Ncert Solutions Maths class 12th

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,  

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, t

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

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 elimin

 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 $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

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

$\therefore$ Therefore, AB and C