Respuesta :
Refer to the diagram shown below.
Define
[tex]\hat{i} [/tex] = unit vector in the eastern direction
[tex]\hat{j}[/tex] = unit vector n the northern drection.
Then the displacement vectors are
Vector #1: [tex]1550(cos27^{o} \hat{i} + sin27^{o} \hat{j} ) \, miles = 1381.1\hat{i}+703.7\hat{j}[/tex]
Vector #2: [tex]a(sin41^{o})\hat{i} - cos41^{o}\hat{j} ) = a(0.6561\hat{i} - 0.9873\hat{j} )[/tex]
Vector #3: [tex]b(-cos27^{o} \hat{i} + sin27^{o} \hat{j} ) = b(-0.891\hat{i} + 0.454 \hat{j} )[/tex]
Because the vector sum of all three vector is zero, therefore
1381.1 +0.6561a - 0.891b = 0
703.7 - 0.9873a + 0.454b = 0
That is,
0.6561a - 0.891b = -1381.1 (1)
-0.9873a + 0.454b = -703.7 (2)
From (1), obtain
b = 0.7364a + 1550.1 (3)
Substitute (3) into (2).
-0.9873a + 0.454(0.7364a + 1550.1) = - 703.7
-0.653a = -1407.4
a = 2155.3 mles
From (3), obtain
b = 3137.2 mles
Answer:
The magnitudes of the two displacement vectors are
2155.3 miles and 3137.2 miles.
Define
[tex]\hat{i} [/tex] = unit vector in the eastern direction
[tex]\hat{j}[/tex] = unit vector n the northern drection.
Then the displacement vectors are
Vector #1: [tex]1550(cos27^{o} \hat{i} + sin27^{o} \hat{j} ) \, miles = 1381.1\hat{i}+703.7\hat{j}[/tex]
Vector #2: [tex]a(sin41^{o})\hat{i} - cos41^{o}\hat{j} ) = a(0.6561\hat{i} - 0.9873\hat{j} )[/tex]
Vector #3: [tex]b(-cos27^{o} \hat{i} + sin27^{o} \hat{j} ) = b(-0.891\hat{i} + 0.454 \hat{j} )[/tex]
Because the vector sum of all three vector is zero, therefore
1381.1 +0.6561a - 0.891b = 0
703.7 - 0.9873a + 0.454b = 0
That is,
0.6561a - 0.891b = -1381.1 (1)
-0.9873a + 0.454b = -703.7 (2)
From (1), obtain
b = 0.7364a + 1550.1 (3)
Substitute (3) into (2).
-0.9873a + 0.454(0.7364a + 1550.1) = - 703.7
-0.653a = -1407.4
a = 2155.3 mles
From (3), obtain
b = 3137.2 mles
Answer:
The magnitudes of the two displacement vectors are
2155.3 miles and 3137.2 miles.
