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Mathematics: Post your doubts here!

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Vectors can be challenging sometimes... But still.. vectors in P1 are lots easier than the vectors which I did in O level Add Maths (Which I still haven't understood) :p

To be honest... I find vectors one of the easier topics. Functions is pretty hard.. :unsure:
 
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I have almost done with 10 past papers and still on 60,62/75 in P1.... grade thresholds are to high. I am worried for the final
 
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specially the functions integrated with trigonometry or the functions question which is Q10 or 11......:D
 
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I have a problem in the whole vector chapter for p3 .. does anybodyhave a link for notes I can use to understand a2 vectors? thanks
 
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I have a problem in the whole vector chapter for p3 .. does anybodyhave a link for notes I can use to understand a2 vectors? thanks
If you want to have best understanding than read Pure Mathematics 2&3 OF (Hodder Education)
 
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AB = OB - OA
AB = (2-5)i+(7-1)j+(p-2)k
= -3i + 6j + 2k

Now make it a unit vector, so we can later multiply the magnitude of the vector we need (28) to get the vector itself.
For that you'll need to find the modulus by finding the square root of the sum of i^2, j^2 and k^2...

That gives you
Modulus = sqrt((-3)^2 + 6^2 + 2^2)
= 7

To find the unit vector, we need to divide our vector by the modulus, or multiply it by 1/modulus as so...

Unit vector = 1/7(-3i+6j+2k)

Multiply the magnitude (28) by the unit vector, you'll get
(1/7)(28)(-3i+6j+2k)

4(-3i+6j+2k)
Ans. = -12i+24j+8k
 
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Since you're only asking about D, I'm going to guess you found M... which is (5,2)

M is the midpoint between B and D

The formula for midpoint would be..
(x2-x1)/2 , (y2-y1)/2

So (x2-x1)/2, where x2 is the x value of D, and x1 is the x value of B are equal to 5, the x value of M the midpt. Solve.
So (y2-y1)/2, where y2 is the y value of D, and y1 is the y value of B are equal to 2, the y value of M the midpt. Solve.

You'll get
x2= 7
y2= -2
 
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AB = OB - OA
AB = (2-5)i+(7-1)j+(p-2)k
= -3i + 6j + 2k

Now make it a unit vector, so we can later multiply the magnitude of the vector we need (28) to get the vector itself.
For that you'll need to find the modulus by finding the square root of the sum of i^2, j^2 and k^2...

That gives you
Modulus = sqrt((-3)^2 + 6^2 + 2^2)
= 7

To find the unit vector, we need to divide our vector by the modulus, or multiply it by 1/modulus as so...

Unit vector = 1/7(-3i+6j+2k)

Multiply the magnitude (28) by the unit vector, you'll get
(1/7)(28)(-3i+6j+2k)

4(-3i+6j+2k)
Ans. = -12i+24j+8k
Thanks man :D
 
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