Chapter 3

Analytical Solid Geometry

3.2     Lines

Directed value of a line segment: For a line segment PQ, directed values (l, m, n) of PQ, where P is (x1, y1, z1) and Q is (x2, y2, z2) is defined by
  (PQ) = (l,m,n) = (x2 - x1, y2 - y1, z2 - z1)  

Then the length of the segment PQ is

  PQ = √ (x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2 = √ l2 + m2 + n2  


Real numbers and points on the line: For every real number k, there is a point R on the line PQ with respect to point P and k vice versa.

Case 1. 0 ≤ k ≤ 1

Point R is between P and Q such that PR = kPQ.

Draw PA, RC and QB perpendicular to xy-plane:    AB is projection of PQ on xy-plane.

Draw PT ∥ AB and RS ∥ AB.

fig 3.7

Then TS ⁄ TQ = PR ⁄ PQ
z - z1⁄ z2 - z1 = k
          z = z1 + k(z2 - z1)
          z = z1 + kl
Similarly, by drawing the projection of PQ on yz-plane,
x - x1 ⁄ x2 - x1 = k
          x = x1 + k(x2 - x1)
          x = x1 + kl
and by drawing the projection of PQ on a zx-plane,
y - y1 ⁄ y2 - y1 = k
          y = y1 + k(y2 - y1)
          y = y1 + kl

Case 2. k > 1

Point R is after Q such that PR = kPQ.

Draw PT ∥ AC and QS ∥ AC.
fig 3.8

TR⁄ TS = PR⁄PQ
z - z1 ⁄ z2 - z1 = k
        z = z1 + k(z2 - z1)
        z = z1 + kn

Case 3. k < 0.

Point R is before P such that RP = -kPQ.
Draw PT ∥ CB and RS ∥ CB.
fig 3.9

ST ⁄ TQ = RP⁄ PQ
z1 - z ⁄ z2 - z1 = k
        z = z1 + k(z2 - z1)
        z = z1 + kn

Similarly, for Case 2 and Case 3,
x - z1 ⁄ x2 - x1 = k
        x = x1 + k(x2 - x1)
        x = x1 + kl

y - y1 ⁄ y2 - y1 = k
        y = y1 + k(y2 - y1)
        y = y1 + km
Therefore the coordinates of point R on the line PQ with respect to the point P and a real number k are

    (x, y, z) = (x1 + kl, y1 + km, z1 + kn)    

This equation is called coordinate form of the equation of line PQ and k is called a parameter.
In general, if a segment of a line through (x, y, z) has directed values (l, m, n) which are not equal to 0, then the equation of the line can be written as

    x - x1 ⁄ l = y - y1 ⁄ m = z - z1 ⁄ n    

Since
(PR) = (x1 + kl - x1,   y1 + km - y1,   z1 + kn - z1)
        = (kl, km, kn)

and R is any point on the line PQ with (PQ) = (l, m, n), we may define directed values of the line PQ as (kl, km, kn) where k is a real number.

Example 2.

Given P(1, 2, 3) and Q(3, 6, 5), find the coordinates of point R(x, y, z) on the line PQ with rspect to the point P and the following parameters.

(a) k = 1 ⁄ 2       (b) k = 2       (c) k = -2

Solution

Given P(1, 2, 3) and Q(3, 6, 5), we have (PQ) = (l, m, n) = (2, 4, 2)

(a) k = 1⁄ 2
(x, y, z) = (1 + 1⁄ 2(2),   2 + 1 ⁄ 2(4),   3 + 1 ⁄ 2(2) ) = (2, 4, 4)

(b) k = 2
(x, y, z) = (1 + 2(2),   2 + 2(4),   3 + 2(2) ) = (5, 10, 7)

(c) k = -2
(x, y, z) = (1 + (-2)(2),   2 + (-2)(4),   3 + (-2)2 ) = (-3, -6, -1)

Example 3.
Given P(-1, 2, 3) and Q(3, 5, -2), determine whether or not the following points are on the line PQ. If the point is on the line PQ, Find the corresponding parameter with respect to the point P.

