81.1 Vector Properties Problem Set

VECTOR PROPERITES | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: The general term for a vector that represents the sum of two vectors is: A. Scalar B. Resultant C. Tensor D. Moment



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PROBLEM 2: The sum of the vectors, 𝐹! , 𝐹! , and 𝐹! , is most close to: 𝐹! = 4𝑖 + 7𝑗 + 6𝑘 𝐹! = 9𝑖 + 2𝑗 + 11𝑘 𝐹! = 5𝑖 − 3𝑗 − 8𝑘 A. 𝐹 = 18𝑖 + 6𝑗 + 9𝑘 B. 𝐹 = −18𝑖 − 6𝑗 − 9𝑘 C. 𝐹 = 18𝑖 + 12𝑗 + 25𝑘 D. 𝐹 = 18𝑖 − 6𝑗 − 25𝑘



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PROBLEM 3: The displacement experienced by an object pushed from a point (5,2) to the point (6,10) by a force represented by the vector 𝐹 = (5,4), is most close to: A. (2, 4) B. (3, 5) C. (1,8) D. (5, 5)



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PROBLEM 4: The vector sum 𝐶, if vector 𝐴 = 8𝑖 + 13𝑗 is added to vector 𝐵 = 26𝑖 + 7𝑗, is most close to: A. 𝐶 = 21𝑖 + 33𝑗 B. 𝐶 = 34𝑖 + 20𝑗 C. 𝐶 = 20𝑖 + 34𝑗 D. 𝐶 = 15𝑖 + 39𝑗



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PROBLEM 5: Vector C, which is the vector resulting from adding vector A to vector B, is most closely written as: Given: 𝐴 = 3𝑖 + 7𝑗 + 4𝑘 𝐵 = 2𝑖 + 9𝑗 + 11𝑘 A. 𝐶 = 5𝑖 + 16𝑗 + 15𝑘 B. 𝐶 = 14𝑖 + 22𝑗 + 0𝑘 C. 𝐶 = 16𝑖 + 15𝑗 + 5𝑘 D. 𝐶 = 12𝑖 + 9𝑗 + 5𝑘



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PROBLEM 6: !

Multiply the vector 𝑇 = 3𝑖 + 5𝑗 + 7𝑘 by the scalar − !. !

!

A. −𝑖 − ! 𝑗 − ! 𝑘 !

B. −𝑖 − 𝑗 + 7𝑘 !

!

!

!

!

!

!

!

!

C. – 𝑖 + 𝑗 + 𝑘 D. −𝑖 − 𝑗 − 𝑘



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PROBLEM 7: Scaling the vector difference when Vector O is subtracted from Vector D by a factor of 4 is most close to: Given: 𝑂 = 9𝑖 + 10𝑗 𝐷 = 16𝑖 + 4𝑗 A. 24𝑖 − 28𝑗 B. −28𝑖 + 24𝑗 C. 24𝑖 − 6𝑗 D. −24𝑖 + 20𝑗



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PROBLEM 8: Vector C, which is the vector resulting from adding vector A to vector B, is most closely written as: Given: 𝐴 = 7𝑗 + 4𝑘 𝐵 = 2𝑖 + 9𝑗 A. 𝐶 = 2𝑖 + 16𝑗 + 4𝑘 B. 𝐶 = 2𝑖 + 22𝑗 C. 𝐶 = 15𝑗 + 5𝑘 D. 𝐶 = 2𝑖 + 9𝑗 + 5𝑘



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VECTOR PROPERITES | SOLUTIONS SOLUTION 1: The TOPIC of VECTORS can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. By definition, the SUM of two vectors is known as the RESULTANT VECTOR. The correct answer choice is B. 𝐑𝐞𝐬𝐮𝐥𝐭𝐚𝐧𝐭

SOLUTION 2: The TOPIC of VECTOR ADDITION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Right off the bat, we want to start with this MAJOR TIP…this problem is lengthy, yet extremely doable by hand, as we will show you. However, it can be hacked in a fraction of the time using the VECTOR CALCULATOR HACK that you can find in the CALCULATOR WORKSHOP section within the Prepineer Program. The ADDITION of VECTORS involves breaking down each element of a given vector and combining those elements as they relate to a common dimension, or direction.



