8-2 Vectors in the Coordinate Plane
37. ROWING Nadia is rowing across a river at a speed of 5 miles per hour perpendicular to the shore. The river has a
current of 3 miles per hour heading downstream.
a. At what speed is she traveling?
b. At what angle is she traveling with respect to the shore?
SOLUTION: a. Nadia’s rowing can be represented by the vector
and the current can be represented by the vector .
Add the vectors representing r and c to find the resultant vector, v .
The speed at which Nadia is traveling is the magnitude of v .
Thus, she is traveling at about 5.8 miles per hour.
b. Find θ by constructing a right triangle with the given vectors and using the tangent ratio.
Thus, Nadia is traveling at an angle of about 59° with respect to the shore.
Find the component form of v with the given magnitude and direction angle.
39. | v | = 4, θ = 135°
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8-2 Vectors in the Coordinate Plane
Thus, Nadia is traveling at an angle of about 59° with respect to the shore.
Find the component form of v with the given magnitude and direction angle.
39. | v | = 4, θ = 135°
SOLUTION: 41. | v | = 16, θ = 330°
SOLUTION: 43. | v | = 15, θ = 125°
SOLUTION: Find the direction angle of each vector to the nearest tenth of a degree.
45. –2i + 5j
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Since the vector lies in Quadrant II, θ = 180° + (−68.2°) or about 111.8°.
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8-2 Vectors in the Coordinate Plane
Find the direction angle of each vector to the nearest tenth of a degree.
45. –2i + 5j
SOLUTION: Since the vector lies in Quadrant II, θ = 180° + (−68.2°) or about 111.8°.
47. –4i – 3j
SOLUTION: Since the vector lies in Quadrant III, θ = 180° + 36.9° or about 216.9°.
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8-2 Vectors in the Coordinate Plane
49. SOLUTION: So, the direction angle of the vector is 45°.
51. SOLUTION: Since the vector lies in Quadrant IV, θ = 360° + (−69.4)° or about 290.6°.
53. NAVIGATION An airplane is traveling due east with a speed of 600 miles per hour. The wind blows at 85 miles per hour at an angle of S59°E.
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8-2 Vectors in the Coordinate Plane
53. NAVIGATION An airplane is traveling due east with a speed of 600 miles per hour. The wind blows at 85 miles per hour at an angle of S59°E.
a. Determine the speed of the airplane’s flight.
b. Determine the angle of the airplane’s flight.
SOLUTION: a. Since the airplane is traveling due east with a speed of 600 miles per hour, the component form of the speed v 1 is
. Use the magnitude and the direction of the wind v 2 to write this vector in component form. Let θ = −31° since θ is the direction angle that v 2 makes with the positive x-axis.
Add the algebraic vectors representing v 1 and v 2 to find the resultant velocity, vector r.
Find the magnitude of the resultant.
The speed of the airplane’s flight is about 674.3 miles per hour.
b. Find the resultant direction angle θ.
Since r lies in Quadrant IV, θ = 360 + (−3.7) or 356.3°. Therefore, the angle of the airplane’s flight is about S86°E.
Determine whether
and with the initial and terminal points given are equivalent. If so, prove that
=
. If not, explain why not.
55. A(3, 5), B(6, 9), C(–4, –4), D(–2, 0)
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Find the magnitude and direction of each vector.
For
, find the component form.
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Since r lies in Quadrant IV, θ = 360 + (−3.7) or 356.3°. Therefore, the angle of the airplane’s flight is about S86°E.
Determine whether
and with the initial and terminal points given are equivalent. If so, prove . If not,
explain whyPlane
not.
8-2 that
Vectors= in the
Coordinate
55. A(3, 5), B(6, 9), C(–4, –4), D(–2, 0)
SOLUTION: Find the magnitude and direction of each vector.
For
, find the component form.
Use the component form of the vector to find the magnitude. Substitute x2 − x1 = 3 and y 2 − y 1 = 4 into the formula
for the magnitude of a vector in the coordinate plane.
Next, find the direction angle of
For
.
, find the component form.
Use the component form of the vector to find the magnitude. Substitute x2 − x1 = 2 and y 2 − y 1 = 4 into the formula
for the magnitude of a vector in the coordinate plane.
Next, find the direction angle of
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.
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8-2 Vectors in the Coordinate Plane
Next, find the direction angle of
No;
and
equivalent.
.
are not equivalent. The magnitude and direction are not the same for both vectors, so they are not
57. A(1, –3), B(0, –10), C(11, 8), D(10, 1)
SOLUTION: Find the magnitude and direction of each vector.
For
, find the component form.
Use the component form of the vector to find the magnitude. Substitute x2 − x1 = −1 and y 2 − y 1 = −7 into the
formula for the magnitude of a vector in the coordinate plane.
Next, find the direction angle of
For
.
, find the component form.
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Use the component form of the vector to find the magnitude. Substitute x2 − x1 = −1 and y 2 − y 1 = −7 into the
formula for the magnitude of a vector in the coordinate plane.
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8-2 Vectors in the Coordinate Plane
Use the component form of the vector to find the magnitude. Substitute x2 − x1 = −1 and y 2 − y 1 = −7 into the
formula for the magnitude of a vector in the coordinate plane.
Next, find the direction angle of
Yes;
and
equivalent.
.
are equivalent. The magnitude and direction are the same for both vectors, so they are
59. NAVIGATION A jet is flying with an air speed of 480 miles per hour at a bearing of N82°E. Because of the wind,
the ground speed of the plane is 518 miles per hour at a bearing of N79°E.
a. Draw a diagram to represent the situation.
b. What are the speed and direction of the wind?
c. If the pilot increased the air speed of the plane to 500 miles per hour, what would be the resulting ground speed
and direction of the plane?
SOLUTION: a. Sample answer: Draw a diagram to represent the situation. If the initial bearing of the jet is N82°E but is flying at a bearing of N79°E because of the wind, the angle created by the two vectors is 3°.
b. The resultant vector r for the ground speed of the jet is the sum of the vector representing the air speed of the jet
v 1 and the vector representing the wind v 2, or r = v 1 + v 2.
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b. The resultant vector r for the ground speed of the jet is the sum of the vector representing the air speed of the jet
8-2 vVectors
in the Coordinate Plane
v
r v
v
1 and the vector representing the wind 2, or = 1 + 2.
Use the air speed and bearing of the jet to write v 1 in component form. Let θ = 8° since the bearing is N82°E.
Use the ground speed and bearing of the jet to write r in component form. Let θ = 11° since the bearing is N79°E.
Substitute the component forms of r and v 1 into r = v 1 + v 2 and solve for v 2.
The component form of the wind is
.
Find the magnitude of v 2.
Find the resultant direction angle θ.
The directed angle of the wind is about 44°. Thus, the wind is blowing about 46.1 miles per hour at a bearing of about N46°E.
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c. Use the air speed and bearing of the jet to write v 1 in component form. Let θ = 8° since the bearing is N82°E.
8-2 The directed angle of the wind is about 44°. Thus, the wind is blowing about 46.1 miles per hour at a bearing of Vectors in the Coordinate Plane
about N46°E.
c. Use the air speed and bearing of the jet to write v 1 in component form. Let θ = 8° since the bearing is N82°E.
Substitute the component forms of v 1 and v 2 into r = v 1 + v 2 and solve for r.
Find the magnitude of r.
Find the resultant direction angle θ.
The directed angle of the jet is about 11°. Thus, the jet is traveling at a resulting ground speed of about 538 miles per
hour at a directed angle of about N79°E.
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