₦airaland Forum

Mikebis: time to shw ma talent..here is d solutn
1.four vectors a,b,c nd d are said to b coplanar if nd only if there xist scalars p,q,v,s(not all zero)
such dat pa+qb+vc+sd=0
where p+q+v+s=0
since we are nt gvn d equatn of d vector,to shw dat d equatn is equal to zero.we slv 4 dat
3i-2j+4k=a
6i+3j+k=b
5i+3j+3k=c
2i+2j+6k=d
d vector eqn is
-a+b-c+d=0
sub 4 a,b,c nd d
-(3i-2j+4k)+(6i+3j+k)-(5i+7j+3k)+(2i+2j+6j)=0
-3i+2j-4k+6i+3j+k-5i-7j-3k+2i+2j+6k=0
(-3i+6i-5i+2i)+(2j+3j-7j+2j)+(-4k+k-3k+6k)=0
(0+0+0)=0
hence a,b,c nd d are complanar.
2.loadin tinz.......
3.prvn dat d identity of(a^b).(c^d)=(a.c)(b.d)-(a.d)(b.c)
the identity of
(a^b).(c^d) are knwn as commutatv
so let assume dat
p.q=p+q-pq
where p=(a^b)
q=(c^b)
(a^b).(c^b)=(a^b)+(c^b)-(a^b)(c^b)
(a^b).e=(a^b)+e-(a^b)e
(a^b)-(a^b)=e[1-(a^b)]
0=e[1-(a^b)]
e=0
d identity of LHS is equal to zero
(a.c)(b.d)-(a.d)(b.c)
(a.c)(b.d)^(a.d)(b.c)=(a.c)(b.d)-(a.d)(b.c)
(a.c)(b.d)^(b.d)^e=(a.c)(b.d)-e
(a.c)(b.d)=(a.c)(b.d)-e
(a.c)(b.d)-(a.c)(b.d)=-e
0=-e
e=0
d identity RHS is equal to zero
dazzal thank u

smurfy: Let me welcome you to this outstanding thread by giving you something to chew and digest.

Find the integral of x * root(x-1) dx.

rhydex 247: Question 1.
Show that the four points whose position vectors are 3i-2j+4k, 6i+3j+k, 5i+7j+3k and 2i+2j+6k are coplanar.
Question 2.
If a ^ r=b+¥a and a.r=3 where a=2i+j-k and b=-i-2j+k. Then find r and ¥.
Question 3.
Prove the identity. (a ^ b).(c ^d)=(a.c)(b.d)-(a.d)(b.c). Where ^ means cap or cross

godofphysics: pls who solved d questn so dat i can view d person's posts. It wil save me d tyme of going 2ru all d pages. Thanks....
(I'm stil sweating wit d last few pages coz it's like i've not rly understood some of d solutions)

godofphysics: Pls, any reasonable workings to dis questn?
Find x&y if x+y=5 &x^x+y^y=31 (i tink d answers r 2&3, i'd b glad if i can get d workings)

benbuks: Find the anti-derivative of (cosx +1)^-2

benbuks: Find the anti-derivative of (cosx +1)^-2

statusnet: Integrate ^{(tan^2) x} / _tan2x dx

rhydex 247: Question 1.
Show that the four points whose position vectors are 3i-2j+4k, 6i+3j+k, 5i+7j+3k and 2i+2j+6k are coplanar.
Question 2.
If a ^ r=b+¥a and a.r=3 where a=2i+j-k and b=-i-2j+k. Then find r and ¥.
Question 3.
Prove the identity. (a ^ b).(c ^d)=(a.c)(b.d)-(a.d)(b.c). Where ^ means cap or cross

Fetus: Differentiate ((2) xlog~e3x....where ~= subscript i.e log base e

Laplacian: ...sin2x+sinx=1...find x...