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The result of combining the elements of a pair of vectors provides us with a new unique vector. VECTOR ADDITION is defined as calculating the sum between the scalars of each component with respect to each component or dimension. The GENERAL FORMULA for VECTOR ADDITION can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors, 𝐴 and 𝐵: 𝐴 = 𝑎! 𝑖 + 𝑎! 𝑗 + 𝑎! 𝑘 𝐵 = 𝑏! 𝑖 + 𝑏! 𝑗 + 𝑏! 𝑘 We can write the SUM of the two vectors as: 𝐴 + 𝐵 = 𝑎! + 𝑏! 𝑖 + 𝑎! + 𝑏! 𝑗 + 𝑎! + 𝑏! 𝑘 In this problem, we are given three defined VECTORS, which are: 𝐹! = 4𝑖 + 7𝑗 + 6𝑘 𝐹! = 9𝑖 + 2𝑗 + 11𝑘 𝐹! = 5𝑖 − 3𝑗 − 8𝑘



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The fact that we are dealing with three vectors, as opposed to the standard two that we are given a template for, changes nothing…although originally, it might seem more complicated. Recall the COMMUTATIVE LAW states that the order of addition does not matter, we could plug in the elements in whatever order, such that: 𝐴+𝐵+𝐶 =𝐶+𝐵+𝐴 We can write the SUM of the three vectors as: 𝐹! + 𝐹! + 𝐹! = 𝐹!! + 𝐹!! + 𝐹!! 𝑖 + 𝐹!! + 𝐹!! + 𝐹!! 𝑗 + 𝐹!! + 𝐹!! + 𝐹!! 𝑘 As you can see, we have our VECTORS broken down in to their individual ELEMENTS, ( i, j and k), and we are just combining them using the ADDITION property. Plugging in the appropriate values we have: 𝐹! + 𝐹! + 𝐹! = 4 + 7 + 6 𝑖 + 9 + 2 + 11 𝑗 + 5 − 3 − 8 𝑘 Which gives us the ELEMENTS of our new vector in VECTOR NOTATION as: 𝐹! = 17𝑖 + 22𝑗 − 6𝑘 The correct answer choice is A. 𝐅 = 𝟏𝟕𝐢 + 𝟐𝟐𝐣 − 𝟔𝐤



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SOLUTION 3: This problem originally presents itself as extremely complicated. We are talking in context of vectors and points within a Cartesian coordinate axis, but without other laws established, how are we supposed to translate that information in to a DISPLACEMENT. In this problem, we are given that their exists a FORCE: 𝐹 = (5,4) And we are told that this FORCE moves an unidentified OBJECT from: Point A: (5,2) To: Point (6,10) So our duty now is determine how to relate the force to the movement of the object, or is it? We are after the DISPLACEMENT, and if we recall, this is the DISTANCE in which an object moves without reference to any variable other than two defined points.



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So no external effects are considered, it’s all built in to the change in distance of the OBJECT, or otherwise, the DISPLACEMENT between POINT A and POINT B…and we have those points. This is simply a VECTOR SUBTRACTION problem. The TOPIC of VECTOR SUBTRACTION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. VECTOR SUBSTRACTION is defined as calculating the DIFFERENCE between the scalars of each component with respect to the dimension in which they reside. In this problem, we are given the two points, which can be written in VECTOR NOTATION as: 𝐴 = 5𝑖 + 2𝑗 𝐵 = 6𝑖 + 10𝑗 And we are asked to determine the DISPLACEMENT given these two measurements, or the DIFFERENCE between the two, therefore, we are dealing with VECTOR SUBTRACTION. The SUBTRACTION of VECTORS is the reverse operation of the additional of vectors. The only difference is that you actually convert the vector being subtracted to negative vector and then add the vectors.



