Maths

$\frac{1-x}{3}=\frac{7y-14}{2\rho }=\frac{z-3}{2}and\frac{7-7x}{3\rho }=\frac{y-5}{1}=\frac{6-z}{5}$

The standard form of a pair of Cartesian lines is;

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}and\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}-----(1)$

So,

$\begin{array}{l}\frac{-\left(x-1\right)}{3}=\frac{7\left(y-5\right)}{2\rho }=\frac{z-3}{2}and\frac{-7\left(x-1\right)}{3\rho }=\frac{y-5}{1}=\frac{-\left(z-6\right)}{5}\\ \frac{x-1}{-3}=\frac{y-2}{2\rho /7}=\frac{z-3}{2}and\frac{x-1}{-3\rho /7}=\frac{y-5}{1}=\frac{z-6}{-5}-----(2)\end{array}$

Comparing (1) and (2) we get

$\begin{array}{l}{a}_1=-3,b_1=\frac{2\rho }{7},c_1=2\\ a_2=\frac{-3\rho }{7},b_2=1,c_2=-5\end{array}$

Now, both the lines are at right angles

So, $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\begin{array}{l}\Rightarrow \left(-3\right)*\frac{\left(-3\rho \right)}{7}+\frac{2\rho }{7}*1+2*\left(-5\right)=0\\ \Rightarrow \frac{9\rho }{7}+\frac{2\rho }{7}+\left(-10\right)=0\\ \Rightarrow \frac{9\rho +2\rho }{7}=10\\ \Rightarrow 11\rho =70\\ \rho =\frac{70}{11}\end{array}$

$\therefore$ The value of $\rho$ is $\frac{70}{11}$

(i) Let  ${\overrightarrow{b}}_1$  and  ${\overrightarrow{b}}_2$  be the vectors parallel to the pair of lines,  $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\&\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$ , respectively.

$\begin{array}{l}{\overrightarrow{b}}_1=2\widehat{i}+5\widehat{j}-3\widehat{k}\&{\overrightarrow{b}}_2=-\widehat{i}+8\widehat{j}+4\widehat{k}\\ \therefore \left|{\overrightarrow{b}}_1\right|==\\ \left|{\overrightarrow{b}}_2\right|===9\\ {\overrightarrow{b}}_1.{\overrightarrow{b}}_2=\left(2\widehat{i}+5\widehat{j}-3\widehat{k}\right).\left(-\widehat{i}+8\widehat{j}+4\widehat{k}\right)\\ =2(-1)+5*8+(-3).4\\ =-2+40-12\\ =26\end{array}$

32. Given, f (x) = sin $x$

Let g (x) = sin x and h (x) = $x$ then as sine f x and modulus f x are continuous in x e R

g and h are continuous.

So, (goh) (x) = g (h (x) = g (|x|) = sin |x| = f (x)

Is a continuous f x being a competitive f x of two continuous f x.

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,  $cosQ=\left|\frac{{\overrightarrow{b}}_1.{\overrightarrow{b}}_2}{\left|{\overrightarrow{b}}_1\right|\left|{\overrightarrow{b}}_2\right|}\right|$

The given lines are parallel to the vectors,  ${\overrightarrow{b}}_1=3\widehat{i}+2\widehat{j}+6\widehat{k}\&{\overrightarrow{b}}_2=\widehat{i}+2\widehat{j}+2\widehat{k}$ , respectively.

$\begin{array}{l}\therefore \left|{\overrightarrow{b}}_1\right|==7\\ \left|{\overrightarrow{b}}_2\right|==3\\ {\overrightarrow{b}}_1.{\overrightarrow{b}}_2=\left(3\widehat{i}+2\widehat{j}+6\widehat{k}\right).\left(\widehat{i}+2\widehat{j}+2\widehat{k}\right)\\ =3*1+2*2+6*2\\ =3+4+12\\ =19\\ \Rightarrow cosQ=\frac{19}{7*3}\\ \Rightarrow Q=co{s}^{-1}\left(\frac{19}{21}\right)\end{array}$

31. Given, f (x) = $\left|cosx\right|.$

Let g (x) = cos x and h (x) = $x$

Hence, as cosine function and modulus f x are continuous $\forall x\in ?,g$ h are continuous.

Then, (hog) x = h (g (x)

= h (cos x)

$=\left|cosx\right|.$

= f (x) is also continuous being

A composites fxn of two continuous f x $\forall x\in ?$

Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by,

$\overrightarrow{a}=3\widehat{i}-2\widehat{j}-5\widehat{k}$

The direction ratios of PQ are given by,

$\left(3-3\right)=0,\left(-2+2\right)=0,\left(6+5\right)=11$

The equation of the vector in the direction of PQ is

$\overrightarrow{b}=0.\widehat{i}-0.\widehat{j}+11\widehat{k}=11\widehat{k}$

The equation of PQ in vector form is given by,  $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\lambda \in R$

$\Rightarrow \overrightarrow{r}=\left(5\widehat{i}-2\widehat{j}-5\widehat{k}\right)+11\lambda \widehat{k}$

The equation of PQ in Cartesian form is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ i.e,

$\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}$

30. Given f (x) = cos (x2)

Let g (x) = cos x is a lregononuie fa (cosine) which is continuous function

and let h (x) = x2 is a polynomial f xn which is also continuous

Hence (goh) x = g (h (x)

= g (x)2

= cos (x2)

= f (x)

is also a continuous f x being a composite fxn of how continuous f x $\forall x\in ?$

Given,

Cartesian equation,

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$

The given line passes through the point $\left(5,-4,6\right)$

i.e. position vector of $\overrightarrow{a}=5\widehat{i}-4\widehat{j}+6\widehat{k}$

Direction ratio are 3, 7 and 2.

Thus, the required line passes through the point $\left(5,-4,6\right)$ and is parallel to the vector $3\widehat{i}+7\widehat{j}+2\widehat{k}$ .

Let $\overrightarrow{r}$ be the position vector of any point on the line, then the vector equation of the line is given by,

$\overrightarrow{r}=\left(5\widehat{i}-4\widehat{j}+6\widehat{k}\right)+\lambda \left(3\widehat{i}+7\widehat{j}+2\widehat{k}\right)$

Given,

The point $\left(-2,4,-5\right)$ .

The Cartesian equation of a line through a point $\left(x_1,y_1,z_1\right)$ and having direction ratios a, b, c is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Now, given that

$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ is parallel

to point $\left(-2,4,-5\right)$

Here, the point $\left(x_1,y_1,z_1\right)$ is $\left(-2,4,-5\right)$ and the direction ratio is given by $a=3,b=5,c=6$

$\therefore$ The required Cartesian equation is

$\begin{array}{l}\frac{x-\left(-2\right)}{3}=\frac{y-4}{5}=\frac{z-\left(-5\right)}{6}\\ \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\end{array}$