Analytical Solid Gepmetry

Chapter 4

Vector Algebra


We have learned vectors in two-dimensional rectangular coordinates system. In this chapter, we will extend the system to three dimensions and we will use vectors to calculate angles, distances and areas. We will also apply vectors to geometrical problems of lines and planes.

4.1   Vectors in Three Dimensions

In the plane, each point is associated with an ordered pair of real numbers. In space, each point is associated with an ordered triple of real numbers. Through a fixed point, called the origin O, draw three mutually perpendicular lines:   the x-axis, the yaxis and the z-axis. A point P in space is determined by an ordered triple (x, y, z) of real numbers as shown in the diagram. These numbers x, y, z are called the coordinates of P.

fig 4.1-1

Example 1.
Illustrate the points (a) A(0, 3, 0)   (b) B(4, 0, 2)   (c) C(-1, 2, 2)
Solution
fig 4.1-2
(a)
fig 4.1-3
(b)
fig 4.1-4
(c)


To represent vectors in space, we introduce the standard unit vectors î, ĵ and k̂ where
î = 1 0 0 0 1 0 0 0 1 ĵ = k̂ = represent the unit vector in the positive y direction, represents the unit vector in the positive x direction, represent the unit vector in the positive z direction,
fig 4.1-5
Position Vectors in Three Dimensions
If OA⟶ is a vector with initial point at the origin O and terminal point at point A (2, 3, 4), then we can represent OA→ in terms of the vector î, ĵ and k̂ as
OA⟶   = 2î + 3ĵ + 4k̂ .
We can also represent vector in three dimensions using column vector as like in two dimensions.
OA = 2 3 4 ⟶
fig 4.1-6
The numbers in each column are called the components of the vector.

A vector whose initial point at the origin is called a position vector. For example, the position of point P(a, b, c) in the diagram can be represented by its position vector, OP⟶. So,
    p→   = OP→
= a b c
          = a î + b ĵ + c k̂   is the position vector of the point P.
fig 4.1-7


The magnitude of a vector OP→   =     p→  
= a b c is given by
p |     | ⟶ √ = a   +   b   + c 2 2 2
In general, for the position vectors of two points A(x1, y1, z1) and B(x2, y2, z2), we can find the vector AB→   as follows:
AB→   =   OB→   - OA→
= x   -   x 2 1 y   -   y 2 1 z   -   z 2 1
And AB→   is called the position vector of B relative to A.

Example 2.
If P is (-3, 1, 2) and Q is (1, -1, 3), find:
(a) OP→           (b) PQ→           (c) |PQ→ |
(d) QP→           (e) |QP→ |

Solution
(a)
OP = -3 1 2 ⟶

(b) PQ→   = OQ→   - OP→  
- = 1 -1 3 -3 2 1 = 1   +   3 -1   -   1 3   -   2 = 4 -2 1

(c)
|       | → PQ √ = 4   +   (-2)   + 1 2 2 2 = √ 21
(d)   QP⟶   = OP⟶   - OQ⟶
- = -3 1 2 1 3 -1 = -3   -   1 1   +   1 2   -   3 = -4 2 -1

(e)
|       | → QP √ = (-4)   +   2   + (-1) 2 2 2 = √ 21

Algebraic Operation with Vectors

Negative Vector
In the diagram, the vector   - a⟶   has the same length as the vector   a⟶   but the opposite direction.
a → → -a

When   a⟶   is given in component form, the components of   - a⟶   are the same as those for a⟶   but with their sign reversed. So
= - 11 -6 5 -11 -5 6

Zero Vector
In three dimensional space, the zero vector is denoted by
0 = 0 0 0 ⟶

The rule for algebra with vectors extend from two dimensions to three dimensions:
If ⟶ a   = x y z 1 1 1 and ⟶ b = x y z 2 2 2 , then ⟶ a   + ⟶ b   = x y z 1 1 1 + x y z 2 2 2 = x   +   x y   +   y z   +   z 1 1 1 2 2 2 , ⟶ a   - ⟶ b   = x y z 1 1 1 - x y z 2 2 2 = x   -   x y   -   y z   -   z 1 1 1 2 2 2 and k   a ⟶ = kx ky kz 1 1 1 for any scalar k.

