Ncert Solutions Maths class 12th

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 =

Given,

A (2,3,4), B (-1, -2,1), C (5,8,7)

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

Where, a₁=3, b₁=-5, c₁=-3

Direction ratio of BC= $\begin{array}{l}\left(5-\left(-1\right)\right),\left(8-\left(-2\right)\right),\left(7-1\right)\\ =\left(6,10,6\right)\end{array}$

Where, a₂=6, b₂=10, c₂=6

Now,

$\begin{array}{l}\frac{a_2}{a_1}=\frac{6}{-3}=-2\\ \frac{b_2}{b_1}=\frac{10}{-5}=-2\\ \frac{c_2}{c_1}=\frac{6}{-3}=-2\end{array}$

Here, direction ratio of two-line segments are proportional.

So, A, B, C are collinear.

Direction cosine are 

$\begin{array}{l}\frac{-18}{},\frac{+12}{},\frac{-4}{}\\ =\frac{-18}{22},\frac{12}{22},\frac{-4}{22}\\ =\frac{-9}{11},\frac{6}{11},\frac{-2}{11}\end{array}$

26. Given f (x) = $\{\begin{array}{ll}k{x}^2\hfill & if x\le 2.\hfill \\ 3\hfill & if x>2.\hfill \end{array}$

For continuous at x = 2,

f (2) = k (2)2 = 4x.

L.H.L. = $\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}{x}^2=4x$

R.H.L. = $\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}3=3$

Then, L.H.L = R.H.L. = f (2)

i e, 4x = 3

$\Rightarrow x=\frac{3}{4}.$

Let the angles be α, β, r which are equal  

Let the direction cosines of the line be l, m, n.

$\begin{array}{l}l=cos\alpha ,m=cos\beta ,n=cosr\\ \alpha =\beta =r\\ l^2+m^2+n^2=1\\ co{s}^2\alpha +co{s}^2\beta +co{s}^{2}r=1\\ co{s}^2\alpha +co{s}^2\alpha +co{s}^2\alpha =1\\ 3co{s}^2\alpha =1\\ co{s}^2\alpha =\frac{1}{3}\\ cos\alpha =\frac{1}{}\end{array}$

Let the direction cosine of the line be l, m, n.

Then,

$\begin{array}{l}l=90^{?}=cos90^{?}=0\\ m=cos135=\frac{-1}{}\\ n=cos45^{?}=\frac{1}{}\end{array}$

 25. Given, f(x) = $\{\begin{array}{cc}\hfill \frac{\pi cosx}{\pi -2x}\hfill & \hfill if x\ne \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 3\hfill & \hfill if x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \end{array}$

For continuity at $x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\underset{x\to \raisebox{1ex}{${\pi }^{-}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}f(x)=\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}f(x)=f\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right).$

$\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{\pi -2x}=\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2^{+}$}\right.}{lim}\frac{xcosx}{\pi -2x}=3.$

Take $\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{x-2x}=3$ .

Putting x = $\frac{\pi }{2}+h$ such that as $x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,h\to 0.$

So $\underset{h\to 0}{lim}\frac{xcos(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+h)}{x-2(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+n)}=\underset{h\to 0}{lim}\frac{x(-sinx)}{-2h}$

$=\underset{h\to 0}{lim}\frac{xsinh}{2h}$

$=\frac{k}{2}\underset{h\to 0}{lim}\frac{sinh}{h}=\frac{k}{2}.$

i e, $\frac{k}{2}=3$

$\Rightarrow$ k = 6

Similarly from $\underset{x\to \raisebox{1ex}{${\pi }^{+}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{\pi -2x}=3$

$\Rightarrow \underset{h\to 0}{lim}\frac{hcos\left(\frac{\pi }{2}+h\right)}{\pi -2\left(\frac{\pi }{2}+h\right)}=\underset{h\to 0}{lim}\frac{-hsinh}{-2h}$

$=\frac{}{2}\underset{h\to 0}{lim}\frac{sinh}{h}$

$=\frac{x}{2}$

So, $\frac{x}{2}=3$

$\Rightarrow$ k = 6

24. Given, f(x) = $\{\begin{array}{cc}\hfill sinx-cosx,\hfill & \hfill if x\ne 0\hfill \\ \hfill -1,\hfill & \hfill if x=0.\hfill \end{array}$

For x = c $\cancel{=}$ 0,

f(c) = sin c cos c.

$\underset{x\to c}{lim}$ f (x) = $\underset{x\to c}{lim}$ (sin x cos x) = sin c cos c = f(c)

So, f is continuous at $x\ne 0$

For x = 0,

f(0) = 1

$\underset{x\to 0}{lim}$ f (x) = $\underset{x\to 0}{lim}$ (sin x cos x) = sin 0 cos 0 = 0 1 = 1

$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}(sinx-corx)=sin0-cos0=-1.$

∴ $\underset{x\to 0-}{lim}$ f(x) = $\underset{x\to 0+}{lim}$ f (x) = f (0)

So, f is continuous at x = 0.

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29.

23. Given f (x) = $\{\begin{array}{ll}{x}^{2}sin\frac{1}{x}, if x\ne 0.\hfill & 0 if x=0.\hfill \end{array}$

For x = c $\cancel{=}$ 0,

f (c) = $c^{2}sin\frac{1}{c}$

$\underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}{x}^{2}sin\frac{1}{x}=c^{2}sin\frac{1}{c}.$

So, f is continuous for $x\ne 0.$

For x = 0,

f (0) = 0

$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}\left(x^{2}sin\frac{1}{x}\right)$

As we have sin $\frac{1}{x}\in [-1,1]$

$\underset{x\to 0}{lim}$ f (x) = 02 a where $a\in [-1,1]$

= 0 = f (0).

∴ f is also continuous at x = 0.