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3-1.Vectors
hard
If the sum of two unit vectors is a unit vector, then magnitude of difference is
A
$\sqrt 2 $
B
$\sqrt 3 $
C
$1/\sqrt 2 $
D
$\sqrt 5 $
Solution
(b) Let ${\hat n_1}$ and ${\hat n_2}$ are the two unit vectors, then the sum is
${\overrightarrow n _s} = {\hat n_1} + {\hat n_2}$ or
$n_s^2 = n_1^2 + n_2^2 + 2{n_1}{n_2}\cos \theta $
$ = 1 + 1 + 2 \cos \theta $
Since it is given that $n_s$ is also a unit vector, therefore $1 = 1 + 1 + 2\cos \theta $
$⇒$ $\cos \theta = – \frac{1}{2}$
$\therefore \theta = 120^\circ $
Now the difference vector is ${\hat n_d} = {\hat n_1} – {\hat n_2}$ or
$n_d^2 = n_1^2 + n_2^2 – 2{n_1}{n_2}\cos \theta $
$ = 1 + 1 – 2\cos (120^\circ )$
$n_d^2 = 2 – 2( – 1/2) = 2 + 1 = 3$
$ \Rightarrow \,\,{n_d} = \sqrt 3 $
Standard 11
Physics