Laplacian: let x, y€R then for each w€W...we have
w=xu1+yu2
=(x+y, x-3y, x+4y, x-2y)...so that
w-v
=(x+y-1, x-3y-3, x+4y-5, x-2y-7)...
Now define the norm of v-w to be....//w-v//=|w-v|=d...from calculus we know that if an abscissa optimizes the square of a function, then it also optimizes that function: proof y^2=f(x), then 2yy'=f'(x)...it follows that, from
d^2=(x+y-1)^2+(x-3y-3)^2+(x+4y-5)^2+(x-2y-7)^2, differentiatin partially wrt x & y and solvin gives d solution when their values are substituted in d vector eqn for w...

Alpha Maximus: .....alright, I'm not really conversant with trigonometric equations but I'll give it a shot!
Tan ! =sin ! /cos !
Thus Tan$=sin$/cos$=((a-1)/x))/((1-a)/x))
=a-1/x*(x/1-a)
=a-1/1-a
Tan$=a-1/1-a
Multipky both sides by -1
-tan$=-a+1/1-a
-tan$=1
Divide both sides by -1
Tan$=-1
Am I right?

benbuks: ...here is the solution,

y^3 -y^2 +y -1
=(y^3-y^2) + (y-1)
=y^2(y-1) +(y-1)

==>y^3-y^2 +y-1
=(y^2 +1)(y-1)

got that?

benbuks: ...re-try.

Prinzoladimeji: Plz i want know aw to do addition and subtraction of Number bases

benbuks: Re-solve. y^3 -y^2 ^2 +y-1

benbuks: Factorise x^4 +x^2b^2 +b^4

rhydex 247: @ laplacian.
Here is the soln.
a^(x^a) implies
(a.x)a-(a.a)x. Where x=(x1,x2,x3) let a=(a1,a2,a3).
a.a=(a1,a2,a3)(a1,a2,a3).
a.a=a^21^a^22+a^23.
a.x=(a1,a2,a3)(x1,x2,x3).
a.x=a1x1+a2x2+a3x3.
a^(x^a)=(a1x1+a2x2+a3x3)a-(a^21+a^22+a^23)x. Recall that a=(a1,a2,a3) nd x=(x1,x2,x3).
a^(x^a)=(a1,a2,a3)(a1x1+a2x2+a3x3)-(x1,x2,x3)(a^21+a^22+a^23). When u expand nd open d bracket u get 0.
Hence a^(x^a)=0. Buh 4rm d questn we av x+a^(x^a)=b. Put a^(x^a)=0 in d eqn we av
x+0=b. x=b. Where x=x1,x2,x3. Hence x=(x1,x2,x3)=b. Note ^ means cap but ^ in a^21+a^22+a^23 means raise to power. All is well.

Laplacian: wat is there 2 finish in no.1..biko....
Just dat he made a HARMLESS mistake in d genral solution...
y=(Ax+B)e^-x
...for number 2, d numer of digits in any given number n, say, is given by Logn (approx.)
number of digits in
2004^2004=2004*Log2004 (approx.)

benbuks: Sum the series cos@ + cos 2@ + cos3@ +...

Laplacian: i ve very high esteem 4 ur questns
If x+a^(x^a)=b, from d structure of x we can infer dat d above eqn is a vector eqn, applyin vector tripl produt to d 2nd term on d L.H.S,
x+x(a.a)-a(a.x)=b....eqn1, multiply d given eqn by a. to get,
a.x+a.[a^(x^a)]=a.b, scalar tripl prodct givs
a.x-(x^a).[a^a]=a.b, recall dat
a^a=0, so a.x=a.b, substitt in eqn1, to get; x(1+|a|^2)-a(a.b)=b, so
x=[b+a(a.b)]/(1+|a|^2)

Laplacian: i ve very high esteem 4 ur questns
If x+a^(x^a)=b, from d structure of x we can infer dat d above eqn is a vector eqn, applyin vector tripl produt to d 2nd term on d L.H.S,
x+x(a.a)-a(a.x)=b....eqn1, multiply d given eqn by a. to get,
a.x+a.[a^(x^a)]=a.b, scalar tripl prodct givs
a.x-(x^a).[a^a]=a.b, recall dat
a^a=0, so a.x=a.b, substitt in eqn1, to get; x(1+|a|^2)-a(a.b)=b, so
x=[b+a(a.b)]/(1+|a|^2)