(a) (1, 7 ⁄ 2, 1 ⁄2)         (b) (7, 8, -7)
(c) (-5, -1, 8)         (d) (7, 8, -2)
Solution
Given P(-1, 2, 3) and Q(3, 5, -2), we have (PQ) = (l, m, n) = (4, 3, -5). The equation of the line PQ is
x + 1 ⁄ 4 = y - 2 ⁄ 3 = z - 3 ⁄ -5

(a) If (x, y, z) = (1, 7 ⁄ 2, 1 ⁄2),
x + 1 ⁄4 = 1 + 1 ⁄4 = 1 ⁄2
y - 2 ⁄3 = 7 ⁄2 - 2 ⁄3 = 1 ⁄2
z - 3 ⁄-5 = 1 ⁄2 - 3 ⁄-5 = 1 ⁄2
∴ x + 1 ⁄4 = y - 2 ⁄3 = z - 3 ⁄-5 for (x, y, z) = (1, 7 ⁄2, 1 ⁄2).

Therefore the point (1, 7 ⁄2, 1 ⁄2) is on the line PQ, corresponding parameter is 1 ⁄2.

(b) If (x, y, z) = (7, 8, -7),
x + 1 ⁄4 = 7 + 1 ⁄4 = 2
y - 2 ⁄3 = 8 - 2 ⁄3 = 2
z - 3 ⁄-5 = -7 - 3 ⁄-5 = 2
x + 1 ⁄4 = y - 2 ⁄3 = z - 3 ⁄-5 for (x, y, z) = (7, 8. -7)
Therefore the point (7, 8, -7) is on the line PQ, corresponding parameter is 2.

(c) If (x, y, z) = (-5, -1, 8),
x + 1 ⁄4 = -5 + 1 ⁄4 = -1
y - 2 ⁄3 = -1 - 2 ⁄3 = -1
z - 3 ⁄-5 = 8 - 3 ⁄-5 = -1
x + 1 ⁄4 = y - 2 ⁄3 = z - 3 ⁄-5 for (x, y, z) = (-5, -1, 8).
Therefore the point (-5, -1, 8) is on the line PQ, corresponding parameter is -1.

(d) If (x, y, z) = (7, 8, -2),
x + 1 ⁄4 = 7 + 1 ⁄4 = 2
y - 2 ⁄3 = 8 - 2 ⁄3 = 2
z - 3 ⁄-5 = -2 - 3 ⁄-5 = 1
x + 1 ⁄4 = y - 2 ⁄3 ≠ z - 3 ⁄-5 for (x, y, z) = (7, 8, -2)
Therefore the point (7, 8, -2) is not on the line PQ.

Example 4.
Given P(2, 1, 3) and Q(6, -5, 3), determine whether or not the following points are on the line PQ. If the point is on the line PQ, find the corresponding parameter with respect to the point P.

(a) (4, -2, 3)         (b) (-2, 7, 3)
(c) (10, -11, 3)         (d) (1, 1, 3)

Solution
Given P(2, 1, 3) and Q(6, -5, 3), we have (PQ) = (l, m, n) = (4, -6, 0).
Then the coordinates of the point (x, y, z) on the line PQ are
(x, y, z) = (2 + 4k, 1 - 6k, 3)
This means that the line PQ is on the plane z = 3.
(a) If (x, y, z) = (4, -2, 3), then
(4, -2, 3) = (2 + 4k, 1 - 6k, 3)
we have k = 1 ⁄2. Therefore the point (4, -2, 5) is on the line PQ with corresponding parameter 1 ⁄2.

(b) If (x, y, z) = (-2, 7, 3), then
(-2, 7, 3) = (2 + 4k, 1 - 6k, 3)

We have k = -1. Therefore the point (-2, 7, 3) is on the line PQ with corresponding parameter -1.
(c) If (x, y, z) = (10, -11, 3), then
(10, -11, 3) = (2 + 4k, 1 - 6k, 3)
We have k = 2. Therefore the point (10, -11, 3) is on the line PQ with corresponding parameter 2.

(d) If (x, y, z) = (1, 1, 3), then
(1, 1, 3) = (2 + 4k, 1 - 6k , 3)
There is no value k that satisfies this condition. Therefore the point (1, 1, 3) is not on the line PQ.

