11mod(x+y+z)2>=0
3+2(x.y+y.z+z.x)>=0
2(x.y+y.z+z.x)>=-3
E=given expression=25+49-2.35(x.y+y.z+z.x)
<=25+49+35.3
<=179
[5x-7y]2+[5y-7z]2+[5z-7x]2
the maximum value of the above expression
where[] denotes modulus and x y z are unit vectors
i am sorry for confusing u all i have edited it once again
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25 Answers
11
3it shud be 109 and also the minimum value for the exp (x.y+y.z+z.x) shud be -1 not -3
11not exactly i+j+k≠0 even though the three points form vertices of a triangle
1mod(x+y+z)2>=0
3+2(x.y+y.z+z.x)>=0
2(x.y+y.z+z.x)>=-3
E=given expression=3*(25+49)-2.35(x.y+y.z+z.x)
<=3*(25+49)+35*3
<=327
3any three vectors can be made to be the vertices of a triangle
and so the x+y+z=0
1take all the vectors at 120 degrees u may get the correct answer which i think is 327.
11guyz check this one out
this was the solution available to me
i know the answer is wrong
21i'm gettin 292....
subhash plz chk da ans its gotta be wrong...
take x= i
y =i
z = -i
u'll get 292.........
3i think this hint really helps u
x+y+z=0
and mod(x)=mod(y)=mod(z)=1
and simplify the given expression
use x=a1i+a2j+a3k
similarly for other two vectors.
i think i was correct with my method.
can anyone post if i was wrong...
11the answer given is 179 but ihave got doubts can some one post ur solutions
11x,y,z are unit vectors i have edited the question but 12 is not the correct answer