Vector. A VECTOR can describe anything that has both MAGNITUDE and DIRECTION The MAGNITUDE describes the size of the vector. The DIRECTION tells you where.

Presentasi berjudul: "Vector. A VECTOR can describe anything that has both MAGNITUDE and DIRECTION The MAGNITUDE describes the size of the vector. The DIRECTION tells you where."— Transcript presentasi:

1 Vector

2 A VECTOR can describe anything that has both MAGNITUDE and DIRECTION The MAGNITUDE describes the size of the vector. The DIRECTION tells you where the vector is pointing.

3 Here’s a typical example: An airplane is flying East at a velocity of 600 kilometers per hour. Here the magnitude (speed) is 600 km/hr and the direction is East. A diagram of this vector might look like this: V= 600 km/hr

4 Here’s a typical example: An airplane is flying East at a velocity of 600 kilometers per hour. Here the magnitude (speed) is 600 km/hr and the direction is East. A diagram of this vector might look like this: V= 600 km/hr Vectors will be written in BOLDFACE or with an arrow above the letter: = 600 km/hr EastorV=600 km/hr East

5 Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY?

6 Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: 100 km/hr 600 km/hr

7 Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: So how do we find the new SPEED of the plane? 100 km/hr 600 km/hr

8 Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: So how do we find the new SPEED of the plane? Answer: Add the vectors together Is the new speed just 600 km/hr +100 km/hr = 700 km/hr? 100 km/hr 600 km/hr

9 Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: So how do we find the new SPEED of the plane? Answer: Add the vectors together Is the new speed just 600 km/hr +100 km/hr = 700 km/hr? NO Note: If the wind blows East we can add them – do you see why? 100 km/hr 600 km/hr

10 Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: So how do we find the new SPEED of the plane? 100 km/hr 600 km/hr

11 100 km/hr 600 km/hr Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram:

12 100 km/hr 600 km/hr Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: Ervian Arif Muhafid, M.Pd V total ≈ 608 km/hr

13 100 km/hr 600 km/hr Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: Now we have the magnitude, but what is the direction? V total ≈ 608 km/hr

14 100 km/hr 600 km/hr Let’s try a quick example with our airplane. Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? The plane gets blown off course, of course. We need to find the new VELOCITY. Like any vector, it will have a MAGNITUDE (speed) and a DIRECTION Here is a diagram: V total ≈ 608 km/hr Now we have the magnitude, but what is the direction? We use our right-triangle rules for this: tan(θ) = opposite/adjacent = 100/600 θ = tan -1 (1/6) θ ≈ 9.5°

15 Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? V total ≈ 608 km/hr 100 km/hr 600 km/hr θ θ ≈ 9.5° Finally! We have our answer! The plane will speed up to 608 km/hr, but be pushed off-course by an angle of 9.5°. Note: the angle may be described as 9.5° North of East.

16 Suppose that this plane is flying 600 km/hr East, and it encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY? V total ≈ 608 km/hr 100 km/hr 600 km/hr θ θ ≈ 9.5° You may see a vector represented in a few different ways: Next we will work out a (slightly harder) problem from a textbook.

17 A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

18 The first step is to set up a coordinate system. The most convenient thing to do is just call East the x-direction and North the y-direction, as shown. y x A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

19 Now we look at each leg of the journey and find the components: First leg: 2.00 km in the x-direction; 0.00 km in the y-direction We could label everything to keep it all organized: If the first leg is represented by the vector V 1, we can write the components separately. V 1x = +2.00 km(note: positive indicates East – negative would be West) V 1y = 0.00 km y x A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

20 y x The 2 nd leg has components in both x- and y-directions, so we can find the components from our triangle rules: V 2x = V 2y = A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

21 y x The 2 nd leg has components in both x- and y-directions, so we can find the components from our triangle rules: V 2x = +3.50*cos(45°) ≈ 2.47 km(positive for East) V 2y = -3.50*sin(45°) ≈ -2.47 km(negative for South) A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

22 y x The 3 rd leg has unknown components, but we can write them as unknowns: V 3x = ? V 3y = ? A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

23 y x The next step is the only one where you have to think about the problem. Since we know that the total journey from start to finish is 5.80 km East, we know the components: V total,x = V total,y = A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

24 y x The next step is the only one where you have to think about the problem. Since we know that the total journey from start to finish is 5.80 km East, we know the components: V total,x = +5.80 km V total,y = 0.00 km A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

25 y x Finally we can just add everything up and solve for our unknowns. We will have 2 equations – one for each direction: V 1x + V 2x + V 3x = V total,x V 1y + V 2y + V 3y = V total,y A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

26 y x Finally we can just add everything up and solve for our unknowns. We will have 2 equations – one for each direction: V 1x + V 2x + V 3x = V total,x 2.00 + 2.47 + V 3x = 5.80 V 3x = +1.33 km (positive is East) V 1y + V 2y + V 3y = V total,y 0.00 – 2.47 + V 3y = 0.00 V 3y = +2.47 km (positive is North) A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

27 y x Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3 rd leg. A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

28 y x Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3 rd leg. Use the Pythagorean Theorem to find the magnitude: V 3 = 2.81 km For extra fun, find the angle of the vector too! A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.

29 y x Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3 rd leg. Use tan(θ) = V y /V x to find the direction tan(θ) = 2.47/1.33 Θ = tan -1 (1.86) Θ = 61.7° North of East (this angle is measured above the x-axis) A sailor in a boat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey. Use the Pythagorean Theorem to find the magnitude: V 3 = 2.81 km

30 18° 60° Here are a few vectors. Find their x and y components. y x 22° 14cm

31 60° Here are a few vectors. Find their x and y components. y x AxAx AyAy

32 y x 22° BxBx ByBy

33 18° Here are a few vectors. Find their x and y components. y x 14cm CxCx CyCy

34 18° 60° y x 22° 14cm

35 y x Group the components together:

36 85° 95° y x Group the components together: Re-combine to find magnitude and direction:

37 18° 60° Ervian Arif Muhafid, M.Pd Dot Product y x 22° 14cm We already found the components of these vectors

38 18° 60° Dot Product y x 22° 14cm We already found the components of these vectors

39 18° 60° Dot Product y x 22° 14cm We already found the components of these vectors We can use the dot product to find the angle between any two vectors – here is a formula

40 18° 60° Dot Product y x 22° 14cm We already found the components of these vectors We can use the dot product to find the angle between any two vectors – here is a formula Applying this to our vectors:

41 18° 60° Ervian Arif Muhafid, M.Pd Cross Product y x 22° 14cm We already found the components of these vectors

42 18° 60° Ervian Arif Muhafid, M.Pd Cross Product y x 22° 14cm We already found the components of these vectors Our cross product formula is typically written in terms of a 3-dimensional xyz coordinate system: 3x3 determinant

43 18° 60° Ervian Arif Muhafid, M.Pd Cross Product y x 22° 14cm We already found the components of these vectors Since our vectors are in the x-y plane the calculation will be simpler. We will get a vector in the z-direction.

44 18° 60° Ervian Arif Muhafid, M.Pd Cross Product y x 22° 14cm We already found the components of these vectors Since our vectors are in the x-y plane the calculation will be simpler. We will get a vector in the z-direction. This is a vector in the +z direction (out of the page). Note that if we did the cross-product in reverse order we would get the opposite vector (into the page). We can use a right-hand rule to find the direction.