Exercise 3.2
  1. Given P(3, 1, 5) and Q(-3, 7, -2), find the coordinates of point R(x, y, z) on the line PQ with respect to the point P and the following parameters.
    2. (a) k = 1 ⁄ 2         (b) k = 3         (c) k = -2

  2. Given P(-2, 1, 3) and Q(4, 4, -3), determine whether of not the following points are on the line PQ. If the point is on the line PQ, find the corresponding parameter with respect to the point P.
    3. (a) (6, 3, -6)         (b) (6, 5, -5)
    (c) (-4, 0, 5)         (d) (7, 8, -2)

  3. Given P(3, 2, -1) and Q(4, 2, 5), determine whether or not the following points are on the line PQ. If the point is on the line PQ, find the corresponding real number with respect to the point P.
    4. (a) (5, 2, 11)         (b) (2, 2, -7)
    (c) (7 ⁄ 2, 2, 2)         (d) (6, 2, 10)

  4. Find the points of intersection of the line joining the two points (2, 4, 5) and (3, 5, -4) with the following planes.
    5. (a) xy-plane         (b) yz-plane         (c) zx-plane


Analytical Solid Geometry

Answers

Exercise 3.2
1.   Given P(3, 1, 5) and Q(-3, 7, -2), find the coordinates of point R(x, y, z) on the line P2 with respect to the point P and the following parameters.
(a) k = 1 ⁄ 2         (b) k = 3         (c) k = -2

Solution
Given P(3,1, 5) and   Q(-3, 7, -2), we have
(PQ) = (l, m, n) = (x2 - x1,   y2 - y1,   z2 - z1) = (-6, 6, -7)
The coordinate of point R on the line PQ with respect to the point P and a real number k are
(x, y, z) = (x1 + kl,   y1 + km,   z1 + kn)
(a)   k = 1 ⁄ 2
(x, y, z) = (3 + 1⁄ 2 (-6),   1 + 1⁄2 (6),   5 + 1⁄2(-7) )
= (3 + (-3),   1 + 3,   5 - 7 ⁄ 2 )
= (0,   4,   3⁄ 2 )
(b)   k = 3
(x, y, z) = (3 + 3(-6),   1 + 3(6),   5 + 3(-7) )
= (-15, 19, -16)

(c)   k = -2
(x, y, z) = (3 + (-2)(-6),   1 + (-2)(6),   5 + (-2)(-7))
= (3+12,   1-12,   5+14)
= (15, -11, 19)

2.   Given P(-2, 1, 3) and Q(4, 4, -3), determine whether of not the following points are on the line PQ. If the point is on the line PQ, find the corresponding parameter with respect to the point P.
(a) (6, 3, -6)         (b) (6, 5, -5)
(c) (-4, 0, 5)         (d) (7, 8, -2)

Solution
Given P(-2, 1, 3) and Q(4, 4, -3), we have
PQ = (l, m, n) = (x2 - x1,   y2 - y1,   z2 - z1) = (6, 3, -6)
The equation of the line PQ is
x - x1⁄ l = y - y1⁄ m = z - z1⁄ n
x - (-2)⁄ 6 = y - 1⁄ 3 = z - 3⁄ -6
x + 2⁄ 6 = y - 1⁄ 3 = z - 3⁄-6
(a)   If (x, y, z) = (6, 3, -6),
x + 2⁄ 6 = 6 + 2⁄ 6 = 8⁄ 6 = 4⁄3
y - 1⁄ 3 = 3 - 1⁄ 3 = 2⁄ 3
z - 3⁄ -6 = -6 -3 ⁄ -6 = -9⁄ -6 = 3⁄2
x + 2⁄ 6 ≠ y - 1⁄ 3 ≠ z - 3⁄ -6 for (x, y, z) = (6, 3, -6).
∴ The point (6, 3, -6) is not on the line PQ.

(b)   If (x, y, z) = (6, 5, -5)
x + 2⁄ 6 = 6 + 2⁄ 6 = 8⁄ 6 = 4⁄3
y - 1⁄ 3 = 5 - 1⁄ 3 = 4⁄3
z - 3⁄ -6 = -5 -3 ⁄ -6 = -8⁄ -6 = 4⁄3
x + 2⁄ 6 = y - 1⁄ 3 = z - 3⁄ -6 for (x, y, z) = (6, 5, -5).
∴ The point (6, 5, -5) is on the line PQ , the corresponding parameter is 4⁄3.

(c)   If (x, y, z) = (-4, 0, 5)
x + 2⁄ 6 = -4 + 2⁄ 6 = -2⁄ 6 = -1⁄3
y - 1⁄ 3 = 0 - 1⁄ 3 = -1⁄3
z - 3⁄ -6 = 5 -3 ⁄ -6 = 2⁄ -6 = - 1⁄3
x + 2⁄ 6 = y - 1⁄ 3 = z - 3⁄ -6 for (x, y, z) = (-4, 0, 5).
∴ The point (-4, 0, 5) is on the line PQ , the corresponding parameter is -1⁄3.

