1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector Questions

  1. Aug 21, 2005 #1
    Just had some general questions, and some more specific problems regarding vectors:

    1. Can a vector be zero if one of its components is not zero? Let us assume that r = a+b . Then [itex] r_{x} = a_{x} + b_{x} [/itex] and [itex] r_{y} = a_{y}+b_{y} [/itex]. So [itex] r = \sqrt{r_{x}^{2} + r_{y}^{2}} [/itex]. I'm guessing that both components have to be 0, so that [itex] 0 = \sqrt{r_{x}^{2} + r_{y}^{2}} [/itex]

    2. Name some scalar quantities. Is the value of a scalar quantity dependent on the reference frame chosen? I am choosing temperature as a scalar quantity. I dont think the reference frame matters because you can convert between systems.

    3. We can order events in time. There can be a time order between three events a,b,c. Is time a vector then? It has a magnitude, but I would not say a direction because even scalar quantities can be "ordered." Is this correct?

    4. Two vectors a and b are added. Show that the magnitude of the resultant cannot be greater than a+b or smaller than | a-b |. Wouldnt I just use the triangle inequality theorem in which the sum of any two sides has to be greater than the third side, and the difference of any two sides has to less than the third side?

    5. If we have two vectors a and b then when (a) a + b = c and a + b = c does this mean the the vectors are neither parallel nor perpindicular? (b) If a + b = a - b does this mean that b = 0 ? (c) a + b = c and [itex] a^{2} + b^{2} = c^{2} [/itex] this means that a and b are perpindicular?

    Any help is appreciated

    Last edited: Aug 21, 2005
  2. jcsd
  3. Aug 21, 2005 #2
    The answer to your very first question is no. If we take the case of a vector which has 2 components x and y. The resultant is:

    [tex] r = (x^2 + y^2)^\frac {1}{2} [/tex]

    If either of these components is zero the resultant vector r is still non-zero. If both are zero then of course the vector is zero.

    As for your second question. A scalar quantity such as speed would certainly depend upon the reference frame.

    As for the rest, take pen and paper and make some sketches.

    :rofl: :rofl: :rofl: :rofl: :rofl: :rofl: :rofl: :rofl:
  4. Aug 21, 2005 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    As you guessed, the answer is no. Another way to see it is to square both sides of of [itex]0=\sqrt{r_{x}^{2} + r_{y}^{2}} [/itex]. You get [itex]0= {r_x}^2+ {r_y}^2[/itex]. Since squares are never negative, the two terms can't cancel. The only way the sum can be 0 is if both terms are 0.

    Very good. temperature can be written as a single number and so is a scalar. You are correct that one of the thing saying that temperature is a scalar means is that the temperature at a given point does not depend upon the reference frame. Of course, the numbers used to "label" that point will so the answer to the question "What is the temperature at (1, 0, 1)?" may depend on the reference frame.

    This is a bit ambiguous. I would accept your answer as correct- although since there exist + and - directions, you can sometimes think of time as vector.

    Yes, you can do that. (I presume you are using a to mean the length of a.) You would also want to look at the "extreme" examples: where a and b are in the same direction and where they are in opposite directions. What kind of triangle do you have then?

    No, it doesn't. Look at what I just said in (4).

    Well, you can do arithmetic on vectors as easily as numbers. What happens if you subtract a and add b to both sides of the equation?

    Draw a picture! Think "Pythagorean Theorem".

    You're welcome.
    Last edited: Aug 21, 2005
  5. Aug 21, 2005 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right, there is some ambiguity... any field of scalars can also be said to be a one-dimensional vector space, so each scalar can be said to be a one dimensional vector.

    However, we can look at how we're using the objects. I think, based on usage, that we would say time is a scalar, but duration is a vector.
  6. Aug 21, 2005 #5
    That means b = 0

    6. Generalize the analytical method of resolution and addition to the case of three or more vectors.

    Let r = a + b + c . In a given coordinate system, three vectors such as r and a+b+c can only be equal if their corresponding components are equal, or [itex] r_{x} = a_{x} + b_{x} + c_{x} [/itex] and [itex] r_{y} = a_{y}+b_{y}+c_{y} [/itex] and [itex] r_{z} = a_{z}+b_{z}+c_{z} [/itex] Hence [itex] r = \sqrt{(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}) }[/itex]. Thus for n vectors we would have [itex] r = \sqrt{(r_{x}^{2} + r_{y}^{2} + r_{z}^{2} + ... + r_{n}^{2})} [/itex]. Is this correct? I am guessing when they want the analytical method of resolution and addition of three or more vectors they want the angle also. You cant work with the tangent function, so how would you find the angle?

