Acute angles between 2 vectors Watch

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jjr14
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#1
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(This question is off the edexcel C4 specimen paper, question 5c)

How do I find the acute angle between 2 vector lines? They are 3D vectors and their equations are:

r=5i+3j-2k+(lamda)(i-2j+2k)
r=2i-11j+14k+(mew)(-3i-4j+5k)


Also, how do I integrate:
1/((1+tanx^2)^2)

Be grateful for any help!
jjr14
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Gaz031
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(Original post by jjr14)
(This question is off the edexcel C4 specimen paper, question 5c)

How do I find the acute angle between 2 vector lines? They are 3D vectors and their equations are:

r=5i+3j-2k+(lamda)(i-2j+2k)
r=2i-11j+14k+(mew)(-3i-4j+5k)


Also, how do I integrate:
1/((1+tanx^2)^2)

Be grateful for any help!
jjr14
For the vector question you want to use the scalar product. a.b=(moda).(modb).cost.

For the second question use the identity 1+(tanx)^2=(secx)^2 which is derived from dividing (sinx)^2 + (cosx)^2 =1 by (cosx)^2.
You mean then want to use the double angle formulae cos2A=2(cosA)^2 - 1 to put the expression in terms of multiple powers that are raised only to the power of 1.
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nas7232
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question 1 - issit the direction vectors you use?
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Gaz031
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(Original post by nas7232)
question 1 - issit the direction vectors you use?
yes.
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jjr14
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sorry i don't quite get this. i use the identidy and get
1/(1+sec^2 x - 1)^2 which is really 1/(sec^2 x)^2 but where do I go from here? I don't get how the double angle formula comes into it.

here's an idea to get your head round:
is 1/(sec^2 x)^2 not the same as 1/(sec^4 x)?
i know that if i integrate sec^2 x i will get tan x so would it not follow that if i had sec^4 x i'll get tan^2 x when i integrate. might this help me in some way or have i gone completly wrong!
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dvs
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No! Just because the integral of sec²x is tanx, doesn't mean the integral of (sec²x)^n is (tanx)^n.

As Gaz said, use the double angle formula for cosx:
cos2x = 2cos²x - 1
=> cos²x = (cos2x + 1)/2

Sub this in your expression:
(cos²x)² = [(cos2x + 1)/2]² = (cos²(2x) + 2cos2x + 1)/4

Now use the double angle formula again for cos²(2x):
cos²(2x) = (cos4x + 1)/2
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jjr14
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but where does the cos^2 x come into it - i've only got tan and sec so far!
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dvs
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1/secx = cosx
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jjr14
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#9
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see where your coming from now - will try and work it through now. thanx )
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