Maths

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.

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 $\lambda$ ,

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

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

19. $$

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

$=\frac{1}{2}+1$ $=\frac{3}{2}$

= R.H.S.

 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 = $\frac{9}{5}.$

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

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.

Let AB be the line joining the points $\left(1,-1,2\right)$ and $\left(3,4,-2\right)$ and CD be the line joining the point $\left(0,3,2\right)$ and $\left(3,5,6\right)$ .

The direction ratios, a, b, c of AB are $\begin{array}{l}\left(3-1\right),\left(4-\left(-1\right)\right),\left(-2-2\right)\\ =\left(2,5,-4\right)\end{array}$

The direction ratios $a_2,b_2,c_2$ of CD are $\begin{array}{l}\left(3-0\right),\left(5-3\right),\left(6-2\right)\\ =\left(3,2,4\right)\end{array}$

AB and CD will be perpendicular to each other, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\begin{array}{l}=2*3+5*2+\left(-4\right)*4\\ =6+10-16\\ =0\end{array}$

$\therefore$ Therefore, AB and CD are perpendicular to each other.

18. $$

$\begin{array}{l}{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-1^2\\ =\frac{1}{4}+\frac{1}{4}-1\\ =\frac{1+1+4}{4}\\ =\frac{-2}{4}\\ =\frac{-1}{2}\end{array}$

=R.H.S

Two lines with direction cosines l, m, n and l₂, m₂, n₂ are perpendicular to each other, if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$ .

Now, for the 3 lines with direction cosine,

$\begin{array}{l}\frac{12}{13},\frac{-3}{13},\frac{-4}{13}and\frac{4}{13},\frac{12}{13},\frac{3}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{12}{13}*\frac{4}{13}+\frac{\left(-3\right)}{13}*\frac{12}{13}+\frac{\left(-4\right)}{13}*\frac{3}{13}\\ =\frac{48}{169}-\frac{36}{169}-\frac{12}{169}\\ =0\end{array}$

Hence, the lines are perpendicular.

For lines with direction cosines,

$\begin{array}{l}\frac{4}{13},\frac{12}{13},\frac{3}{13}and\frac{3}{13},\frac{-4}{13},\frac{12}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{4}{13}*\frac{3}{13}+\frac{12}{13}*\frac{\left(-4\right)}{13}+\frac{3}{13}*\frac{12}{13}\\ =\frac{12}{169}-\frac{48}{169}+\frac{36}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

For the lines with direction cosines,

$\begin{array}{l}\frac{3}{13},\frac{-4}{13},\frac{12}{13}and\frac{12}{13},\frac{-3}{13},\frac{-4}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{3}{13}*\frac{12}{13}+\frac{\left(-4\right)}{13}*\frac{\left(-3\right)}{13}+\frac{12}{13}*\frac{\left(-4\right)}{13}\\ =\frac{36+12-48}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

Therefore, all the lines are perpendicular.

The vertices of ABC are A (3,5, -4), B (-1,1,2) and C (-5, -5, -2)

Direction ratio of side AB $\begin{array}{l}=\left(-1-3\right)\left(1-5\right)\left(2-(-4)\right)\\ =\left(-4,-4,6\right)\end{array}$

 

Direction cosine of AB,

Direction ratios of BC= $\begin{array}{l}\left(-5-\left(-1\right)\right),\left(-5-1\right),\left(-2-2\right)\\ =\left(-4,-6,-4\right)\end{array}$

Direction cosine of BC =

 

 

 

Direction of CA= $\begin{array}{l}\left(-5-3\right)\left(-5-5\right)\left(-2-\left(-4\right)\right)\\ =\left(-8,-10,2\right)\end{array}$

Direction cosine of CA =