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c4 vectors

hello. need help with part 2 of this question, i thought you would dot product the two vectors, divide it by the product of their respective magnitude and then inverse cos the hwole thing. i don not get the right answer and am not sure what else to do

ie,i was using this formula
cosθ=abab\cos \theta = \dfrac{a{\cdot}b}{|a|{\cdot}|b|}

and got

cosθ=2129310\cos \theta = \dfrac{-21}{\sqrt{29}{\cdot}3\sqrt{10}}

(edited 11 years ago)
Reply 1
Original post by DopeSk1llz
hello. need help with part 2 of this question, i thought you would dot product the two vectors, divide it by the product of their respective magnitude and then inverse cos the hwole thing. i don not get the right answer and am not sure what else to do

cosθ=abab\cos \theta = \frac{a{\cdot}b}{|a|{\cdot}|b|}


What's the question?
what question?
Reply 3
Original post by joostan
What's the question?



Original post by Philippamalko
what question?


posted it now!
Reply 4
Original post by DopeSk1llz
hello. need help with part 2 of this question, i thought you would dot product the two vectors, divide it by the product of their respective magnitude and then inverse cos the hwole thing. i don not get the right answer and am not sure what else to do

ie,i was using this formula
cosθ=abab\cos \theta = \frac{a{\cdot}b}{|a|{\cdot}|b|}

and got

cosθ=2129310\cos \theta = \frac{-21}{\sqrt{29}{\cdot}3\sqrt{10}}


Your magnitudes are incorrect.
Edit: i realise you got the correct ones, your notation confused me :tongue:
Is the problem that you have an obtuse angle?
(edited 11 years ago)
Reply 5
Original post by joostan
Your magnitudes are incorrect.
Edit: i realise you got the correct ones, your notation confused me :tongue:


haha! can you identify where else i may have gone wrong then? or am i approaching the question in the wrong way?
Reply 6
Original post by DopeSk1llz
haha! can you identify where else i may have gone wrong then? or am i approaching the question in the wrong way?

To find the acute angle from the obtuse angle - you simply use 180 or pi - (STUFF FROM YOUR CALCULATOR)
Original post by joostan
To find the acute angle from the obtuse angle - you simply use 180 or pi - (STUFF FROM YOUR CALCULATOR)


Sorry to intrude but wouldn't that still give you the wrong answer because the question wanted the angle OAB so wouldn't you have to use the result from the first part to help work out your angle?


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Reply 8
Original post by joostan
To find the acute angle from the obtuse angle - you simply use 180 or pi - (STUFF FROM YOUR CALCULATOR)


strange. i still seem to get the wrong answer. the mark scheme states the answer is 43 to one degree, or 0.75 in radians.

my answer when doing this gives me 65.7 degrees
Reply 9
Original post by MathsNerd1
Sorry to intrude but wouldn't that still give you the wrong answer because the question wanted the angle OAB so wouldn't you have to use the result from the first part to help work out your angle?
Posted from TSR Mobile



Original post by DopeSk1llz
strange. i still seem to get the wrong answer. the mark scheme states the answer is 43 to one degree, or 0.75 in radians.

my answer when doing this gives me 65.7 degrees



Apparently so. . . I haven't actually done this question yet, I'll take a look :smile:
Given that you know all three lengths of the sides of the triangle I wouldn't bother using a dot product. But it seems to want you to.
(edited 11 years ago)
Reply 10
Original post by joostan
Apparently so. . .
Given that you know all three lengths of the sides of the triangle I wouldn't bother using a dot product. But it seems to want you to.


the question states i must use a scalar product! plus my other trigonmetry is a bit rusty hahah
Reply 11
Original post by DopeSk1llz
the question states i must use a scalar product! plus my other trigonmetry is a bit rusty hahah


Cos rule never hurt anyone :tongue:
Though it seems to be giving the same result as the dot product. :s-smilie:
Reply 12
Original post by DopeSk1llz
the question states i must use a scalar product! plus my other trigonmetry is a bit rusty hahah



Original post by joostan
Cos rule never hurt anyone :tongue:
Though it seems to be giving the same result as the dot product. :s-smilie:


You guys need to read the question a bit more carefully :smile:

You are asked for angle OAB...the dot product you are working out is relevant to angle AOB.

You will find that the work you've done finding the length of vector AB will help you out :smile:
Reply 13
Original post by davros
You guys need to read the question a bit more carefully :smile:

You are asked for angle OAB...the dot product you are working out is relevant to angle AOB.

You will find that the work you've done finding the length of vector AB will help you out :smile:


Ahhh, yeah. Cheers, saved me from some serious head scratching :smile:

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