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This result demonstrates that the method for adding algebraic vectors in is the
same as in R 2 . Adding two vectors means adding their respective components. It
should also be noted that the result for the subtraction of vectors in is analogous
to the result in If and then
OA ! OB ! OA ! OB ! R 3
R 2 . 1a 1 , a 2 , a 3 2 1b 1 , b 2 , b 3 2,
1a 1 , a 2 , a 3 2 1b 1 , b 2 , b 3 2 1a 1 b 1 , a 2 b 2 , a 3 b 3 2.
In R 3 , the shape that was used to generate the result for the addition of three vectors
was not a parallelogram but a parallelepiped, which is a box-like shape with pairs of
opposite faces being identical parallelograms. From our diagram, it can be seen that
parallelograms ODAB and CEPQ are copies of each other. It is also interesting to note
that the parallelepiped is completely determined by the components of the
three position vectors OB ! , OC ! , and OD ! . That is to say, the coordinates of all the vertices
of the parallelepiped can be determined by the repeated application of the
Triangle Law of Addition.
For vectors in R 2 , we showed that the multiplication of an algebraic vector by a
scalar was produced by multiplying each component of the vector by the scalar.
In R 3 , this result also holds, i.e., mOP ! m1a, b, c2 1ma, mb, mc2, mR.
R 3
EXAMPLE 2 Selecting a strategy to determine a combination of vectors in R 3
Given a ! i ! 2j ! k ! , b ! 2j ! 3k ! , and c ! i ! 3j ! 2k ! , determine each of the
following:
a. 2a ! b ! c ! b. a ! b ! c !
Solution
a. Method 1 (Standard Unit Vectors)
2a ! b ! c ! 21i !
2i ! 2j !
4j ! k !
2k ! 2 12j !
2j ! 3k !
3k ! 2
i ! 1i !
3j ! 3j !
2k ! 2k ! 2
2i !
i !
j ! i ! 4j !
7k ! 2j ! 3j ! 2k ! 3k ! 2k !
11, 1, 72
Method 2 (Components)
Converting to component form, we have
and
c ! a ! 11, 2, 12, b ! 10, 2, 32,
11, 3, 22.
Therefore, 2a ! b ! c ! 211, 2, 12 10, 2, 32 11, 3, 22
12, 4, 22 10, 2, 32 11, 3, 22
12 0 1, 4 2 3, 2 3 22
11,
i ! 1,
j ! 72
7k !
NEL CHAPTER 6 329