Question

The vectors $a.b$ and $c$ are such that:
(i) $\left|a\right|=\left|b\right|=1;\left|c\right|=2$
(ii) $a\times (a\times c)+b=0$
find the possible angles between $a$ and $c$

Answer 1

Given

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1$---------(1)

$\left|\overrightarrow{c}\right|=2$

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

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

$(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{a}-{\left|\overrightarrow{a}\right|}^2\overrightarrow{c}+b=0$

$(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{a}-\overrightarrow{c}+b=0$ from eq (1)

on squaring both sides

$(\overrightarrow{a}\cdot \overrightarrow{c}{)}^2{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{c}\right|}^2-2(\overrightarrow{a}\cdot \overrightarrow{c}\left)\right(\overrightarrow{a}\cdot \overrightarrow{c})={\left|\overrightarrow{b}\right|}^2$

from given

$(\overrightarrow{a}\cdot \overrightarrow{c}{)}^2(1)+4-2(\overrightarrow{a}\cdot \overrightarrow{c}{)}^2=1$

$-(\overrightarrow{a}\cdot \overrightarrow{c}{)}^2=1-4$

$-(\overrightarrow{a}\cdot \overrightarrow{c}{)}^2=-3$

$(\overrightarrow{a}\cdot \overrightarrow{c})=\sqrt{3}$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|\cos \theta =\sqrt{3}$

$1\times 2\cos \theta =\sqrt{3}$

$2\cos \theta =\sqrt{3}$

$\cos \theta =\frac{\sqrt{3}}{2}$

$\theta =30$