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Reply 1
Anyone :frown:
Reply 2
not very hard, but your thread was looking lonely... ergo:

A = 3,-1,2
B = 2,0,2

a) find the equation of the line (l1) passing through a and b

b) l2 = (4,1,-1) + d(1,0,-1)

show l1 and l2 intersect, and find point of intersection

c) show point C = (9,1,-6) lies on l2

d) find point d on l1 such that cd is perpendicular to l1
How do you show that a point lies on a line?
Ah do you equate the i's,j's and k's and show that the constant will be the same for all of them?
Reply 5
Here are a few questions. They aren't from any book in particular and they aren't very difficult but you'll have to go by consensus for the answers.

Q1: Integrate e^x.sinx with respect to x.

Q2: Find the shortest distance of the point P(2,3,1) from the line which passes through A(5,8,1) and B(2,0,3).

Q3: A number 'a' is such that when added to its reciprocal the result is 5/2. A number 'b' is such that it is equal to the distance between the points representing (1,3,4) and (1,0,8) in the x,y,z plane. A number 'c' is such that 2^(c+2)=4^c. Find a cartesian equation of the circle with centre (a,b) and radius c.

Q4: Integrate x(cosx)^4 dx with respect to x.

Q5: Find the equations of the tangents to the circle (x-4)^2+(y-5)^2=9 that pass through P(0,1).

Q6: f(x)=x^3+x^2-x-1. g(x)=x^2. Prove that the equation fg(x) has one, and only one, root.

Q7: a^2+2a-I=0. I=Integral of (cosx)^2-0.5 dx with limits pi/2,0. Find a to 3 significant figures.

Q8: A statement by a student was as follows:
(1) INT a/ax dx = lnax+C (as INT f'(x)/f(x)=lnf(x)+C)
(2) INT a/ax dx = INT 1/x dx = lnx+C (as INT 1/x dx = lnx+C).
But (1)=(2) so lnax+c=lnx+c, lnax=lnx, lnx+lna=lnx, lna=0.
Hence lna=0.
Identify the error in the statement

Q9: By expressing sect in the form sect[tant+sect]/[sect+tant] or otherwise find I=INT sect dt with lower limit 0 and upper limit x. Hence find the value(s) of x [In radians to 3sf] for which I=2ln2.
Reply 6
wont question 1 keep going on and on using integration by parts?
Reply 7
Freeway
wont question 1 keep going on and on using integration by parts?


Yep, i think it will, i think it needs a substitution but i can figure what.
Gaz031
Here are a few questions. They aren't from any book in particular and they aren't very difficult but you'll have to go by consensus for the answers.

Q1: Integrate e^x.sinx with respect to x.
Q2: Find the shortest distance of the point P(2,3,1) from the line which passes through A(5,8,1) and B(2,0,3).


Help with these two please :-(
Reply 9
a)

l1 = (3,-1,2) + n[(2,0,2)-(3,-1,2)]
l1 = (3,-1,2) + n(-1,1,0)

b)

(4,1,-1) + d(1,0,-1) = (3,-1,2) + n(-1,1,0)

[1] 4 + d = 3 - n
[2] 1 = -1 + n
[3] -1 - d = 2

[2] => n = 2
[3] => d = -3

putting into [1] gives 4 + (-3) = 3 - (2) => 1=1

intersection = (4,1,-1) + (-3)(1,0,-1)
= (1,1,-2)

c.)

(9,1,-6) = (4,1,-1) + n(1,0,-1)
n(1,0,-1) = (5,0,-5)
n = 5

d)

CD.AB = 0

(3,-1,2) + n(-1,1,0) - (9,1,-6).(-1,1,0)
(-6-n)(-1) + (-2+n) = 0
6 + n - 2 + n = 0
n = -2

D = (3,-1,2) -2(-1,1,0)
D = (5,-3,2)
Reply 10
Gaz031
Here are a few questions. They aren't from any book in particular and they aren't very difficult but you'll have to go by consensus for the answers.

Q1: Integrate e^x.sinx with respect to x.
.


done this 1.. u introduce the symbol I for integration.
so.. u = e^x du/dx = e^x
dv/dx= sin x v= -cos x
I = integral of e^x sin x.
i = -e^x cosx - integral -cosx e^x dx
i = -e^x cos x + integral e^x cosx.
i = -e^x cosx + [e^x sinx - integral e^x sin x dx.]

i= -e^x + e^x sin x - i so....
2i = -e^x cosx + e^x sin x.

i = (e^x)/2 (sinx - cosx) :eek: i no...we have an answer!!
Is there any other of doing it except forming 2 equations like vishpatel did?
Reply 12
sequence123
Is there any other of doing it except forming 2 equations like vishpatel did?


dont think so!
im sure solving parts questions how you did isnt covered by the edexcel p3 text book, which kinda makes it a difficult technique to learn
Reply 14
Vish's answer is right.
As far as I know it's the only way to do it. The trick is ensuring that you choose the same type of function (ie exponential or trigonometric) to differentiate or integrate each time.
You do have the ability to do it from what's on the P3 syllabus though in an exam they'd probably give you an idea to do it. I think an integration similar to that was one of the parts to a question on an AEA paper[which require only P1-P3/C1-C4 knowledge].
For the second question. Consider a general point on the line and find the distance between the given point and the general point. You can work out your answer from there.
I dont really understand what vshy is doing

Can anymore give me a clearer explanation?

thanks
Gaz031
Vish's answer is right.
As far as I know it's the only way to do it. The trick is ensuring that you choose the same type of function (ie exponential or trigonometric) to differentiate or integrate each time.
You do have the ability to do it from what's on the P3 syllabus though in an exam they'd probably give you an idea to do it. I think an integration similar to that was one of the parts to a question on an AEA paper[which require only P1-P3/C1-C4 knowledge].

This is just crazy
I need help on this vectors q:


Find the angle between the vectors a and b given that |a|=3, |b|=5 and |a-b|=7.




Any tips/hints much appreciated.
Reply 18
sequence123
I dont really understand what vshy is doing

Can anymore give me a clearer explanation?

thanks

If we say the original integral is 'I'.
He performs integration by parts once on I and then again on the second integral generated by using the integration by parts formula.
He then has I = 'string of functions' - I and so 2I = 'string of functions' and so on.
Reply 19
endeavour
I need help on this vectors q:


Find the angle between the vectors a and b given that |a|=3, |b|=5 and |a-b|=7.




Any tips/hints much appreciated.

Draw out a triangle. Pick an arbitrary point on your paper and draw lines representing 'a' and 'b' from it[it doesn't matter where they go or what direction in]. The line representing 'a-b' then represents the line joining other ends of these lines. The angle you want is the angle between 'a' and 'b' that were drawn from your original arbitrary point. You can then use the cosine rule: a^2=b^2+c^2-2bccosA to find the required angle.
Edit: Sorry for the crap explanation. My drawings are bad enough - nevermind my attempted explanations of my drawings in words.