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 :
$( S 1): 2|\hat{ a } \times \hat{ b }|=|\hat{ a }-\hat{ b }|$
$( S 2)$ : 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
$2 t^{2}-t-1=0$
$2 t^{2}-2 t+t-1=0$
$2 t(t-1)+(t-1)=0$
$(2 t+1)(t-1)=0$
$t=-\frac{1}{2}$ 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 $(\hat{a}+\hat{b})$.
$\frac{\hat{a} \cdot(\hat{a}+\hat{b})}{|\hat{a}+\hat{b}|}=\frac{1+\cos \left(\frac{2 \pi}{3}\right)}{\sqrt{2+2 \cos \frac{2 \pi}{3}}}$
$=\frac{1-\frac{1}{2}}{\sqrt{1}}$
$=\frac{1}{2}$