Question

Match the statements in List $1$ with those in List $2$

Answer 1

Let the required line through the origin be $L:xa=yb=zc$

This line intersects the two given lines. Hence, the shortest distance between $L$ and $L_1$ as well as the shortest distance between $L$ and $L_2$ is zero.

The shortest distance between $L$ and $L_1$ is zero. Hence,

$(\overline{b}\times \overline{b_1}).(\overline{a}-\overline{a_1})=0$, ($\overline{b}$ denotes the direction ratios along the line and $\overline{a}$ denotes the position vector of a point on the line )

$\Rightarrow \left|\begin{array}{ccc}a& b& c\\ 1& -2& 1\\ 2& 1& -1\end{array}\right|=0$

On simplifying we get, $5c+3b+a=0$

Similarly, the shortest distance between $L$ and $L_2$ is zero.

Hence,

$\left|\begin{array}{ccc}a& b& c\\ 2& -1& 1\\ \frac{8}{3}& -3& 1\end{array}\right|=0$

This leads us to : $3a+b-5c=0$

On simplifying the equations we get,

$a:b:c=5:-5:2$

Using this information, we can find the points of intersection between $L$ and$L_1$ as

$P\equiv (5,-5,2)$ and between $L$and $L_2$ as

$Q\equiv (\frac{10}{3},-\frac{10}{3},\frac{8}{3})$

$\Rightarrow P{Q}^2=d^2=6$.

$\left(B\right)\left(1\right),\left(3\right)$

${\tan }^{-1}(x+3)-{\tan }^{-1}(x-3)={\sin }^{-1}\left(3/5\right)$

$\Rightarrow {\tan }^{-1}\frac{(x+3)-(x-3)}{1+(x^2-9)}={\tan }^{-1}\frac{3}{4}\Rightarrow \frac{6}{x^2-8}=\frac{3}{4}$

$\therefore x^2-8=8$

or $x=\pm 4$

$\left(C\right)\left(2\right),\left(4\right)$

As

$\overline{a}=\mu \overrightarrow{b}+4\overrightarrow{c}\Rightarrow \mu \left(|\overrightarrow{b}|\right)=-4\overrightarrow{b}\cdot \overrightarrow{c}$

and $|\overrightarrow{b}{|}^2=4\overrightarrow{a}\cdot \overrightarrow{c}$ and

$|\overrightarrow{b}{|}^2+\overrightarrow{b}\cdot \overrightarrow{c}-\overrightarrow{d}\cdot \overrightarrow{c}=0$ Again, as

$2|\overrightarrow{b}+\overrightarrow{c}|=|\overrightarrow{b}-\overrightarrow{a}|$

Solving and eliminating $\overrightarrow{b}\cdot \overrightarrow{c}$ and eliminating $|\overrightarrow{a}{|}^2$

we get $(2{\mu }^2-10\mu )|\overrightarrow{b}{|}^2=0\Rightarrow \mu =0$ and 5.

$\left(D\right)\left(3\right)$

$I=\frac{2}{\pi }{\int }_{-\pi }^{\pi }$ $\frac{sin9\left(x/2\right)}{sin\left(x/2\right)}dx=\frac{2}{\pi }\times 2$

${\int }_0^{\pi }\frac{sin9\left(x/2\right)}{sin\left(x/2\right)}dx$

$x/2=\theta \Rightarrow dx=2d\theta$

$x=0,\theta =0$

$x=\pi \theta =\pi /2$

$I=\frac{8}{\pi }{\int }_0^{\pi /2}\frac{sin9\theta }{sin\theta }d\theta$

$=\frac{8}{\pi }{\int }_0^{\pi /2}$ $\frac{(\sin 9\theta -\sin 7\theta )}{\sin \theta }$ $+\frac{(\sin 7\theta -\sin 5\theta )}{\sin \theta }$ $+\frac{(\sin 5\theta -\sin 3\theta )}{\sin \theta }$ $+\frac{(\sin 3\theta -\sin \theta )}{\sin \theta }$ $+\frac{\sin \theta }{\sin \theta }d\theta$

$=\frac{16}{\pi }{\int }_0^{\pi /2}$ $(\cos 8\theta$ $+\cos 6\theta$ $+\cos 4\theta$ $+\cos 2\theta$ $+1$ )$d\theta$

$=4$