    Last edited: Aug 21, 2005
  7. Aug 21, 2005 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    NO! As Hurkyl said, any scalar can be considered a "one-dimensional" vector so, as a check, see if it works for numbers. If a and b are numbers such that a+ b= a- b, try a= b= 2.

    What I said before was "subtract a and add b to both sides of the equation".
    a+ b= a- b and subtracting a from both sides give b= -b: a is now gone completely. Adding b to both sides gives -2b= 0. What does that tell you about b?

    Yes, that is correct.
  8. Aug 21, 2005 #7
    7. A golfer takes three strokes to get his ball into the hole once he is on the green. The first stroke displaces the ball 12 ft north, the second stroke 6.0 ft southeast, and the third stroke 3.0 ft southwest. What displacement was needed to get the ball into the hole on the first stroke?

    Ok so I have three vectors a = 12j , b = 6cos45i + 6sin45j , and c = 3cos45i + 3sin45j . So would I just add these vectors and get r = (6cos45 + 3 cos45)i + (12+ 6sin45 + 3sin45) j . Am I correct in assuming that the angle is 45 degrees?

  9. Aug 21, 2005 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    You always have to have a coordinate system before you can separate a vector in to components. Since your first vector is 12j I assume you are using a kind of standard coordinate system: i is east, j is north.
    With that coordinate system, yes, "12 ft north" is 12j. Yes, "southeast" is exactly halfway between e and s. Now you COULD think of that as "45 degrees s of e". Then the x-component of "6 ft, se" is 6 cos 45 but because it is south and your j is north, the y-component would be -6 sin 45: the vector is b= 6 cos(45)i- 6 sin(45)j. Notice the "-"!!!
    c is "3.0 ft southwest". Again sw is 1/2 way between s and w or 45 degrees from s. However, notice that west is "-i" and south is "-j" since we have define i as pointing east and j north. c= -3 cos(45)- 3 sin(45).

    Now add those to find the "resultant" vector.
  10. Aug 22, 2005 #9
    8. Generalize the analytical method of resolving and adding two vectors in three dimensions.

    So let r = a+b be the sum of two vectors a and b lying in the x-y-z plane. In a given coordinate system, two vectors such as r and a+b can only be equal if their corresponding components are equal, or [itex] r_{x} = a_{x} + b_{x} [/itex], [itex] r_{y} = a_{y} + b_{y} [/itex] and [itex] r_{z} = a_{z} + b_{z} [/itex]. Thus [itex] r = \sqrt{(r_{x}^{2} + r_{y}^{2} + r_{z}^{2})} [/itex]. The only problem is how would you find the angle between the vectors ? Would you just equate [itex] tan\theta = \frac{r_{z}}{r_{y}} = \frac{r_{y}}{r_{x}} [/itex]?
    9. (a) A man leaves his front door, walks 1000 feet east, 2000 feet north, and then takes a penny from his pocket and drops it from a cliff 500 feet high. Set up a coordinate system and write down the expression, using unit vectors, for the displacement of the penny.

    (a) I am using i,j,k to mean x,y,z coordinates respectively. So a = 1000i + 2000j + 500k . Would this be correct?

    (b) The man then returns to his front door, following a different path on the return trip. What is the resultant displacement for the round trip?

    Wouldnt it just be 0, because he ends up where he started?
    10. A room has dimensions 10 [itex] \times [/itex] 12 [itex] \times [/itex] 14 ft. If a fly crawls what is the shortest path it can take in crossing at one corner to a diametrically opposite corner.

    I already figured out the magnitude of its displacement if the fly flew (using Pythogrean Theorem). Would the shortest distance be when the fly goes down 14 ft, and then across the diagonal ([itex] \sqrt{(10^{2} + 12^{2})} [/itex]?
    11. A man flies from Washington to Manila. Describe the displacement vector. What is its magnitude if the latitude and longitude of the two cities are 39 N, 77 W and 15 N, 121 E?

