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Vector Questions

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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

Thanks
 
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Answers and Replies

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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:
 
HallsofIvy
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courtrigrad said:
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]
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.

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.
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.

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?
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.

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?
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?

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

(b) If a + b = a - b does this mean that b = 0 ?
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?

(c) a + b = c and [itex] a^{2} + b^{2} = c^{2} [/itex] this means that a and b are perpindicular?
Draw a picture! Think "Pythagorean Theorem".

Any help is appreciated

Thanks
You're welcome.
 
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Hurkyl
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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.
 
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(b) If a + b = a - b does this mean that b = 0 ?
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?

Thanks
 
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HallsofIvy
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courtrigrad said:
That means a = b
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?

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?

Thanks
Yes, that is correct.
 
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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?

Thanks
 
HallsofIvy
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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.
 
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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]?
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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?
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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]?
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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
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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?
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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?
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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?
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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:
 
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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?

Thanks
 
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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?
 
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Any help/feedback would be appreciated
 
HallsofIvy
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Dude! You're overwhelming us here! I'm calling for backup!

courtrigrad said:
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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]?
***********************************
You are fine until the question about the angle. You can't find the angle between two vectors with just two components- and in general [itex]\frac{r_z}{r_y}[/itex] is not equal to [itex]\frac{r_y}{r_x}[/itex]. The best way to find the angle between two vectors is to use the formula you cite later: [itex] x\dot y= |x||y|cos\theta[/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?
Do you have something against negative numbers? As I said in a previous post, before you can talk about components, you have to have a coordinate system- and you need to say what it is! I surmise that you are taking your coordinate system to have (0,0,0) at the man's front door, positive x-axis east, positive y-axis north, and positive z-axis up. In that case, since he dropped the penny down what should the k component be.
(Your answer would be completely correct if you specified the positive z-axis was downward that is so unusual (among other things it is a left-hand coordinate system) that you had better say that clearly!)

(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?
Of course! The resultant displacement is 0 on any "round trip"!

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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]?
That's really a complicated problem! Here's my suggestion: draw a "blown up" picture of the room (i.e. cut along the edges of the walls and "open" the picture into connected rectangles). Mark the starting and ending points and draw a straight line connecting them (going outside the walls, etc. if neccessary. Now "close" the picture (you might want to actually cut and fold to do this) and see where the lines meet on the edges.

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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
Convert from spherical coordinates (θ= longitude, φ= 90- Latitude, and ρ= radius of the earth) to x,y,z coordinates.

***********************************************
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?
Here a and b are the lengths of the vectors. I would recommend choosing your coordinate system so that the x-axis is along vector a. That way, a has components ai+ 0j. Write b as xi+ yj where x2+ y2= b2. Now look at the right angle formed. By the way- that formula is simply the cosine law.

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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?
Yes, of course.

(b)The direction of [itex] b\times a [/itex] would be in the -z direction and perpendicular to both a and b .
Yes. You don't need to say "perpendicular to both ...".

(c) The direction of [itex] \frac{b}{d} [/itex] would depend on the scalar d. So would the direction be +y or - y?
Yes. Another way to say that is "parallel or anti-parallel to y depending on whether d is positive or negative. Oh- there's one other possibility: if d is 0, there is no direction.

(d) I am assuming [itex] a\bullet b = ab\cos\theta [/itex]. Would this be correct?
Yes

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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?
I'm tempted to say "because they are defined that way". Yes, dot product is defined using |a||b| cos(θ) and multiplication of numbers is commutative. cross products are defined using the right-hand-rule and that depends on the order of the vectors.

*********************************************
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?
I don't know what you mean by "there is no direction" in the real world! (Actually I'm not to clear on what the "real world" is any more!)

The associative law for any products says a(bc)= (ab)c. If a, b, and c are vectors and this is dot product, then bc is a number, not a vector so the we can't have a dot product of a(bc)!

Whew! I'm going to take a (long) break! I do have a life you know!
 
courtrigrad said:
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?

Thanks
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.

courtrigrad said:
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?
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.
 

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