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(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.

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

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,

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

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The required line passes through the origin. Therefore, its position vector is given by,

$\overrightarrow{a}=0.........(1)$

The direction ratios of the line through origin and $\left(5,-2,3\right)$ are

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

The line is parallel to the vector given by the equation,  $\overrightarrow{b}=5\widehat{i}-2\widehat{j}+3\widehat{k}$

The equation of the line in vector form through a point with position vector  $\overrightarrow{a}$ and parallel to  $\overrightarrow{b}$ is,  $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\lambda \in R$

$\begin{array}{l}\Rightarrow \overrightarrow{r}=\overrightarrow{0}+\lambda \left(5\widehat{i}-2\widehat{j}+3\widehat{k}\right)\\ \Rightarrow \overrightarrow{r}=\lambda \left(5\widehat{i}-2\widehat{j}+3\widehat{k}\right)\end{array}$

The equation of the line through the point $(x_1, y_1, z_1)$ and direction ratios  $a, b, c$ is given by, 

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

Therefore, the equation of the required line in the Cartesian form is

$\begin{array}{l}\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\\ \Rightarrow \frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\end{array}$

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

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

29. Given, f(x) = $\{\begin{array}{ccc}\hfill 5\hfill & \hfill if x\le 2\hfill & \hfill ax+b\hfill \\ \hfill if 2<x<10\hfill & \hfill 21\hfill & \hfill if x\ge 10\hfill \end{array}$

For continuity at x = 2,

$\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{+}}{lim}f(x)=f(2)$

$\Rightarrow \underset{x\to 2^{-}}{lim}5=\underset{x\to 2^{+}}{lim}ax+b=5.$

$\Rightarrow$ 5 = 2a + b (i)

For continuous at x = 10,

$\underset{x\to 10^{-}}{lim}f(x)=\underset{x\to 10^{+}}{lim}f(x)=f(10)$

$\Rightarrow \underset{x\to 10^{-}}{lim}ax+b=\underset{x\to 10^{+}}{lim}21=21.$

$\Rightarrow$ 10a + b = 21 (2).

So, e q (2) 5 e q (1) we get,

10a + b 5 (2a + b) = 21 5 5.

$\Rightarrow$ 10a + b 10a 5b = 21 25.

$\Rightarrow$ 4b = 4

$\Rightarrow$ b = 1.

And putting b = 1 in e q (1),

$\Rightarrow$ 2a = 5 b = 5 1 = 4

$\Rightarrow a=\frac{4}{2}=2.$

Hence, a = 2 and b = 1.

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The line passes through the point with position vector, $\overrightarrow{a}=2\widehat{i}-\widehat{j}+4\widehat{k}-----(1)$

The given vector: $\overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k}-----(2)$

The line which passes through a point with position vector $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$ is given by,

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \overrightarrow{r}=2\widehat{i}-\widehat{j}+4\widehat{k}+\lambda \left(\widehat{i}+2\widehat{j}-\widehat{k}\right)\end{array}$

$\therefore$ This is required equation of the line in vector form.

Now,

Let $\begin{array}{l}\overrightarrow{r}=x\widehat{i}-y\widehat{j}+z\widehat{k}\\ \Rightarrow x\widehat{i}-y\widehat{j}+z\widehat{k}=\left(\lambda +2\right)\widehat{i}+\left(2\lambda -1\right)\widehat{j}+\left(-\lambda +4\right)\widehat{k}\end{array}$

Comparing the coefficient to eliminate $\lambda$ ,

$\begin{array}{l}x=\lambda +2,x_1=2,a=1\\ y=2\lambda -1,y_1=-1,b=2\\ z=-\lambda +4,z_1=4,c=-1\end{array}$

$\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}$

Given,

The line passes through the point $A\left(1,2,3\right)$ .

Position vector of A,

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

Let $\overrightarrow{b}=3\widehat{i}+2\widehat{j}-2\widehat{k}$

The line which passes through point $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$ is given by,

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ =\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda \left(3\widehat{i}+2\widehat{j}-2\widehat{k}\right)\end{array}$ , where $\lambda$ is constant

 28. Given, f (x) $\{\begin{array}{ll}kx+1,\hfill & if x\le 5\hfill \\ 3x-5,\hfill & if x>5.\hfill \end{array}$

For continuity at x = 5,

$\underset{x\to 5^{-}}{lim}f(x)=\underset{x\to 5^{-}}{lim},kx+1=5x+1.$

$\underset{x\to 5^{+}}{lim}f(x)=\underset{x\to 5^{+}}{lim}3x-5=15-5=10$

f (5) = 5k + 1

So,  $\underset{x\to 5^{-}}{lim}f(x)=\underset{x\to 5^{+}}{lim}f(x)=f(5).$

i e, 5k + 1 = 10

$\Rightarrow$ 5k = 10 1

$\Rightarrow$ k = $\frac{9}{5}.$

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19. $$

$=\frac{1}{2}+4.\frac{1}{4}$

$=\frac{1}{2}+1$ $=\frac{3}{2}$

= R.H.S.

Let AB be the line through the point $\left(4,7,8\right)$ and $\left(2,3,4\right)$ and CD be line through the point $\left(-1,-2,1\right)$ and $\left(1,2,5\right)$

Direction cosine, $a_1,b_1,c_1$ of AB are

$\begin{array}{l}=\left(2-4\right),\left(3-7\right),\left(4-8\right)\\ =\left(-2,-4,-4\right)\end{array}$

Direction cosine, $a_2,b_2,c_2$ of CD are

$\begin{array}{l}=\left(1-\left(-1\right)\right),\left(2-\left(-2\right)\right),\left(5-1\right)\\ =\left(2,4,4\right)\end{array}$

AB will be parallel to CD only

If

$\begin{array}{l}\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\ \Rightarrow \frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}\\ \Rightarrow -1=-1=-1\end{array}$

Here, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, AB is parallel to CD.