(d)   If (x, y, z) = (7, 8, -2)
x + 2⁄ 6 = 7 + 2⁄ 6 = 9⁄ 6 = 3⁄2
y - 1⁄ 3 = 8 - 1⁄ 3 = 7⁄3
z - 3⁄ -6 = -2 -3 ⁄ -6 = -5⁄ -6 = 5⁄6
x + 2⁄ 6 ≠ y - 1⁄ 3 ≠ z - 3⁄ -6 for (x, y, z) = (7, 8, -2).
∴ The point (7, 8, -2) is not on the line PQ.

3.   Given P(3, 2, -1) and Q(4, 2, 5), determine whether or not the following points are on the line PQ. If the point is on the line PQ, find the corresponding real number with respect to the point P.
(a) (5, 2, 11)         (b) (2, 2, -7)
(c) (7 ⁄ 2, 2, 2)         (d) (6, 2, 10)

Solution
Given P(3, 2, -1) and Q(4, 2, 5), we have
PQ = (l, m, n) = (x2 - x1,   y2 - y1,   z2 - z1) = (1, 0, 6)
The coordinate of the point (x, y, z) on the line PQ are
(x, y, z) = (3+1k,   2+0k,   -1+6k)
= (3+k,   2,   -1+6k
This means that line PQ is on the plane y = 2.
(a)   If (x, y, z) = (5, 2, 11),
(3+k,   2,   -1+6k) = (5, 2, 11)
3 + k = 5
k = 5 - 3
k = 2
-1 + 6k = 11
6k = 11 + 1
6k = 12
k = 2
The point (5, 2, 11) is on the line PQ with corresponding parameter 2.

(b)   If (x, y, z) = (2, 2, -7),
(3+k,   2,   -1+6k) = (2, 2, -7)
3 + k = 2
k = 2 - 3
k = -1
-1 + 6k = -7
6k = -7 + 1
6k = -6
k = -1

The point (2, 2, -7) is on the line PQ with corresponding parameter -1.

(c)   If (x, y, z) = (7⁄ 2, 2, 2),
(3+k,   2,   -1+6k) = ( 7⁄ 2 , 2, 2)
3 + k = 7⁄ 2
k = 7⁄ 2 - 3
k = 7 - 6⁄ 2
= 1⁄ 2
-1 + 6k = 2
6k = 2 + 1
6k = 3
k = 3⁄ 6
= 1⁄ 2

The point ( 7⁄ 2, 2, 2) is on the line PQ with corresponding parameter 1⁄ 2 .

(d)   If (x, y, z) = (6, 2, 10),
(3+k,   2,   -1+6k) = (6, 2, 10)
3 + k = 6
k = 6 - 3
k = 3
-1 + 6k = 10
6k = 10 + 1
6k = 11
k = 11⁄6

The point (6, 2, 10) is not on the line PQ.

4.   Find the points of intersection of the line joining the two points (2, 4, 5) and (3, 5, -4) with the following planes.
(a) xy-plane         (b) yz-plane         (c) zx-plane

Solution
Let P = (2, 4, 5) and Q = (3, 5, -4)
PQ = (l, m, n) = (x2 - x1,   y2 - y1,   z2 - z1) = (1, 1, -9)

(a)   The point of intersection of the line PQ with xy-plane is
(x, y, z) = (2+k, 4+k, 5-9k)
For xy-plane, z = 0
5 - 9k = 0
-9k = -5
k = -5⁄-9
= 5⁄9

(x, y, z) = (2 + 5⁄9,   4 + 5⁄9 ,  5 - 9 ⋅ 5⁄9 )
= (18 + 5⁄9,   36 + 5⁄9 ,   0)
= (23⁄9,   41⁄9 ,   0)


(b)   The point of intersection of the line PQ with yz-plane is
(x, y, z) = (2+k, 4+k, 5-9k)
For yz-plane, x = 0
2 + k = 0
k = -2
(x, y, z) = (2 + (-2),   4 + (-2),   5 - 9(-2))
= (2 - 2,   4 - 2,   5 + 18)
= (0, 2, 23)


(c)   The point of intersection of the line PQ with zx-plane is
(x, y, z) = (2+k, 4+k, 5-9k)
For zx-plane, y = 0
4 + k = 0
k = -4
(x, y, z) = (2 + (-4),   4 + (-4),   5 - 9(-4))
= (2 - 4,   4 - 4,   5 + 36)
= (-2, 0, 41)


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