The following figures give the geometric interpretation of two vectors for addition, subtraction and scalar multiplication of two vectors, respectively.
fig 4.1-8
Example 3.
If ⟶ p   = 1 -1 4 and ⟶ q = -2 0 2 , find: (a) ⟶ ⟶ p   +   q (b) ⟶ ⟶ p   -       q 1 2 (c) ⟶ ⟶ q   -   p 3 2

Solution
(a) ⟶ p   + ⟶ q   = 1 -1 4 + -2 0 2 = 1 - 2 -1 + 0 4 + 2 = -1 -1 6 (b) ⟶ p   - ⟶ q   = 1 2 - 1 2 1 -1 4 - -2 0 2 = 1 + 1 -1 - 0 4 - 1 = 2 -1 3

(c) ⟶ q   - ⟶ p   = 3 2 - 3 2 -2 0 2 1 -1 4 = -3 - 1 0 + 1 3 - 4 = -4 1 -1

Equal Vectors and Parallel Vectors

Two vectors are equal if they have the same magnitude and direction.
So, if arrows are used to represent vectors, then equal vectors are parallel and equal in length. This means that equal vector arrows are translations of one another, but in space.
a → → a → a

If ⟶ a   = x y z 1 1 1 and ⟶ b = x y z 2 2 2 , then ⟶ a   = ⟶ b if and only if
x1 = x2,   y1 = y2,   z1 = z2.
In this diagram, we see that
⟶ ⟶ a   = b   , ⟶ p   = ⟶ q  .
p → → q → a → b

Two vectors ⟶ ⟶ a and b are parallel if and only if ⟶ a   = ⟶ k   b   for some scalar k.

Example 4.
Find u and v given that
$$\text{Find} \, u \, \text{and} \, v\ \text{given} \, \text{that} \, \vec a = \begin{pmatrix}-1\\ -1\\ u\end{pmatrix} \text{is \, parallel \, to} \, \vec b = \begin{pmatrix}v \\ 2 \\ -2 \end{pmatrix}. $$ Solution
$$\text{Since}\, \vec a \, \text{and} \, \vec b \, \text{are \, parallel}\,, \text{we \, have}\, \vec a = k \vec b \, \text{for\, some\, scalar}\, k.$$ $$ \begin{pmatrix}-1\\-1\\u\end{pmatrix} = k \begin{pmatrix}v\\2\\-2\end{pmatrix}.$$ $$-1 = kv, \, \, \, -1=2k \, \, \, and \, \, u=-2k.$$ We have $$k = - \frac{1}{2} $$ Thus u = 1 and v = 2.

Example 5.
ABCD is a parallelogram. A is (-1, 1, 1), B is (2, 0, -2), and D is (3, 1, 4). Find the coordinates of C.
Solution
Let C be (x, y, z).
Since ABCD is a parallelogram, AB \\ DC, and they have the same length, so
$$\overrightarrow{DC} = \overrightarrow{AB}$$ $$\overrightarrow{OC}-\overrightarrow{OD} = \overrightarrow{OB}-\overrightarrow{OA}$$ $$\begin{pmatrix}x\\y\\z\end{pmatrix} - \begin{pmatrix}3\\1\\4\end{pmatrix} = \begin{pmatrix}2\\0\\-2\end{pmatrix} - \begin{pmatrix}-1\\1\\1\end{pmatrix}$$ $$\begin{pmatrix}x - 3\\y - 1\\z - 4\end{pmatrix} = \begin{pmatrix}3\\-1\\-3\end{pmatrix}$$ x - 3 = 3,
x = 6
y - 1 = -1,
y = 0
z - 4 = -3,
z = 1.
Therefore C is (6, 0, 1).
A(-1,1,1) B(2,0,-2) D(3,1,4) C(x, y, z)

Unit Vector
A vector with magnitude 1 is called a unit vector.

