# If the sum of two unit vectors is a unit vector, what is the magnitude of the difference of the two vectors?

Jun 9, 2015

Unit vector: The vector whose magnitude is unity(1) is called unit vector.

#### Explanation:

Let ai+bj and ci+dj be the unit vectors.
From the defination,
$\sqrt{{a}^{2} + {b}^{2}} = 1$
$\implies \left({a}^{2} + {b}^{2}\right) = 1$__(1)
$\sqrt{{c}^{2} + {d}^{2}} = 1$
$\implies \left({c}^{2} + {d}^{2}\right) = 1$
_____(2)

(ai+bj)+(ci+dj) =(a+c)i+(b+d)j__(3)

Also given that,
The magnitude of equation(3) is also unity.
$\sqrt{{\left(a + c\right)}^{2} + {\left(b + d\right)}^{2}} = 1$
$\implies \left({\left(a + c\right)}^{2} + {\left(b + d\right)}^{2}\right) = 1$
$\implies \left({a}^{2} + {c}^{2} + 2 a c\right) + \left({b}^{2} + {d}^{2} + 2 b d\right) = 1$
$\implies 1 + 2 a c + 1 + 2 b d = 1$
$\implies \cancel{1} + 2 a c + 2 b d + 1 = \cancel{1}$
$\implies a c + b d = - \frac{1}{2}$___(4)
the magnitude of their difference is:?
THe difference of two unit vectors is
$\left(a i + b j\right) - \left(c i + \mathrm{dj}\right) = \left(a - c\right) i + \left(b - d\right) j$
Magnitude=sqrt((a-c)^2+(b-d)^2
$= \sqrt{{a}^{2} + {c}^{2} - 2 a c + {b}^{2} + {d}^{2} - 2 b d}$
=sqrt(1+1-2(ac+bd) [From (1) and (2)]
=sqrt(2-2(-1/2) [From equation(4)]
$= \sqrt{2 + 1}$
$= \sqrt{3}$
$= 1.73205$

If the sum of two unit vectors is a unit vector, then the magnitude of their difference is 1.73205

Jun 10, 2015

The magnitude (length) of the difference is $\sqrt{3}$
(Alternate method of seeing this).

#### Explanation:

For those more visually or geometrically oriented; consider the following diagram, noting that if the sum of two unit vectors is a unit vector then the 3 unit vectors can be combined into an equilateral triangle:

By extending the initial unit vector and dropping a perpendicular to that extension from the terminal point of the vector $\vec{a - b}$
we have a right triangle with sides:
$1 + \frac{1}{2}$ and $\frac{\sqrt{3}}{2}$

So
$\left\mid \vec{a - b} \right\mid = \sqrt{{\left(\frac{3}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXXX}}$$= \sqrt{\frac{9 + 3}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3}$