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The OPERATION that we are asked to carry out is: 𝐷 =𝐵−𝐴 Which is equivalent to the OPERATION: 𝐷 = 𝐵 + (−𝐴) This illustrates how the SUBTRACTION of VECTORS is just the reverse operation of the additional of vectors. So let’s create the NEGATIVE VECTOR for VECTOR A. If: 𝐴 = 5𝑖 + 2𝑗 Then: −𝐴 = −5𝑖 − 2𝑗 Creating this NEGATIVE VECTOR isn’t really necessary, but for illustration and edification purposes we did. The GENERAL FORMULA for VECTOR SUBTRACTION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

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Given two vectors, 𝐴 and 𝐵, the difference, or the SUBTRACTION, of the two can be written as: 𝐵 − 𝐴 = 𝑏! − 𝑎! 𝑖 + 𝑏! − 𝑎! 𝑗 + 𝑏! − 𝑎! 𝑘 Plugging in our ELEMENTS we get: 𝐵 − 𝐴 = 6 − 5 𝑖 + 10 − 2 𝑗 Which gives us the ELEMENTS of our new vector in VECTOR NOTATION as: 𝐷 = 𝐵 − 𝐴 = 𝑖 + 8𝑗 The correct answer choice is C. (𝟏, 𝟖)

SOLUTION 4: The TOPIC of VECTOR ADDITION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The ADDITION of VECTORS involves breaking down each element of a given vector and combining those elements as they relate to a common dimension, or direction. The result of combining the elements of a pair of vectors provides us with a new unique vector.

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VECTOR ADDITION is defined as calculating the sum between the scalars of each component with respect to each component or dimension. The GENERAL FORMULA for VECTOR ADDITION can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors, 𝐴 and 𝐵: 𝐴 = 𝑎! 𝑖 + 𝑎! 𝑗 + 𝑎! 𝑘 𝐵 = 𝑏! 𝑖 + 𝑏! 𝑗 + 𝑏! 𝑘 We can write the SUM of the two vectors as: 𝐴 + 𝐵 = 𝑎! + 𝑏! 𝑖 + 𝑎! + 𝑏! 𝑗 + 𝑎! + 𝑏! 𝑘 In this problem, we are given the two VECTORS: 𝐴 = 8𝑖 + 13𝑗 𝐵 = 26𝑖 + 7𝑗 Recall the COMMUTATIVE LAW states that the order of addition does not matter, we could plug in the elements in whatever order, such that: 𝐴+𝐵 =𝐵+𝐴



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We can write the SUM of the two vectors as VECTOR C, which is: 𝐶 = 𝐴 + 𝐵 = 𝑎! + 𝑏! 𝑖 + 𝑎! + 𝑏! 𝑗 As you can see, we have our VECTORS broken down in to their individual i and j ELEMENTS and we are just combining them using the ADDITION property. Plugging in the appropriate values we have: 𝐶 = 𝐴 + 𝐵 = 8 + 26 𝑖 + 13 + 7 𝑗 Which gives us the ELEMENTS of our new vector in VECTOR NOTATION as: 𝐶 = 𝐴 + 𝐵 = 34𝑖 + 20𝑗 The correct answer is B. 𝑪 = 𝟑𝟒𝒊 + 𝟐𝟎𝒋

SOLUTION 5: The TOPIC of VECTOR ADDITION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The ADDITION of VECTORS involves breaking down each element of a given vector and combining those elements as they relate to a common dimension, or direction.