$$\text{For\, a\, nonzero\, vector} \,\vec a, \, \text{the\, unit\, vector}$$ $$ \text{in\, the\, (same)\, direction\, of}\, \vec a,\, \text{denoted\, by}\, \hat a\, \text{is\, given\, by}$$ $$\hat a = \frac{\vec a}{|\vec a|}$$
Example 6.
(a) Find the unit vector in the same direction as
$$\vec a = \begin{pmatrix}2\\-2\\1\end{pmatrix}.$$ $$\text{(b)\, \, Find\, a\, vector\, of\, magnitude\, 5\, that\, is\, parallel\, to}\,\vec a.$$ Solution
(a) $$|\vec a| = \sqrt{2^2 + (-2)^2 + 1^2} = 3.$$ $$\text{The\, unit\, vector\, of}\, \vec a \, \text{is}$$ $$\hat a = \frac{\vec a}{|\vec a|} = \frac{1}{3} \begin{pmatrix}2\\-2\\1\end{pmatrix}$$ $$= \begin{pmatrix}\frac{2}{3}\\ - \frac{2}{3}\\ \frac{1}{3}\end{pmatrix}.$$ $$\text{(b)\, \, Let}\, \vec b \, \text{be\, parallel\, to}\, \vec a \, \text{and}\, |\vec b| = 5.$$ Then $$\vec b = 5\hat a = \begin{pmatrix}\frac{10}{3}\\ - \frac{10}{3}\\ \frac{5}{3}\end{pmatrix}$$ or $$\vec b = -5\hat a = \begin{pmatrix}- \frac{10}{3}\\ \frac{10}{3}\\ - \frac{5}{3}\end{pmatrix}.$$
Collinear Points
Three of more points are said to be collinear if they lie on the same straight line.
A, B and C are collinear if $$\overrightarrow{AB} = k\overrightarrow{BC}$$ for some scalar k.
A B C

Example 7.
Prove that A(8, 2, 2), C (20, 5, 5) and B (12, 3, 3) are collinear.
Solution
$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$$ $$=\begin{pmatrix}12\\3\\3\end{pmatrix} - \begin{pmatrix}8\\2\\2\end{pmatrix} = \begin{pmatrix}4\\1\\1\end{pmatrix}.$$ $$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}$$ $$=\begin{pmatrix}20\\5\\5\end{pmatrix} - \begin{pmatrix}12\\3\\3\end{pmatrix} = \begin{pmatrix}8\\2\\2\end{pmatrix}.$$ $$= 2\begin{pmatrix}4\\1\\1\end{pmatrix}$$ $$= \overrightarrow{AB}.$$ Therefore BC is parallel to AB.
Since B is common to both, it follows that A, B and C are collinear.

Exercise 4.1

1.   Let ⟶ a   = 5 -2 -4 , ⟶ b = 3 -6 1 and ⟶ c   = 0 7 -1 , Find the following vectors: (a) ⟶ 3 a (b) → 4 b (c) ⟶ a   -   b → (d) ⟶ b   +   c → (e) → 2 b   +   c → (f) → a   -   2 b → (g) → a   +   b   -   2 c → → (h) → 3 a   -   b   +   c → →
Solution
(a) ⟶ 3 a   = 3 5 -2 -4 = 15 -6 -12
(b) ⟶ 4 b   = 4 3 -6 1 = 12 -24 4
(c) → a   -   b → = 5 -2 -4 - 3 -6 1 = 2 4 -5
(d) → b   +   c → = 3 -6 1 + 0 7 -1 = 3 1 0
(e) → 2 b   +   c → = 2 3 -6 1 + 0 7 -1 = 6 + 0 -12 + 7 2 - 1 = 6 -5 1
(f) → a   -  2 b → = 2 5 -2 -4 - 3 -6 1 = 5 - 6 -2 + 12 -4 - 2 = -1 10 -6
(g) → a   + b   - 2 c → → = 5 -2 -4 + 3 -6 1 - 2 0 7 -1 = 5 + 3 - 0 -2 - 6 - 14 -4 + 1 + 2 = 8 -22 -1
(h) → 3 a   - b   + c → → =   3 5 -2 -4 - 3 -6 1 + 0 7 -1 = 15 - 3 + 0 -6 + 6 + 7 -12 - 1 - 1 = 12 7 -14

