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
To prove that
Look for the explanation below.
Explanation:
Let
where
Since
Using values of
squaring both sides
Now, the sum of
Let us find magnitude of
Using values of
Since the magnitude of
By definition then
So, if both If
# || 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) ||^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
Refer to the Explanation.
Explanation:
Knowing that,
given,
Enjoy Maths.!