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The result of combining the elements of a pair of vectors provides us with a new unique vector. VECTOR ADDITION is defined as calculating the sum between the scalars of each component with respect to each component or dimension. The GENERAL FORMULA for VECTOR ADDITION can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors, 𝐴 and 𝐵: 𝐴 = 𝑎! 𝑖 + 𝑎! 𝑗 + 𝑎! 𝑘 𝐵 = 𝑏! 𝑖 + 𝑏! 𝑗 + 𝑏! 𝑘 We can write the SUM of the two vectors as: 𝐴 + 𝐵 = 𝑎! + 𝑏! 𝑖 + 𝑎! + 𝑏! 𝑗 + 𝑎! + 𝑏! 𝑘 In this problem, we are given the two VECTORS: 𝐴 = 3𝑖 + 7𝑗 + 4𝑘 𝐵 = 2𝑖 + 9𝑗 + 11𝑘



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Recall the COMMUTATIVE LAW states that the order of addition does not matter, we could plug in the elements in whatever order, such that: 𝐴+𝐵 =𝐵+𝐴 We can write the SUM of the two vectors as VECTOR C, which is: 𝐶 = 𝐴 + 𝐵 = 𝑎! + 𝑏! 𝑖 + 𝑎! + 𝑏! 𝑗 + 𝑎! + 𝑏! 𝑘 As you can see, we have our VECTORS broken down in to their individual i and j ELEMENTS and we are just combining them using the ADDITION property. Plugging in the appropriate values we have: 𝐶 = 𝐴 + 𝐵 = 3 + 2 𝑖 + 7 + 9 𝑗 + 4 + 11 𝑘 Which gives us the ELEMENTS of our new vector in VECTOR NOTATION as: 𝐶 = 𝐴 + 𝐵 = 5𝑖 + 16𝑗 + 15𝑘 The correct answer is A. 𝑪 = 𝟓𝒊 + 𝟏𝟔𝒋 + 𝟏𝟓𝒌

SOLUTION 6: The GENERAL FORMULA for SCALAR MULTIPLICATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We



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must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. A SCALAR is a mathematical quantity that retains a magnitude only, whereas, a vector is one that possesses both magnitude and direction. In this problem, we are given a VECTOR: 𝑇 = 3𝑖 + 5𝑗 + 7𝑘 And a SCALAR of:

𝑎=−

1 3

Carrying out SCALAR MULTIPLICATION with a POSITIVE value other than 1 changes the MAGNITUDE of the vector by that SCALE, but not its direction. Carrying our SCALAR MULTIPLICATION by a NEGATIVE value equal to or less than 1 changes the MAGNITUDE of the vector by that SCALE and REVERSES it direction. In this problem, we will be SCALING DOWN the MAGNITUDE as well as FLIPPING the DIRECTION. We can deploy the DISTRIBUTIVE PROPERTY of SCALAR MULTIPLICATION to define our result, which is expressed as: 𝑎 𝐴 + 𝐵 = 𝑎𝐴 + 𝑎𝐵

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Or if we are dealing with only a single vector: 𝑎 𝐴 = 𝑎 𝑎! 𝑖 + 𝑎! 𝑗 + 𝑎! 𝑘 Plugging in our data, we get:

𝑎 𝑇 =−

1 3 𝑖+ 5 𝑗+ 7 𝑗 3

Note that we changed the NOMENCLATURE on the left side of the expression to match the variable of the VECTOR that we are given, which is VECTOR M. This has zero IMPACT on the overall calculation of the problem, that variable can literally be anything, even a poop emoji if you wanted to throw a little humor in to it…do it. And DISTRIBUTING our SCALAR THROUGH:

𝑎 𝑇 =−

1 1 1 3 𝑖− 5 𝑗− 7 𝑗 3 3 3

Resulting in the SCALED DOWN version of our ORIGNAL VECTOR going the OPPOSITE DIRECTION: 5 7 𝑇 ! = −𝑖 − 𝑗 − 𝑘 3 3 𝟓

