IIT JEE , AIEEE Forum - Mathematics , Vectors - is vectors a b clie along diffrent axis then value of sina+sinB+sinc is_ if a b c are ang

manu.saini

is vectors a b clie along diffrent axis then value of sina+sinB+sinc is_ if a b c are angles along which the vectors lie? the one who solves this will top iit for sure

shinee

if a vector subtends some angle with the x axis, some other angles with y and z axes, then it is impossible to determine the other two if we know only one
so, for simplicity, i have assumed that the vector A subtends the angle a with positive x axis, vector (and 90-a with positive y axis and it lies only in xy plane)
vector B makes an angle of b with positive y xais and it lies completly in yz plane and the vector C too lies completly in the yz plane making an angle of c with positive z xais
as the total angle between the y and z axes is 90, let us assume that the angle between the vectors C and B to be 90-(b+c)
the y component of vector a is Asina and the x component is Acosa
now taking the component of vector Asina along the vector B, we get, Asinacosb
now finding the resultant between Asinacosb+B and C, we get
root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))
and the angle this resultant makes with positive y axis is arctan({[Ccos(b+c)]/[Asinacosb+B+Csin(b+c)] +b})
the angle which this resultant makes with positive z axis is
arc tan((Asinacosb+B)cos(b+c)/[p+(Asinacosb+B)sin(b+c)]n +c)

now let us find the vector sum of this resultant with Acosa (which is along x axis)
root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c)+A^2cos^2a
angle which this new resulatnt makes with x axis is
arctan{root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))}/[Acosa]
the cos of angle which this new resulatnt makes with positive y axis is cos(arctanAcosa/root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))cos(arctan(Pcos(b+c)/[(Asinacosb+B)+Psin(b+c)] +b)
the cos of angle which this new resulatnt makes with positive z axis is
cos(arctanAcosa/root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))cos(arctan(Asinacosb+B)cos(b+c)/[P+(Asinacosb+B)sin(b+c)] + c)
we know that cos^2@ +cos^2#+cos^2$=1
but i ma not understanding how i can get the value of sina+sinb+sinc from this realtion
i have written the eqaution for cos^2@ +cos^2#+cos^2$=1
but it took two pages to write it down
i dont know whether i can solve or not
if possible i will post the soln tomorrow
if u know the soln , plz post it

shinee

i have thought a different method
i have assumed that the vector A subtends the angle a with positive x axis, vector (and 90-a with positive y axis and it lies only in xy plane)
vector B makes an angle of b with positive y xais and it lies completly in yz plane and the vector C too lies completly in the yz plane making an angle of c with positive z xais
as the total angle between the y and z axes is 90, let us assume that the angle between the vectors C and B to be 90-(b+c)
the y component of vector a is Asina and the x component is Acosa
now taking the component of vector Asina along the vector B, we get, Asinacosb and the other component of vector Asina is Asinasinb and it lies along yz palne making an angle 90 with vector B
i have found out the resultant of B and C considering the angle between them as 90-(b+c)
then after finding their resultant , i have found the resultant of this with Acosa
then considering this whole resultant , i taken found out the angles with positive x, y, z axes and cos^2@+cos^#+cos^2$=1
where @,#,$ are the angles of this resultant with x,y,z axes
then we get the equation of the form
B^2cos^c+c^2sin^2c=-2BCsinccosb.................................1
and similarly after taking Asinasinb and C,
we get 2C=A(sinasin^2b)(sin^2c-sin^2b)/[sin^2bsinc+cos^2bcosc-2cosbsinb(sinc+cosc)......2
and after considering Asinacosb with C,
we get 2C=A(sinacosbsinb-1)/cosb(cosc-1)...............3
solving 1, 2 and 3, we get the value of sina, sinb,sinc
add them to get the soln
actually i have gtg
i will post the ans tomorrow

shinee

while solving, we get a eqaution in the form
(B^3CPM^3+BC^3P^3M-B^3CM^2P-BC^3P^3-B^2ACP^4)x^3 +(B^3CP^2M^2+BC^3P^4-B^3CPM^3-BC^3P^3M)x^2+ (ACB^2P^4-B^3CM^3P^3-BC^3P^5M+B^3CP^3M^2+BC^3P^5+B^4M^5+C^4MP^4+2B^2C^2M^3P^2+2B^2C^2M^2P^3+B^4M^4P+C^4P^5)x+B^3CPM^3+BC^3P^3M
where sinc=P and cosc=M and x=sinb
actually , it should have 1 as soln because only two tems have A in them and if we substitute A in place of x, the last two terms wont be cancelled, so 1 should be the soln
so, after substituting 1 in place of x, we get M=-P
and P=0.73 (approx) so, C=180-45 =135
and substituting in the other equations too, we get
sina+sinb+sinc=1/root2+root(6B^2C^2-B^4-C^4)/2root2BC -4BC^2[(2AB-(B^2+C^2)(1-root2)]/A(B^2+C^2)root(6B^2C^2-B^4-C^4)
am i correct

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