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# Maths homework help please watch

1. 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
2. (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.
3. (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
4. (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.
5. (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
6. (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)
7. (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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