Solving Vector Questions 1-7: Tips and Tricks for Success!

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In summary: In this case, [7,9] and [13,2] are the two vectors that span R^2.3) This one I don't really know how to do. Maybe you could give me some feedback on that?4) I don't really know how to do this one either.5) oooh k, this one I know how to do!Q1: so the answer is, [-1, -2, 7]?In summary, vector AB =[-1, -2, 7] and A(-4, 3, 0) have the coordinates (3, -5, 7).
  • #1
xChee
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i'm probably posting in the wrong section but idk where else to post this :/
I need help with a few questions !

Homework Statement



Q1: if vector AB =[3,-5,7] and A(-4, 3, 0), state the coordinates of B

Q2: State a set of vectors that spans [itex]R^{2}[/itex] other than i or j

Q3: let a represent vector a, b vector b, and m be a constant. Given a=[1,2] and b= [2,5], determine m if 2a+mb= [8,19]

Q4: Express vector u as an algebraic vector give the magnitude of vector u = 36 and it has a bearing of 120 degrees.

Q5: state x & y give vector w=[x, 0], vector r=[4, y] and -3w-5r= [2, 1]

Q6: determine the point on the y-axis equidistant from the points (2, -1, 1) and (0, 1, 3)

q7:solve for x given vector u=[3x, 7]; vector v= [5x, x]; the magnitude of vector u plus vector v = 10x


Homework Equations





The Attempt at a Solution



Q1: is it (-7, 8 -7)?
Q2: does [7,9] work?
Q3: this one idk what to do...
Q4: idk how to do this question either..
Q5: i wrote my answer as x= 22/3 and y= -1/5 and got a half out of one... where did i go wrong?
Q6: is started off finding the midpoints my doing (2 + 0)/2, (-1+1)/2, and (1+3)/2. but don't know what to do after..
Q7: i started off by adding 3x and 5x which is 8x. then did x+7. then did the square root of (8x)^2 and (x+7)^2, to solve for x... but it didn't work out nicely...
 
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  • #2
roger Roger, what's your vector Victor!

I'll give you a hand here;
1) no
If AB is the vector pointing from A to B and A is the vector pointing from 0 to A, what happens when we add A and AB?

2)[7,9] is only one vector, the space generated by this is just a line (aka [itex]R^1[/itex]

3)This is just a set of two linear equations for one unknown
if a = [1,2] and b = [2,5]
2a+mb = [8,19] says that
2*[1,2] + m*[2,5] = [8,19]
Which is equivelant to the statement
2*1 + 2m = 8
2*2 + 5m = 19
(Since addition of vectors is componentwise)

4)Picture a triangle with hypotenuse of length 36 at a 120 degree angle from the x axis, how would you find the x and y components of the point at the end of the hypotenuse (what do Cos and Sin do?)

5)This is the same thing as question 3

6)Look at the problem componentwise, what is the midpoint of 2 and 0, what is the midpoint of 1 and -1, what is the midpoint of 1 and 3?

7)Are you sure you copied this correctly, I didn't get a very nice answer for x either, [itex]\frac{1}{65} \left(-2+i \sqrt{3181}\right)[/itex]
 
  • #3
genericusrnme said:
1) no
If AB is the vector pointing from A to B and A is the vector pointing from 0 to A, what happens when we add A and AB?

2)[7,9] is only one vector, the space generated by this is just a line (aka [itex]R^1[/itex]

3)This is just a set of two linear equations for one unknown
if a = [1,2] and b = [2,5]
2a+mb = [8,19] says that
2*[1,2] + m*[2,5] = [8,19]
Which is equivelant to the statement
2*1 + 2m = 8
2*2 + 5m = 19
(Since addition of vectors is componentwise)

4)Picture a triangle with hypotenuse of length 36 at a 120 degree angle from the x axis, how would you find the x and y components of the point at the end of the hypotenuse (what do Cos and Sin do?)

5)This is the same thing as question 3

6)Look at the problem componentwise, what is the midpoint of 2 and 0, what is the midpoint of 1 and -1, what is the midpoint of 1 and 3?

7)Are you sure you copied this correctly, I didn't get a very nice answer for x either, [itex]\frac{1}{65} \left(-2+i \sqrt{3181}\right)[/itex]

Q1: so the answer is, [-1, -2, 7]?

Q2: so to get vectors in two space you need to sets? so... [7,9] and [13,2]?

Q3: m= 3? :D

Q4: do you just express vector u in terms of the sin or cos law?

Q5: oooh k

Q6: the mid points are 1, 0, 2, i got that but idk what to do with those numbers after... or is that the answer?

q7: yea, nvm...
 
  • #4
xChee said:
Q1: so the answer is, [-1, -2, 7]?

Q2: so to get vectors in two space you need to sets? so... [7,9] and [13,2]?

Q3: m= 3? :D

Q4: do you just express vector u in terms of the sin or cos law?

Q5: oooh k

Q6: the mid points are 1, 0, 2, i got that but idk what to do with those numbers after... or is that the answer?

q7: yea, nvm...

1) Yes, I hope you understand WHY this is though, rather than just adding them and hoping for the best

2) Yes, again, I hope you understand WHY this is. For a set of vectors to span a vector space, you need to be able to write ANY vector in the space as a combination of vectors in the set. With just one vector, say [7,9] you can only write vectors of the form [7a,9a]

3)bingo!

4)Yes

6)Where does the point [1,0,2] lie?
 
  • #5
genericusrnme said:
6)Where does the point [1,0,2] lie?

1 on the x axis, 0 on the y and 2 on the z

so the answer is 0 ? :s
 
  • #6
xChee said:
1 on the x axis, 0 on the y and 2 on the z

so the answer is 0 ? :s

you're looking for a point on the y axis
1. What must the x and z components be if the point is to lie on the y axis
2. How would you define a plane that is perpendicular to the line between the two points and passes the midpoint?
3. Using the above, what do you think your answer should be?
 
  • #7
genericusrnme said:
you're looking for a point on the y axis
1. What must the x and z components be if the point is to lie on the y axis
2. How would you define a plane that is perpendicular to the line between the two points and passes the midpoint?
3. Using the above, what do you think your answer should be?

to lie on the y-axis z and x have to be 0. the the coordinates should be [0, y, 0]?

i don't get step 2 though... stuck...

edit; nvm do you just do, (2-0)^2 + (-1 -y)^2 + (1-0)^2 = (0-0)^2 + (1-y)^2 + (3-0)^2

and solve for y? and y should be 1?
 
Last edited:
  • #8
you could also do that, yes :biggrin:

my idea was to find the plane of all equidistant points and then find the one that lies on the y axis
 

FAQ: Solving Vector Questions 1-7: Tips and Tricks for Success!

How can I get help with questions 1-7?

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