Question 31236
vector A=3i+4j is a vector in xy plane,
vector B is a vector perpendicular to vector A,
what will be the vector C equal to,which has projections 1 and 2 along vectors A and B?
LET C BE Pi+Qj....
PROJECTION OF C ALONG  A =1=A.C/|A|=(3i+4j).(Pi+Qj)/|(3i+4j)
=(3P+4Q)/SQRT.(3^2+4^2)=(3P+4Q)/5=1
3P+4Q=5....................................I
BUT PROJECTION OF C ALONG  A =1=|C|*COS(X) WHERE X IS THE ANGLE BETWEEN VECTORS A AND C.
HENCE COS(X)=1/|C|
SINCE B IS PERPENDICULAR TO A WE HAVE ANGLE BETWEEN C AND B =90-X

PROJECTION OF C ALONG  B =2=|C|*COS(90-X)=|C|*SIN(X)
SQUARING WE GET …..4=|C|^2*{SIN(X)}^2 = |C|^2*{1-(COS(X))^2}=|C|^2*{1-1/|C|^2}=|C|^2-1……………
|C|^2=4+1=5
P^2+Q^2=5…………………………..II
SUBSTITUTING FOR Q FROM EQN.1,WE GET
P^2+(5-3P)^2/4^2=5
16P^2+25+9P^2-30P=80
25P^2-30P-55=0
5P^2-6P-11=0
5P^2-11P+5P-11=0
P(5P-11)+1(5P-11)=0
(P+1)(5P-11)=0
P=-1…..OR…….11/5
FROM EQN.I,WE GET 
Q=(5-3P)/4=(5+3)/4=2……..OR………(5-3*11/5)/4=-2
HENCE VECTOR C IS ……
-i+2j…..OR…….11i/5-2j
WE FIND BY CROSS CHECKING THAT 11i/5-2j IS AN EXTRANEOUS SOLUTION NOT COMPATILE WITH |C|=SQRT.5.
HENCE VECTOR C IS ……-i+2j