- $\text{a}=\frac{\pi}{4}$
- $\text{a}=\frac{\pi}{3}$
- $\text{a}=\frac{2\pi}{3}$
- $\text{a}=\frac{\pi}{2}$
Solution:
$\vec{\text{a}}$
and $\vec{\text{b}}$ are unit vectors.$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1\dots(1)$
Now,
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\text{a}$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=\cos\text{a}\dots(2)$
[using (1)]
Given that
$\Big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=1$
$\Rightarrow1+1+2\cos\text{a}=1$
[using (1) and (2)]$\Rightarrow2+2\cos\text{a}=1$
$\Rightarrow2\cos\text{a}=-1$
$\Rightarrow\cos\text{a}=\frac{-1}{2}$
$\Rightarrow\text{a}=\frac{2\pi}{3}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$
where
$f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _e\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _e\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right|$.
Let $p (x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$, and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE ?
$(A)$ $p \left(\frac{3+\sqrt{2}}{4}\right)<0$
$(B)$ $p \left(\frac{1+3 \sqrt{2}}{4}\right)>0$
$(C)$ $p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$
$(D)$ $p \left(\frac{5-\sqrt{2}}{4}\right)<0$
${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$ , ${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$
, ${{c}_{2}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{1}}}{|{{b}_{1}}{{|}^{2}}}{{b}_{1}}$
,${{c}_{3}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{2}}}{|{{b}_{2}}{{|}^{2}}}{{b}_{2}}$,
${{c}_{4}}=a-\frac{c.a}{|a{{|}^{2}}}a$.
Then which of the following is a set of mutually orthogonal vectors is