Page 173 - Schaum's Outline of Theory and Problems of Advanced Calculus

164 VECTORS [CHAP. 7

Angle EOQ ¼ 61:58,using a protractor. Then vector OQ has magnitude 7.4 mi and direction 61.58
north of east.
(b) Analytical Determination of Resultant. From triangle OPQ,denoting the magnitudes of A; B; C by
A; B; C,we have by the law of cosines
2
2
2
2
2
ffiffiffi
p
C ¼ A þ B 2AB cos ff OPQ ¼ 3 þ 5 2ð3Þð5Þ cos 1358 ¼ 34 þ 15 2 ¼ 55:21
and C ¼ 7:43 (approximately).
A C
By the law of sines, ¼ : Then
sin ff OQP sin ff OPQ
A sin ff OPQ
3ð0:707Þ
¼ 0:2855 and ff OQP ¼ 16835 0
C 7:43
sin ff OQP ¼ ¼
Thus vector OQ has magnitude 7.43 mi and direction ð458 þ 16835 Þ¼ 61835 north of east.
0
0
7.4. Prove that if a and b are non-collinear, then xa þ yb ¼ 0 implies x ¼ y ¼ 0. Is the set fa; bg
linearly independent or linearly dependent?
Suppose x 6¼ 0. Then xa þ yb ¼ 0 implies xa ¼ yb or a ¼ ðy=xÞb, i.e., a and b must be parallel to the
same line (collinear), contrary to hypothesis. Thus, x ¼ 0; then yb ¼ 0, from which y ¼ 0. The set is
linearly independent.
7.5. If x 1 a þ y 1 b ¼ x 2 a þ y 2 b, where a and b are non-collinear, then x 1 ¼ x 2 and y 1 ¼ y 2 .
x 1 a þ y 1 b ¼ x 2 a þ y 2 b can be written
x 1 a þ y 1 b ðx 2 a þ y 2 bÞ¼ 0 or ðx 1 x 2 Þa þðy 1 y 2 Þb ¼ 0
Hence, by Problem 7.4, x 1 x 2 ¼ 0; y 1 y 2 ¼ 0; or x 1 ¼ x 2 ; y 1 ¼ y 2 :
Extensions are possible (see Problem 7.49).

7.6. Prove that the diagonals of a parallelogram bisect each B b C
other.
Let ABCD be the given parallelogram with diagonals intersect-
ing at P as shown in Fig. 7-18. a P a
Since BD þ a ¼ b; BD ¼ b a. Then BP ¼ xðb aÞ.
Since AC ¼ a þ b, AP ¼ yða þ bÞ.
But AB ¼ AP þ PB ¼ AP BP, i.e., a ¼ yða þ bÞ xðb aÞ
¼ðx þ yÞa þðy xÞb.
A b D
Since a and b are non-collinear, we have by Problem 7.5,
1 and P is the midpoint of Fig. 7-18
2
x þ y ¼ 1 and y x ¼ 0, i.e., x ¼ y ¼
both diagonals.
7.7. Prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and
has half its length.
From Fig. 7-19, AC þ CB ¼ AB or b þ a ¼ c.
Let DE ¼ d be the line joining the midpoints of sides AC and CB. Then
1
1
1
1
d ¼ DC þ CE ¼ b þ a ¼ ðb þ aÞ¼ c
2
2
2
2
Thus, d is parallel to c and has half its length.
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
A þ A þ A . See Fig.
7.8. Prove that the magnitude A of the vector A ¼ A 1 i þ A 2 j þ A 3 k is A ¼ 1 2 3
7-20.