 - (-2),   1 - 5)
    = (-6, 8, -4)

For R(-1, 1, 4) and S(3, 3, 2)
RS = (3 - (-1),   3 - 1,   2 - 4)
    = (4, 2, -2)

Since -6⁄4 ≠ 8⁄2 ≠ -4⁄-2, directed values of PQ are not multiple of RS.
So the two lines are not parallel.
If point (x, y, z) is on the line PQ and RS, then
x = 4 - 6s           x = -1 + 4t
y = -2 + 8s           y = 1 + 2t
z = 5 - 4s           z = 4 - 2t
for real number s and t.

4 - 6s = -1 + 4t
4 = -1 + 6s + 4t
5 = 6s + 4t
6s + 4t = 5       eq_______ (1)

-2 + 8s = 1 + 2t
-2 + 8s - 2t = 1
8s - 2t = 3       eq_______ (2)

5 - 4s = 4 - 2t
5 - 4s + 2t = 4
-4s + 2t = -1       eq_______ (3)

eq_______(2) + eq_______(3)

  8s - 2t = 3
-4s + 2t = -1
  4s       = 2
  s   = 2⁄4
      = 1⁄2

Substituting s = 1 ⁄ 2 in eq_______(1),
6s + 4t = 5
6 (1 ⁄ 2) + 4t = 5
3 + 4t = 5
  4t = 2
    t = 2⁄4
      = 1⁄2
The value of s and t satisfy the equation.
x = 4 - 6s
= 4 - 6 (1⁄2)
= 4 - 3
= 1

y = -2 + 8s
= 2 + 8 (1⁄2)
= -2 + 4
= 2

z = 5 - 4s
= 5 - 4 (1⁄2)
= 5 - 2
= 3

The point of intersection is (1, 2, 3).
∴ The two lines are intersect.
∴ The two lines are not skew.
l1l2 + m1m2 + n1n2 = (-6) 4   +   8 ⋅ 2   +   (-4)(-2)
= -24 + 16 + 8
= 0
∴ The two lines are perpendicular.

(d)   P(-3, -1, 6),   Q(-1, 3, 0),   R(0, 6, 7),   S(-4, -4, -1)

Solution
For P(-3, -1, 6) and Q(-1, 3, 0)
PQ = (-1 - (-3),   3 - (-1),   0 - 6)
    = (2, 4, -6)

For R(0, 6, 7) and S(-4, -4, -1)
RS = (-4 - 0,   -4 - 6,   -1 - 7)
    = (-4, -10, -8)

Since 2⁄-4 ≠ 4⁄-10 ≠ -6⁄-8, directed values of PQ are not multiple of RS.
So the two lines are not parallel.
If point (x, y, z) is on the line PQ and RS, then
x = -3 + 2s           x = -4t
y = -1 + 4s           y = 6 - 10t
z = 6 - 6s           z = 7 - 8t
for real number s and t.

-3 + 2s = -4t
2s + 4t = 3       eq_______(1)

-1 + 4s = 6 - 10t
4s + 10t = 7       eq_______(2)

6 - 6s = 7 - 8t
-6s + 8t = 1       eq_______(3)

eq_______(1) × 3   +   eq_______(3)

  6s + 12t = 9
-6s +   8t = 1
          20t = 10
          t = 10⁄20
        t = 1⁄2
Substituting t = 1⁄2 in eq_______(1),
2s + 4t = 3
2s + 4 (1⁄2) = 3
2s + 2 = 3
2s = 1
s = 1⁄2

Substituting s = 1⁄2 and t = 1⁄2 in eq_______(2),
4s + 10t = 7
4 (1⁄2) + 10 (1⁄2) = 7
2 + 5 = 7

The value of s and t satisfy the equation.
x = -3 + 2s
= -3 + 2 (1⁄2)
= -2
y = -1 + 4s
= -1 + 4 (1⁄2)
= -1 + 2
= 1

z = 6 - 6s
= 6 - 6 (1⁄2)
= 6 - 3
= 3

The point of intersection is (-2, 1, 3).
∴ The two lines are intersect.
∴ The two lines are not skew.

l1l2 + m1m2 + n1n2 = 2 (-4)   +   4 (-10)   +   (-6) (-8)
= -8 - 40 + 48
= 0
∴ The two lines are perpendicular.

3.   Find the equation of the line passing through the point (8, -1, -10) and perpendicular to the line
(x, y, z) = (1 + 2k,   2 - k,   3 - 7k)
Find also the point of intersection of two lines.

Solution
(x, y, z) = (1 + 2k,   2 - k,   3 - 7k)
= (1 + 2k,   2 - 1k,   3 - 7k)
Directed values of the given line (line1) are (2, -1, -7)
(x, y, z) = (1 + 2k,   2 - k,   3 - 7k)
(x1, y1, z1) = (8, -1, -10)
line2 = (x1 - x,   y1 - y,  z1 - z)
= (8 - (1 + 2k),   -1 - (2 - k),   -10 - (3 - 7k) )
= (8 - 1 - 2k,   -1 - 2 + k,   -10 - 3 + 7k)
= (7 - 2k,   -3 + k,   -13 + 7k)
Directed values of the required line are (7 - 2k,   -3 + k,   -13 + 7k) for some real number k.
If the two lines are perpendicular then
                      (l1l2 + m1m2 + n1n2) = 0
2 (7 - 2k)   +   (-1 (-3 + k))   +   (-7 (-13 + 7k)) = 0
2 (7 - 2k)   -   1 (-3 + k)   -   7 (-13 + 7k) = 0
                  14 - 4k   +   3 - k   +   91 - 49k = 0
                                            108 - 54k = 0
                                                        -54k = -108
                                                                  k = - 108⁄- 54
= 2
So the directed values of the required line (line2) are
(7 - 2k,   -3 + k,   -13 + 7k) = (7 - 2(2),   -3 + 2,   -13 + 7(2))
= 7 - 4,   -1,   -13 + 14)
= (3, -1, 1)

And the equation of the line passing through the point (8, -1, -10) is
(8 + 3t,   -1 - 1t,   -10 + 1t)
= (8 + 3t,   -1 - t,   -10 + t).
∴ The points of intersection of the two line are
Œ
(x, y, z) = (1 + 2k,   2 - k,   3 - 7k)
= (1 + 2(2),   2 - 2,   3 - 7(2) )
= 1 + 4,   0,   3 - 14)
= 5, 0, -11)

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