Question

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $|(\hat{a}+\hat{b})+2(\hat{a}\times \hat{b})|=2$. If $\theta \in (0,\pi )$ is the angle between $\hat{a}$ and $\hat{b}$, then among the statements :
$(S1):2|\hat{a}\times \hat{b}|=|\hat{a}-\hat{b}|$
$(S2)$ : The projection of $\hat{a}$ on $(\hat{a}+\hat{b})$ is $\frac{1}{2}$

• A. Only (S1) is true
• B. Only (S2) is true
• C. Both (S1) and (S2) are true
• D. Both (S1) and (S2) are false

Answer 1

Answer: (C)

$|(\hat{a}+\hat{b})+2(\hat{a}\times \hat{b})|=2,\theta \in (0,\pi )$
$((\hat{a}+\hat{b})+2(\hat{a}\times \hat{b}))\cdot ((\hat{a}+\hat{b})+2(\hat{a}\times \hat{b}))=4$
$|\hat{a}+\hat{b}{|}^2+4|(\hat{a}\times \hat{b}){|}^2+0=4$
Let the angle be $\theta$ between $\hat{a}$ and $\hat{b}$
$2+2\cos \theta +4{\sin }^2\theta =4$
$2+2\cos \theta -4{\cos }^2\theta =0$
Let $\cos \theta =t$ then
$2t^2-t-1=0$
$2t^2-2t+t-1=0$
$2t(t-1)+(t-1)=0$
$(2t+1)(t-1)=0$
$t=-\frac{1}{2}$ or $t=1$
$\cos \theta =-\frac{1}{2}$
$\theta =\frac{2\pi }{3}$
not possible as $\theta \in (0,\pi )$
Now
$S_{1}2|\vec{a}\times \vec{b}|=2\sin \left(\frac{2\pi }{3}\right)$
$|\hat{a}-\hat{b}|=\sqrt{1+1-2\cos \left(\frac{2\pi }{3}\right)}$
$=\sqrt{2-2\times \left(-\frac{1}{2}\right)}$
$=\sqrt{3}$
$S_1$ is correct.
$S_2$ projection of $\hat{a}$ on $(\hat{a}+\hat{b})$.
$\frac{\hat{a}\cdot (\hat{a}+\hat{b})}{|\hat{a}+\hat{b}|}=\frac{1+\cos \left(\frac{2\pi }{3}\right)}{\sqrt{2+2\cos \frac{2\pi }{3}}}$
$=\frac{1-\frac{1}{2}}{\sqrt{1}}$
$=\frac{1}{2}$