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Math Help - Vectors

  1. #1
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    Vectors

    I hope someone can help me with this, I have never posted here before.

    Given vectors a=(3,5) and b=(2,3). Find 5a-b.

    Thanks if you can help me.
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    Quote Originally Posted by gretchen View Post
    I hope someone can help me with this, I have never posted here before.

    Given vectors a=(3,5) and b=(2,3). Find 5a-b.
    Hello,

    5a=(15, 25). Thus

    5a-b=(15, 25) - (2, 3) = (13, 22)

    EB
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  3. #3
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    Quote Originally Posted by gretchen View Post
    I hope someone can help me with this, I have never posted here before.

    Given vectors a=(3,5) and b=(2,3). Find 5a-b.

    Thanks if you can help me.
    Multiplying a vector by a scalar (such as 5a here) multiplies each component
    of the vector by the scalar (so 5a=(5*3, 5*5)=(15, 25)). Adding or
    subtracting two vectors gives a vector whoese components are the sum of
    difference of the corresponding components of the two vectors.

    So:

    5a-b=5(3, 5) - (2, 3) = (15, 25) - (2, 3) = (15-2, 25-3) = (13, 22)

    RonL
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  4. #4
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    Thanks for the help, I see what I had to do now. What about going at it from a different perspective?

    Given vectors a=3i-2j and b=4i+4j. Find 5a-3b.
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    Quote Originally Posted by gretchen View Post
    Thanks for the help, I see what I had to do now. What about going at it from a different perspective?

    Given vectors a=3i-2j and b=4i+4j. Find 5a-3b.
    5a - 3b = 5(3i - 2j) - 3(4i + 4j) = 15i - 10j - 12i -12j = (15i - 12i) + (-10j -12j) = 3i - 22j

    -Dan
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    Thanks guys for your help. I have another one that is confusing me:

    Find the vector that has the same direction as 3i-7j and twice the magnitude.

    Thanks,
    Gretch
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    Quote Originally Posted by gretchen View Post
    Thanks guys for your help. I have another one that is confusing me:

    Find the vector that has the same direction as 3i-7j and twice the magnitude.

    Thanks,
    Gretch
    Just double it,
    6i-14j

    Doubling vector coordinates keeps it in the same direction and twice as long.
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  8. #8
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    What about when asking for a dot product:

    Find the dot product of the vectors i-3j and 4i+2j.
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    Quote Originally Posted by gretchen View Post
    What about when asking for a dot product:

    Find the dot product of the vectors i-3j and 4i+2j.
    Multiply the components.
    Meaning the i with the i and j with the j.

    Thus,
    (1)(4)+(-3)(2)
    Is the dot product.
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  10. #10
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    Thanks, I will work the rest like that. Is there any trick to coming up with angles from two vectors?

    Find the angle between the two vectors (5,2) and (-2, 5).

    Thanks again.
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  11. #11
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    Quote Originally Posted by gretchen View Post
    Thanks, I will work the rest like that. Is there any trick to coming up with angles from two vectors?

    Find the angle between the two vectors (5,2) and (-2, 5).

    Thanks again.
    Hello, Gretchen,

    there isn't any trick to calculate the angle - only a simple formula:

    Let \alpha be the angle between the vectors \vec a and \vec b then you get the angle by:

    \cos(\alpha)=\frac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}

    Use this formula with your vectors:

    \cos(\alpha)=\frac{(5,2)  (-2 ,5)}{\sqrt{5^2+2^2} \cdot \sqrt{(-2)^2+5^2}}=\frac{0}{29}. Thus \alpha = 90^\circ

    By the way: This result shows: Two vectors are perpendicular if their dot-product equals zero.

    EB

    PS: Please do us a favour and start a new thread if you have a new problem to do. Otherwise you risk that nobody will notice that you ask for some help again.
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  12. #12
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    Quote Originally Posted by gretchen View Post
    Thanks, I will work the rest like that. Is there any trick to coming up with angles from two vectors?

    Find the angle between the two vectors (5,2) and (-2, 5).

    Thanks again.
    The formula is,
    \bold{u}\cdot \bold{v} = ||\bold{u} ||\cdot ||\bold{v}|| \cos \theta
    Where,
    \bold{u}\cdot \bold{v}---> Dot product.
    ||\bold{u}||\cdot ||\bold{v}||---> Product of their norms.
    And,
    \cos \theta is the angle between them.
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