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Understanding vectors and gibberish

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data

    For the line y=2x+4

    A, find the posiotion vector of the point A, on the line, Which is 6 units from the origin in the 1st quadrant. Express your answer in:

    i.] Matrix form. [Never done matrix]

    ii.] the form OA [Line with arrow pointing right above it.] = xi +yj

    B, Find the position vector for the point B on the line, which is 12 units from the origin in the 3rd quadrant. Express your answer in,

    i.] Matrix

    ii.] the form OB [Line above it facing right.]

    C, Find the cordinates of the mid point of AB.

    d, what is the vector equation of the line y=2x+4 in parametric form?

    3. The attempt at a solution

    Now I know I am nearly 19. Finished school 2 years ago and am now working a 5am-5pm job manuel not pen pushing an just find it hard to study and remeber when I am working.
    So forgive me for F%%$ing up.

    Right apart from just drawing the graph of y=2x+4 that is all I get.
    I think they are meaning to add a point 6 units from the 0 middle point on graph. But not sure.

    Which would give me a rectangle. Point starting from -2,0 0,6 14,6. That is if I have done it right. I have done it x,y. That they right way around.

    What on earth do I do from here if I am right?


  2. jcsd
  3. Apr 27, 2010 #2
    First off it might make it easier to graph y = 2x + 4.

    A. 6 units is the distance from the origin (0,0) to the point. So if you drew a straight line from (0,0) to the point A the length of that would = 6. That distance can be found using the Pythagorean theorem. Just draw a dot at a random point in the first quadrant (top right) and draw a line from the origin to it. Then make a right triangle. You will see the sides of the triangle are of size x and y. Then use Pythagorean theorem to get the distance.

    I think the two parts is just notation. The line above the OB means it is a vector [tex] \vec{OB} [/tex] Once you solve for x and y from Pythagorean put those into xi +yj and leave the i and j there. The i and j just mean they are vectors going in the x and y directions respectively. i stands for the x direction and j stands for the y direction. For matrix form either [tex]

    x & y

    either one for whatever your x and y values are.
  4. Apr 27, 2010 #3


    User Avatar
    Homework Helper

    so lets start with A i.)

    first, always start by drawing a picture

    Now draw the line y=2x+4
    Then draw a circle of radius 6, with centre (0,0)
    You're looking for where the line and circle intersect.
    There will be 2 intersection points.
    The question asks for the first quandrant, that is just the upper right section of the coordinate axes where both x & y are positive.

    You can solve for this by writing the equation of a circle of radius 6
    [tex] x^2 + y^2 = 6^2 [/tex]

    use the equation of a line to substitute in and solve for x or y
    if you find the solution is:
    x=a, y = b
    i) then in matrix form it will be (a,b) (or vertically with a & b on top of each other)
    ii) write the same numbers as ai+bj
    Last edited: Apr 28, 2010
  5. Apr 28, 2010 #4
    Thank you all very much. That makes much clearer sence.

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