Given #a^2+b^2+c^2=25 ;x^2+y^2+z^2=36 and ax+by+cz=30# for a,b,c being real. How will you prove #a/x=b/y=c/z=5/6# ?

2 Answers
May 31, 2016

Answer:

See demo below

Explanation:

Suppose you have three points in #R^3#

#p_1 = (a,b,c)#
#p_2 = (x,y,z)#
#p_0=(0,0,0)#

The respective lengths of sides #p_1-p_0# and #p_2-p_0#
are

#norm(p_1-p_0)=sqrt(a^2+b^2+c^2)#
#norm(p_2-p_0)=sqrt(x^2+y^2+z^2)#

and their scalar product

#(p_1-p_0).(p_2-p_0) = ax + by +cz#

but #(p_1-p_0).(p_2-p_0) = norm(p_1-p_0)norm(p_2-p_0) cos(hat(p_1p_0p_2)) #

so

#cos(hat(p_1p_0p_2))= (ax+by+cz)/(sqrt(a^2+b^2+c^2)sqrt(x^2+y^2+z^2)) = 30/(5times 6)=1#

Then the sides #p_1-p_0# and #p_2-p_0# are aligned so
#(p_1-p_0) = lambda(p_2-p_0)# and of course

#lambda = 5/6#

Jun 1, 2016

Answer:

Another way

Explanation:

For those who know less

#a^2+b^2+c^2=25….(1)#
#x^2+y^2+z^2=36…….(2)#
#ax+by+cz=30…..(3)#

Dividing equation(1)by #5^2# we have
#(a/5)^2+(b/5)^2+(c/5)^2=1….(4)#

Dividing equation(2)by #6^2#we have
#(x/6)^2+(y/6)^2+(z/6)^2=1…(5)#

Dividing equation(3)by 30 we have
#(a/5*x/6)+(b/5*y/6)+(c/5*z/6)=1…..(6)#

Adding equation (4) ;(5) and subtracting twice of equation(6) from their sum ,we get

#(a/5-x/6)^2+(b/5-y/6)^2+(c/5-z/6)^2=0#

a,b,c,x,y,z are real and sum of three squared exrpressions are zero,so each of them should be zero.

Hence #a/5-x/6=0=>a/x= 5/6 #
Similarly, #b/y=5/6 and c/z =5/6 #
#:.a/x=b/y=c/z= 5/6#
Proved.