# Acute angles between 2 vectorsWatch

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Thread starter 13 years ago
#1
(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
0
13 years ago
#2
(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.
0
13 years ago
#3
question 1 - issit the direction vectors you use?
0
13 years ago
#4
(Original post by nas7232)
question 1 - issit the direction vectors you use?
yes.
0
Thread starter 13 years ago
#5
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!
0
13 years ago
#6
No! Just because the integral of secÂ²x is tanx, doesn't mean the integral of (secÂ²x)^n is (tanx)^n.

As Gaz said, use the double angle formula for cosx:
cos2x = 2cosÂ²x - 1
=> cosÂ²x = (cos2x + 1)/2

Sub this in your expression:
(cosÂ²x)Â² = [(cos2x + 1)/2]Â² = (cosÂ²(2x) + 2cos2x + 1)/4

Now use the double angle formula again for cosÂ²(2x):
cosÂ²(2x) = (cos4x + 1)/2
0
Thread starter 13 years ago
#7
but where does the cos^2 x come into it - i've only got tan and sec so far!
0
13 years ago
#8
1/secx = cosx
0
Thread starter 13 years ago
#9
see where your coming from now - will try and work it through now. thanx )
0
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