1. [tex](-5,6)(-3,3)[/tex]
a. Draw the Coordinates
b. Solve the distance
c. Solve the midpoint
d. From the origin connect the midpoint
e. Solve the vector.
f. Solve the angle of Direction
2. [tex]50 mN , 60 mE , 90 m S , 150 m, E , 75 m N[/tex]
a. Draw the vector diagram
b. Solve the vector
c. Solve the angle of direction
Answer (2)
Calculate the distance between the points ( − 5 , 6 ) and ( − 3 , 3 ) using the distance formula: 13 ≈ 3.61 .
Determine the midpoint between the points ( − 5 , 6 ) and ( − 3 , 3 ) using the midpoint formula: ( − 4 , 4.5 ) .
Compute the vector from ( − 5 , 6 ) to ( − 3 , 3 ) by subtracting the initial point from the terminal point: ( 2 , − 3 ) .
Find the angle of direction of the vector ( 2 , − 3 ) using the arctangent function: − 56.3 1 ∘ .
Calculate the magnitude of the resultant vector from the series of displacements: 212.90 m .
Determine the angle of direction of the resultant vector: 9.4 6 ∘ .
D i s t an ce : 3.61 , M i d p o in t : ( − 4 , 4.5 ) , V ec t or : ( 2 , − 3 ) , A n g l e : − 56.3 1 ∘ , R es u lt an t V ec t or M a g ni t u d e : 212.90 m , R es u lt an t V ec t or A n g l e : 9.4 6 ∘
Explanation
Problem Overview Let's break down this physics problem step by step. We'll start with the coordinate geometry and vector calculations, and then move on to vector addition and direction finding.
Part 1a: Coordinates First, we are given two points: ( − 5 , 6 ) and ( − 3 , 3 ) . We need to find the distance between them.
Part 1b: Distance Calculation The distance d between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by the distance formula: d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 In our case, ( x 1 , y 1 ) = ( − 5 , 6 ) and ( x 2 , y 2 ) = ( − 3 , 3 ) . Plugging in the values: d = (( − 3 ) − ( − 5 ) ) 2 + (( 3 ) − ( 6 ) ) 2 = ( 2 ) 2 + ( − 3 ) 2 = 4 + 9 = 13 So, the distance between the points is 13 ≈ 3.61 .
Part 1c: Midpoint Next, we need to find the midpoint M between the points ( − 5 , 6 ) and ( − 3 , 3 ) .
Midpoint Calculation The midpoint M between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by the midpoint formula: M = ( 2 x 1 + x 2 , 2 y 1 + y 2 ) In our case, ( x 1 , y 1 ) = ( − 5 , 6 ) and ( x 2 , y 2 ) = ( − 3 , 3 ) . Plugging in the values: M = ( 2 ( − 5 ) + ( − 3 ) , 2 ( 6 ) + ( 3 ) ) = ( 2 − 8 , 2 9 ) = ( − 4 , 4.5 ) So, the midpoint is ( − 4 , 4.5 ) .
Part 1e: Vector Now, we need to find the vector V from ( − 5 , 6 ) to ( − 3 , 3 ) .
Vector Calculation The vector V from point A ( x 1 , y 1 ) to point B ( x 2 , y 2 ) is given by: V = ( x 2 − x 1 , y 2 − y 1 ) In our case, A = ( − 5 , 6 ) and B = ( − 3 , 3 ) . Plugging in the values: V = (( − 3 ) − ( − 5 ) , ( 3 ) − ( 6 )) = ( 2 , − 3 ) So, the vector is ( 2 , − 3 ) .
Part 1f: Angle of Direction Finally, we need to find the angle of direction θ of the vector V = ( 2 , − 3 ) .
Angle Calculation The angle of direction θ of a vector ( V x , V y ) is given by: θ = arctan ( V x V y ) In our case, V = ( 2 , − 3 ) . Plugging in the values: θ = arctan ( 2 − 3 ) Using a calculator, we find that θ ≈ − 56.3 1 ∘ .
Part 2a: Vector Diagram Now let's move on to the second part of the problem, which involves vector addition.
Part 2b: Resultant Vector We are given a series of displacements: 50 m N , 60 m E , 90 m S , 150 m E , 75 m N . We need to find the resultant vector.
Magnitude Calculation First, let's resolve the vectors into North-South and East-West components: North component: 50 m + 75 m − 90 m = 35 m East component: 60 m + 150 m = 210 m Now, we can find the magnitude of the resultant vector R using the Pythagorean theorem: R = ( 35 ) 2 + ( 210 ) 2 = 1225 + 44100 = 45325 ≈ 212.90 m So, the magnitude of the resultant vector is approximately 212.90 m .
Part 2c: Angle of Direction Finally, we need to find the angle of direction α of the resultant vector R = ( 210 , 35 ) .
Angle Calculation The angle of direction α of a vector ( R E , R N ) is given by: α = arctan ( R E R N ) In our case, R = ( 210 , 35 ) . Plugging in the values: α = arctan ( 210 35 ) = arctan ( 6 1 ) Using a calculator, we find that α ≈ 9.4 6 ∘ .
Final Results In summary:
For the points ( − 5 , 6 ) and ( − 3 , 3 ) :
Distance: 13 ≈ 3.61
Midpoint: ( − 4 , 4.5 )
Vector: ( 2 , − 3 )
Angle of direction: − 56.3 1 ∘
For the series of displacements:
Resultant vector magnitude: 212.90 m
Angle of direction: 9.4 6 ∘
Examples
Understanding vectors and coordinate geometry is crucial in many real-world applications. For instance, GPS navigation systems use vectors to calculate the shortest path between two points. Similarly, architects and engineers use coordinate geometry to design and construct buildings, ensuring precise measurements and structural integrity. The principles we've applied here are fundamental to these advanced technologies.
We calculated the distance and midpoint between the points (-5, 6) and (-3, 3), found the vector and its direction angle. We also analyzed several vector displacements, determining the resultant vector's magnitude and direction angle. The distance is approximately 3.61 units, the midpoint is (-4, 4.5), and the final vector results in a magnitude of approximately 212.90 m and direction angle of about 9.46 degrees.
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