𝟕

The correct answer is D. – 𝐢 – 𝟑 𝐣 – 𝟑 𝐤



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SOLUTION 7: This problem originally presents itself as complicated due to the fact that it is telling us to SCALE a DIFFERENCE rather than an individual VECTOR. We can do this in a number of different ways, but whatever way we choose, it will take a few additional steps. Let’s hack this problem, first finding the RESULTING VECTOR created when Vector O is subtracted from Vector D and then applying our SCALAR to that new vector. So our first step is simply to make this a VECTOR SUBTRACTION problem. The TOPIC of VECTOR SUBTRACTION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. VECTOR SUBSTRACTION is defined as calculating the DIFFERENCE between the scalars of each component with respect to the dimension in which they reside. In this problem, we are given the two points, which can be written in VECTOR NOTATION as: 𝑂 = 9𝑖 + 10𝑗 𝐷 = 16𝑖 + 4𝑗 And we are asked to determine the DIFFERNCE between these two given VECTORS when VECTOR O is subtracted from VECTOR D.



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The SUBTRACTION of VECTORS is the reverse operation of the additional of vectors. The only difference is that you actually convert the vector being subtracted to negative vector and then add the vectors. The OPERATION that we are asked to carry out is: 𝑁 =𝐷−𝑂 Which is equivalent to the OPERATION: 𝑁 = 𝐷 + (−𝑂) This illustrates how the SUBTRACTION of VECTORS is just the reverse operation of the additional of vectors. So let’s create the NEGATIVE VECTOR for VECTOR O. If: 𝑂 = 9𝑖 + 10𝑗 Then: −𝑂 = −9𝑖 − 10𝑗 Creating this NEGATIVE VECTOR isn’t really necessary, but for illustration and edification purposes we did.

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The GENERAL FORMULA for VECTOR SUBTRACTION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors, 𝑂 and 𝐷, the difference, or the SUBTRACTION, of the two can be written as: 𝐷 − 𝑂 = 𝑑! − 𝑜! 𝑖 + 𝑑! − 𝑜! 𝑗 + 𝑑! − 𝑜 𝑘 Plugging in our ELEMENTS we get: 𝑁 = 𝐷 − 𝑂 = 9 − 16 𝑖 + 10 − 4 𝑗 Which gives us the ELEMENTS of our new vector in VECTOR NOTATION as: 𝑁 = 𝐷 − 𝑂 = −7𝑖 + 6𝑗 We now have a single VECTOR and asked to SCALE it, a scenario we are familiar and comfortable with. The GENERAL FORMULA for SCALAR MULTIPLICATION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. A SCALAR is a mathematical quantity that retains a magnitude only, whereas, a vector is one that possesses both magnitude and direction.

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In this problem, we now have a VECTOR: 𝑁 = −7𝑖 + 6𝑗 And a SCALAR of: 𝑎=4 Carrying out SCALAR MULTIPLICATION with a POSITIVE value other than 1 changes the MAGNITUDE of the vector by that SCALE, but not its direction. Carrying our SCALAR MULTIPLICATION by a NEGATIVE value equal to or less than 1 changes the MAGNITUDE of the vector by that SCALE and REVERSES it direction. In this problem, we will be SCALING UP the MAGNITUDE, the DIRECTION will remain the same. We can deploy the DISTRIBUTIVE PROPERTY of SCALAR MULTIPLICATION to define our result, which is expressed as: 𝑎 𝐴 + 𝐵 = 𝑎𝐴 + 𝑎𝐵 Or if we are dealing with only a single vector: 𝑎 𝐴 = 𝑎 𝑎! 𝑖 + 𝑎! 𝑗 + 𝑎! 𝑘



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Plugging in our data, we get: 𝑎 𝑁 = 4 −7𝑖 + 6𝑗 Note that we changed the NOMENCLATURE on the left side of the expression to match the variable of the VECTOR that we are given, which is VECTOR M. This has zero IMPACT on the overall calculation of the problem, that variable can literally be anything, even a poop emoji if you wanted to throw a little humor in to it…do it. And DISTRIBUTING our SCALAR THROUGH: 𝑎 𝑁 = 4 −7 𝑖 + 4 6 𝑗 Resulting in the SCALED DOWN version of our ORIGNAL VECTOR going the OPPOSITE DIRECTION: 𝑁 ! = −28𝑖 − 24𝑗 It is important to make sure that any values or expressions that we derive in a multistep solution are correct…otherwise, we will experience a Domino Effect that will turn catastrophic at the end of the road. In this problem, we first needed to determine a DIFFERENCE, define a new VECTOR to be SCALED. If we failed at that step, it wouldn’t matter how solid we were in SCALING the VECTOR, if it is wrong, the end result is wrong.