2.   Given vectors
⟶ a   = 1 2 7 , ⟶ b = -3 4 2 and ⟶ c   = -2 p q . (a) Find the values of p and q such that → c is parallel to → a . (b) Find the value of scalar k such that → a + k b → is parallel to vector 0 10 23 .
Solution
(a)   c⟶   is parallel to   a⟶   .
∴   c⟶   = k   a →   for some scalar k.
= k -2 p q 1 2 7 = -2 p q 1k 2k 7k
k = -2,
p = 2k
    = 2 (-2)
    = -4

q = 7k
    = 7 (-2)
    = -14

(b) → a   +  k b → = k 1 2 7 + -3 4 2 = 1 - 3k 2 + 4k 7 + 2k → a   +  k b → is parallel to 0 10 23
∴ → a   +  k b → = t 0 10 23 = 0 10t 23t = 1 - 3k 2 + 4k 7 + 2k 0 10t 23t
1 - 3k = 0 ,     2 + 4k = 10t ,     7 + 2k = 23t
-3k   = -1
k   =     1/3

Substituting k =   1/3 in   2 + 4k = 10t

2 + 4 1 3 = 10 t 6 + 4 3 = 10t 10 3 ⋅ 10 = t t = 1 3
Substituting k = 1 3 in 7 + 2k = 23t 7 + 2 1 3 = 23 t 21 + 2 3 = 23t 23 3 ⋅ 23 = t t = 1 3 ∴ k = 1 3
3.   Points A, B, C and D have position vectors
⟶ a   = 3 -1 1 , ⟶ b = 5 0 3 , ⟶ c   = 7 8 -3 and ⟶ d = 4 3 -2 ,
respectively.
Point E is the midpoint of BC.
(a)   Find the position vector of E.
(b)   Show that ABED is a parallelogram.
Solution
(a)   Point E is the midpoint of BC.
⟶ OE   = 1 2 ⟶ (OB   +   OC) ⟶
= 1 2 5 0 3 + 7 8 -3 = 1 2 12 8 0 = 6 4 0
(b)   AB⟶   = AO⟶   + OB⟶
          = OB⟶   -   OA →
= 5 0 3 - 3 -1 1 = 2 1 2
DE⟶   = DO⟶   + OE⟶
          = OE⟶   -   OD →
= 6 4 0 - 4 3 -2 = 2 1 2
DE = AB and DE ∥ AB
Hence ABED is a parallelogram.

4.   Points A, B and C have position vectors
⟶ a   = 2 -1 4 , ⟶ b = 5 1 2 and ⟶ c   = 3 1 4 , respectively.
Find the position vector of point D such that ABCD is a parallelogram.
Solution
ABCD is a parallelogram.
AB⟶   = DC⟶  
AO⟶   +   OB⟶   =   DO →   +   OC⟶  
OB⟶   - OA⟶   =   OC⟶   -   OD →  
5 1 2 - 2 -1 4 = 3 1 4 - x y z = 3 - x 1 - y 4 - z 3 2 -2
3 = 3 - x
x = 0

2 = 1 - y
y = -1

-2 = 4 - z
z = 6
∴ The position vector of point D is
⟶ d   = 0 -1 6
5.   K(1, -1, 0), L(4, -3, 7) and M(a, 2, b) are collinear. Find a and b.
Solution
K, L and M are collinear.
∴ KL⟶   = t LM⟶  
KO⟶   +   OL →   = t (LO⟶   +   OM⟶  )
OL⟶   -   OK⟶   = t (OM →   - OL →  )
= t 4 -3 7 1 -1 0 - a 2 b - 4 -3 7 t 3 -2 7 = a - 4 2 + 3 b - 7
3 = (a - 4)t,   -2 = 5t,   7 = (b - 7)t
  t = -2/5

(a - 4)t = 3
(a - 4) (-2/5) = 3
a - 4 = 3 × (-5/2)
a - 4 = -15/2
a = 4 - (15/2)
    = (8 - 15)/2
    = -7/2

(b - 7)t = 7
(b - 7) (-2/5) = 7
b - 7 = 7 × (-5/2)
b - 7 = -35/2
b = 7 - (35/2)
    = (14 - 35)/2
    = -21/2