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    The vector p has a magnitude 5 and direction due north and the vector q has magnitude 7 and direction on a bearing of 125°
    A) Find the magnitude and direction of vectors p+q and p-2q
    B)the vector p+kq where k is a constant has direction due east. Find the value of the constant k

    Okay so I did part A and got p+q to be 5.82 and p-2q to be 17.35

    I don't understand how to do part B
    Any help would be appreciated
    Thanks
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    (Original post by Amberinho)
    The vector p has a magnitude 5 and direction due north and the vector q has magnitude 7 and direction on a bearing of 125°
    A) Find the magnitude and direction of vectors p+q and p-2q
    B)the vector p+kq where k is a constant has direction due east. Find the value of the constant k

    Okay so I did part A and got p+q to be 5.82 and p-2q to be 17.35

    I don't understand how to do part B
    Any help would be appreciated
    Thanks
    If it's due East, the j-component (north-south component) will be 0. Write out what the vector p+kq is in terms of k and see what value of k makes the j component 0.
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    (Original post by StayWoke)
    If it's due East, the j-component (north-south component) will be 0. Write out what the vector p+kq is in terms of k and see what value of k makes the j component 0.
    Wait so would it be k=-p/q
    So k=-5/7
    Or have I misunderstood
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    (Original post by Amberinho)
    Wait so would it be k=-p/q
    So k=-5/7
    Or have I misunderstood
    I don't understand what you're doing. Add the vectors p and kq together to get a new vector, which is just in terms of k. You look at the j-component of this vector and see which value of k makes the j-component 0.
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    (Original post by StayWoke)
    I don't understand what you're doing. Add the vectors p and kq together to get a new vector, which is just in terms of k. You look at the j-component of this vector and see which value of k makes the j-component 0.
    I'm really sorry I'm still a little Confused I drew a right angled triangle where the vertical is P=5 and the horizontal is kq =7k
    So 7k^2 +5^2 then square rooted gives the hypotenuse which is root 74k am I right so far
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    (Original post by Amberinho)
    I'm really sorry I'm still a little Confused I drew a right angled triangle where the vertical is P=5 and the horizontal is kq =7k
    So 7k^2 +5^2 then square rooted gives the hypotenuse which is root 74k am I right so far
    no. What year are you in? Do you know how to add vectors together (without drawing diagrams)?

    If you want to make a right angled triangle: Your first vector is P, which is correct, it's 5 units up, but then next vector is kq, whose magnitude is 7k, but you have to make sure you get it's direction right, it's at a 125deg bearing and you have to join P and kq together "tip to tail". With the tail of kq on the tip of P. This makes kq the hypotenus. Your resultant vector (from the tail of P to the tip of kQ) should be a horizontal line (which is the other leg of the right angled triangle)
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    (Original post by StayWoke)
    no. What year are you in? Do you know how to add vectors together (without drawing diagrams)?

    If you want to make a right angled triangle: Your first vector is P, which is correct, it's 5 units up, but then next vector is kq, whose magnitude is 7k, but you have to make sure you get it's direction right, it's at a 125deg bearing and you have to join P and kq together "tip to tail". With the tail of kq on the tip of P. This makes kq the hypotenus. Your resultant vector (from the tail of P to the tip of kQ) should be a horizontal line (which is the other leg of the right angled triangle)
    Okay yeah I understand thank you 😊
 
 
 
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