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So stay ACTIVE in your approach in each problem, do these simple regurgitation of your solution at every step and you will be golden. The correct answer is B. – 𝟐𝟖𝒊 + 𝟐𝟒𝒋

SOLUTION 8: The TOPIC of VECTOR ADDITION can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The ADDITION of VECTORS involves breaking down each element of a given vector and combining those elements as they relate to a common dimension, or direction. The result of combining the elements of a pair of vectors provides us with a new unique vector. VECTOR ADDITION is defined as calculating the sum between the scalars of each component with respect to each component or dimension. The GENERAL FORMULA for VECTOR ADDITION can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.



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Given two vectors, 𝐴 and 𝐵: 𝐴 = 𝑎! 𝑖 + 𝑎! 𝑗 + 𝑎! 𝑘 𝐵 = 𝑏! 𝑖 + 𝑏! 𝑗 + 𝑏! 𝑘 We can write the SUM of the two vectors as: 𝐴 + 𝐵 = 𝑎! + 𝑏! 𝑖 + 𝑎! + 𝑏! 𝑗 + 𝑎! + 𝑏! 𝑘 In this problem, we are given the two VECTORS: 𝐴 = 7𝑗 + 4𝑘 𝐵 = 2𝑖 + 9𝑗 The game can be over right here at this point, if we don’t realize something, and realize it fast…that being, the uniformity of the components given. Each of our two vectors have two individual elements defined. It’s common, while under timed pressures, we may be quick to overlook these elements, and just assume that since two are given, that they are of the same dimension. In this problem, that would be catastrophic. VECTOR A has a j and a k ELEMENT, while VECTOR B has an i and a j ELEMENT.



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VECTOR A doesn’t have an i ELEMENT and VECTOR B doesn’t have a k ELEMENT…we need to account for this to ensure that everything unravels accurately when we go to SUM. We can rewrite our two vectors as: 𝐴 = 0𝑖 + 7𝑗 + 4𝑘 𝐵 = 2𝑖 + 9𝑗 + 0𝑘 We now have VECTORS that are accurately representing how we will be SUMMING moving forward. This step is absolutely not necessary, however, for the purposes of pointing it out, we wanted to make sure that we presented it to you as something to look for when working these types of problems. We can now move forward with solving the problem as if it was a standard VECTOR ADDITION problem. Recall the COMMUTATIVE LAW states that the order of addition does not matter, we could plug in the elements in whatever order, such that: 𝐴+𝐵 =𝐵+𝐴 We can write the SUM of the two vectors as a new VECTOR C, which is: 𝐶 = 𝐴 + 𝐵 = 𝑎! + 𝑏! 𝑖 + 𝑎! + 𝑏! 𝑗 + 𝑎! + 𝑏! 𝑘

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As you can see, we have our VECTORS broken down in to their individual i, j and k ELEMENTS and we are just combining them using the ADDITION property. Plugging in the appropriate values we have: 𝐶 =𝐴+𝐵 = 0+2 𝑖+ 7+9 𝑗+ 4+0 𝑘 Which gives us the ELEMENTS of our new vector in VECTOR NOTATION as: 𝐶 = 𝐴 + 𝐵 = 2𝑖 + 16𝑗 + 4𝑘 The correct answer is A. 𝑪 = 𝟐𝒊 + 𝟏𝟔𝒋 + 𝟒𝒌



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