Question

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`?

Answer 1

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.

Answer 2

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

Answer 3

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.!