Question

$\underset{a}{\overset{}{\to }},\underset{b}{\overset{}{\to }}and\underset{c}{\overset{}{\to }}$ are 3 vector magnitude 1,1 and 2 respectively If $\underset{a}{\overset{}{\to }}\times (\underset{a}{\overset{}{\to }}\times \underset{c}{\overset{}{\to }})+\underset{b}{\overset{}{\to }}=0$ , then acute angle b/w $\underset{a}{\overset{}{\to }}and\underset{c}{\overset{}{\to }}$.

Answer 1

It is give that

$\left|a\right|=1$

$\left|b\right|=1$

$\left|c\right|=2$

Consider,

$\overrightarrow{a}\times (\overrightarrow{a}\times \overrightarrow{c})+\overrightarrow{b}=0$

$\left[\right(a.c)a-(a.a\left)c\right]+b=0$

$\left(2\cos \theta \right)a-c=-b...........\left(1\right)$

$2\cos \theta a.a=(c-b)a$

$2\cos \theta =c.a-b.a$

$2\cos \theta =2\cos \theta a-a.b$

$so,a.b=0$

$takingequation\left(1\right)andtakingdotproductwithb$

$\left(2\cos \theta \right)a.b=(c-b).b=c.b-b.b$

$0=c.b-1\left(from\right(1\left)\right)$

$b.c=\mathrm{1...........}\left(2\right)$

Squaring both sides in equation (1)

$4{\cos }^2\theta a^2=c^2-b^2-2b.c$

$4{\cos }^2\theta =4+1-2\left(from\right(2\left)\right)$

${\cos }^2\theta =\frac{3}{4}$

$\theta =\frac{\pi }{6}$