    I had no idea how to start this problem mainly because of latitudes and longitudes. I know we need to use a three dimensional coordinate system with i,j,k
    12. Two vectors of lengths a and b make an angle [itex] \theta [/itex] with each other when placed tail to tail. Prove, by taking components along two perpendicular axes, that the length of the resultant vector is [itex] \sqrt{a^{2}+b^{2}+2ab\cos\theta}[/itex]

    So [itex] a_{x} = a\cos\theta [/itex], [itex] b_{x} = b\cos\theta [/itex], [itex] a_{y} = a\sin\theta [/itex], [itex] b_{y} = b\sin\theta [/itex]. So [itex] r_{x} = a_{x} + b_{x} [/itex] and [itex] r_{y} = a_{y} + b_{y} [/itex]. So [itex] r = \sqrt{r_{x}^{2} + r_{y}^{2}} [/itex]. Which means that [itex] \sqrt{[(a+b)\cos\theta)^{2} + (a+b)\sin\theta)^{2}]} [/itex]. Would this be correct? How would you simplify it?
    13. Use the standard xyz system of coordinates. Given vector a in he + x-direction, vector b in the +y-direction, and the scalar quantity d: (a) What is the direction of [itex] a\times b [/itex]? (b) What is the direction of [itex] b\times a [/itex]? (c) What is the direction of [itex] \frac{b}{d} [/itex]? (d) What is the magnitude of [itex] a\bullet b [/itex]?

    (a) The direction of [itex] a\times b [/itex] would be perpendicular to both a and b in the +z direction. Is this correct , following from the right hand rule?
    (b)The direction of [itex] b\times a [/itex] would be in the -z direction and perpendicular to both a and b .
    (c) The direction of [itex] \frac{b}{d} [/itex] would depend on the scalar d. So would the direction be +y or - y?
    (d) I am assuming [itex] a\bullet b = ab\cos\theta [/itex]. Would this be correct?
    14. Why doesnt the commutative law apply to vector products? Why does it only apply to dot products?

    I think that because cross products involve direction and magnitude, it cant be commutative. Whereas dot products only involve scalars and magnitude. Is this correct?
    15. I know that the associative law applys to vector products. Would it make sense to talk about an associative law for a scalar product? I am guessing that you can have an associative law, but it doesnt make sense in the real world because there is no direction. Is this correct?

    Thanks :smile:
    Last edited: Aug 22, 2005
  11. Aug 22, 2005 #10
    16. Let two vectors be represented in terms of their coordinates as: a = [itex] ia_{x} + ja_{y} + ka_{z} [/itex] and b = [itex] ib_{x} + jb_{y} + kb_{z} [/itex]. Show analytically that [itex] a\bullet b = a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z} [/itex]. Well I know that [itex] i\bullet i = j\bullet j = k\bullet k = 1 [/tex]. Also [itex] i\bullet j = j\bullet k = k\bullet i = 0 [/itex]. So I then use this and get [itex] (ia_{x}+ja_{y} + ka_{z})(ib_{x} + jb_{y} + kb_{z}) = a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z} [/itex]. Is this correct?

  12. Aug 22, 2005 #11
    17. Use the definition of scalar product [itex] a\bullet b = ab\cos\theta [/itex] and the fact that [itex] a\bullet b = a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z} [/itex] to obtain the angle between to vectors given by a = 3i + 3j - 3k and b = 2i + j + 3k . So [itex] (3\times 2) + (3\times 1) + (-3\times 3) = 0 [/itex]. So [itex] ab\cos\theta = 0 [/itex] so [itex] \cos^{-1} \theta = 0 [/itex] and [itex] \theta = 90 [/itex]. Is this correct?
  13. Aug 23, 2005 #12
    Any help/feedback would be appreciated
  14. Aug 23, 2005 #13


    User Avatar
    Staff Emeritus
    Science Advisor

    Dude! You're overwhelming us here! I'm calling for backup!

  15. Aug 23, 2005 #14
    It's one way to do it, but proving it is only good for memorization purposes. I remember just memorizing the dot product, and using it for whatever was necessary. The definition that [itex] a/bullet b= ab\cos\theta[/itex] was much more useful for me in physics classes, while the defintion you just proved was used more in my math classes. Of course, it all depends on your teacher.

    Yes, you did it correctly.

    You've asked a lot of questions, and I'm just guessing you're nervous about physics in general. The number one thing is to practice so that you can build up a little confidence, so that you can do everything without double checking. Just don't get too confident, and you'll find physics fun, and maybe even easy.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Vector Questions
  1. Vector question (Replies: 4)

  2. Vector question (Replies: 1)

  3. Vectors question. (Replies: 21)

  4. Vector question (Replies: 2)

  5. Vector question (Replies: 6)