Question

Two perpendicular unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are such that $\left[\overrightarrow{r}\overrightarrow{a}\overrightarrow{b}\right]=\frac{5}{4},$ $\overrightarrow{r}\cdot (3\overrightarrow{a}+2\overrightarrow{b})=0$ and $\underset{-2\overrightarrow{r}.\overrightarrow{a}}{\overset{\frac{-4}{3}\overrightarrow{r}.\overrightarrow{b}}{\int }}\frac{x+1}{x^2+1}dx=\frac{\pi }{2}$. Then which of the following is(are) correct ?

• A. $|\overrightarrow{r}{|}^2=\frac{19}{8}$
• B. $|\overrightarrow{r}{|}^2=\frac{7}{4}$
• C. $\overrightarrow{r}=\frac{\overrightarrow{a}}{2}-\frac{3\overrightarrow{b}}{4}+\frac{5}{4}(\overrightarrow{a}\times \overrightarrow{b})$
• D. $\overrightarrow{r}=-\frac{\overrightarrow{a}}{2}+\frac{3\overrightarrow{b}}{2}+\frac{5}{4}(\overrightarrow{a}\times \overrightarrow{b})$

Answer 1

Answer: (A), (C)

The correct options are
A $|\overrightarrow{r}{|}^2=\frac{19}{8}$
C $\overrightarrow{r}=\frac{\overrightarrow{a}}{2}-\frac{3\overrightarrow{b}}{4}+\frac{5}{4}(\overrightarrow{a}\times \overrightarrow{b})$

Any vector $r$ in space can be written as
$\overrightarrow{r}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+k(\overrightarrow{a}\times \overrightarrow{b})$
Taking dot product with $\overrightarrow{a}\times \overrightarrow{b}$, we get $k=\frac{5}{4}$
$\Rightarrow \overrightarrow{r}\cdot \overrightarrow{a}=\lambda ;\overrightarrow{r}\cdot \overrightarrow{b}=\mu$
$\because \overrightarrow{r}\cdot (3\overrightarrow{a}+2\overrightarrow{b})=0\Rightarrow 3\lambda +2\mu =0$

Now, $\underset{-2\lambda }{\overset{-\frac{4}{3}\mu }{\int }}\frac{x+1}{x^2+1}dx=\frac{\pi }{2}$
$\Rightarrow \underset{-2\lambda }{\overset{2\lambda }{\int }}\frac{x+1}{x^2+1}dx=\frac{\pi }{2}$
$\Rightarrow 2\underset{0}{\overset{2\lambda }{\int }}\frac{dx}{x^2+1}=\frac{\pi }{2}$
$\Rightarrow 2{\tan }^{-1}\left(2\lambda \right)=\frac{\pi }{2}$
$\Rightarrow \lambda =\frac{1}{2}$ and $\mu =-\frac{3}{4}$
$\therefore \overrightarrow{r}=\frac{\overrightarrow{a}}{2}-\frac{3}{4}\overrightarrow{b}+\frac{5}{4}(\overrightarrow{a}\times \overrightarrow{b})$
and $|\overrightarrow{r}|=\sqrt{\frac{1}{4}+\frac{9}{16}+\frac{25}{16}}$
$\Rightarrow |\overrightarrow{r}{|}^2=\frac{19}{8}$