If both If #bb(ulhata)# and #bb(ulhat b)# are unit vectors and # || bb(ul hata)-bb(ul hatb) || = sqrt(3) # show that # || bb(ul hata) + bb(ul hatb) || = 1#?

3 Answers
Apr 28, 2017

Answer:

To prove that #|hata+hatb| = 1#.
Look for the explanation below.

Explanation:

Let #hat a = c hati + dhatj + ehatk# and #hat b = fhati+ghatj+hhatk#
where #sqrt(c^2+d^2+e^2) = 1 = sqrt(f^2+g^2+h^2)#

#=>color(green) [c^2+d^2+e^2 = 1 = f^2+g^2+h^2]# ------------- 1.

#=> hata - hatb = (c-f)hati + (d-g)hatj + (e-h)hatk#

Since #|hata-hatb| = sqrt3#

#=> sqrt((c-f)^2 + (d-g)^2 + (e-h)^2) = sqrt3#

#=>sqrt(c^2 + f^2 - 2cf + d^2 + g^2 -2dg + e^2+h^2 - 2eh) = sqrt3#

#=> sqrt(c^2+d^2+e^2 + f^2+g^2+h^2 -2(cf+dg+eh)) = sqrt3#

Using values of #c^2+d^2+e^2# & #f^2+g^2+h^2# from 1.

#=> sqrt(1+1 -2(cf+dg+eh)) = sqrt3#

squaring both sides

#=> 2-2(cf+dg+eh) = 3#

#=>color(green){2(cf+dg+eh) = -1} # ---------- 2.

Now, the sum of #hata and hatb#

#hata + hatb = (c+f)hati + (d+g)hatj + (e+h)hatk#

Let us find magnitude of #hata+hatb#

#|hata + hatb| = sqrt((c+f)^2 + (d+g)^2 + (e+h)^2) #

#=sqrt(c^2 + f^2 + 2cf + d^2 + g^2 +2dg + e^2+h^2 + 2eh) #

#=sqrt(c^2+d^2+e^2 + f^2+g^2+h^2 +2(cf+dg+eh))#

Using values of #c^2+d^2+e^2# & #f^2+g^2+h^2# from 1. and value of #2(cf+dg+eh)# form 2.

#|hata + hatb| = sqrt(1+1-1) = sqrt1 = 1#

#therefore color(red){|hata + hatb| = 1}#

Since the magnitude of #hata+hatb# is #1#, it is a unit vector.
#therefore# sum of #hata# & #hatb# is a unit vector.

By definition then #|| bb(ulA) || ^2= bb(ulA* ulA) #

So, if both If #bb(ulhata)# and #bb(ulhat b)# are unit vectors then:

# || bb(ulhat a) || = 1 iff bb(ula * ula) = 1 #

# || bb(ulhat b) || = 1 iff bb(ulb * ulb) = 1 #

And also we have:

# bb(ula * ulb) = bb(ulb * ula) #

We are given that # || bb(ul hata) - bb(ul hatb) || = sqrt(3)#, and so:

# || bb(ul hata)-bb(ul hatb) ||^2 = 3 #

But using the above definition we also have that:

# || bb(ul hata)-bb(ul hatb) ||^2 = (bb(ul hata)-bb(ul hatb) * (bb(ul hata)-bb(ul hatb)) #

# :. (bb(ul hata)-bb(ul hatb)) * (bb(ul hata)-bb(ul hatb)) = 3 #
# :. bb(ul hata) * bb(ul hata) - bb(ul hata) * bb(ul hatb) - bb(ul hatb) * bb(ul hata) + bb(ul hatb) * bb(ul hatb) = 3 #
# :. bb(ul hata) * bb(ul hata) - 2bb(ul hata) * bb(ul hatb) + bb(ul hatb) * bb(ul hatb) = 3 #
# :. 1 - 2bb(ul hata) * bb(ul hatb) + 1= 3 #
# :. 2 bb(ul hata) * bb(ul hatb) = -1 #

Similarly:

# || bb(ul hata) + bb(ul hatb) ||^2 = (bb(ul hata)+bb(ul hatb)) * (bb(ul hata)+bb(ul hatb)) #
# " " = bb(ul hata) * bb(ul hata) + 2bb(ul hata) * bb(ul hatb) + bb(ul hatb) * bb(ul hatb) #
# " " = 1 -1 + 1 #
# " " = 1 #

And so:

# || bb(ul hata) + bb(ul hatb) || = 1 \ \ \ # QED

Apr 28, 2017

Answer:

Refer to the Explanation.

Explanation:

Knowing that,

#||hata+hatb||^2+||hata-hatb||^2=2[||hata|\^2+||hatb||^2],# we have, by what is

given, #||hata+hatb||^2+(sqrt3)^2=2[1^2+1^2],#

#rArr ||hata+hatb||^2=4-3=1, or,#

# ||hata+hatb||=1,# as desired.

Enjoy Maths.!