Ncert Solutions Maths class 12th

The equations of the planes are  $\overrightarrow{r}.\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)=7,\overrightarrow{r}.\left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)=9$

$\begin{array}{l}\Rightarrow \overrightarrow{r}.\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)-7=0..........(1)\\ \Rightarrow \overrightarrow{r}.\left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)-9=0..........(2)\end{array}$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$\begin{array}{l}\left[\overrightarrow{r}.\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)-7\right]+\lambda \left[\overrightarrow{r}.\left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)-9\right]=0,where,\lambda \in R\\ \overrightarrow{r}.\left[\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)+\lambda \left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)\right]=9\lambda +7\\ \overrightarrow{r}.\left[(2+2\lambda )\widehat{i}+(2+5\lambda )\widehat{j}+(3\lambda -3)\widehat{k}\right]=9\lambda +7..........(3)\end{array}$

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

$\overrightarrow{r}=2\widehat{i}+1\widehat{j}+3\widehat{k}$

Substituting in equation (3), we obtain

$\begin{array}{l}\left(2\widehat{i}+\widehat{j}+3\widehat{k}\right).\left[(2+2\lambda )\widehat{i}+(2+5\lambda )\widehat{j}+(3\lambda -3)\widehat{k}\right]=9\lambda +7\\ \Rightarrow 2(2+2\lambda )+1(2+5\lambda )+3(3\lambda -3)=9\lambda +7\\ \Rightarrow 4+4\lambda +2+5\lambda +9\lambda -9=9\lambda +7\\ \Rightarrow 18\lambda -3=9\lambda +7\\ \Rightarrow 9\lambda =10\\ \Rightarrow \lambda =\frac{10}{9}\end{array}$

Substituting $\lambda =\frac{10}{9}$&nb

The equation of any plane through the intersection of the planes,

$3x - y +2z ­-4=0and x + y + z -2=0,$ is

$\left(3x - y +2z -4\right)+\alpha \left(x + y + z -2\right)=0,where,\alpha \in R..........(1)$

The plane passes through the point $\left(2,2,1\right).$ Therefore, this point will satisfy equation (1).

$\begin{array}{l}\therefore (3\times 2-2+2\times 1-4)+\alpha (2+2+1-2)=0\\ \Rightarrow 2+3\alpha =0\\ \Rightarrow \alpha =-\frac{2}{3}\end{array}$

Substituting $\alpha =-\frac{2}{3}$ in equation (1), we obtain

$\begin{array}{l}(3x-y+2z-4)-\frac{2}{3}(x+y+z-2)=0\\ \Rightarrow 3(3x-y+2z-4)-2(x+y+z-2)=0\\ \Rightarrow (9x-3y+6z-12)-2(x+y+z-2)=0\\ \Rightarrow 7x-5y+4z-8=0\end{array}$

The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form,  y = a

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

$2x+y-z=5 .........\left(1\right)$

Dividing both sides of equation (1) by 5, we obtain

$\begin{array}{l}\frac{2}{5}x+\frac{y}{5}-\frac{z}{5}=1\\ \Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1..........(2)\end{array}$

It is known that the equation of a plane in intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ , where a,  b,  c are the intercepts cut off by the plane at x,  y, and z axes respectively.

Therefore, for the given equation,

$a=\frac{5}{2},b=5andc=-5$

Thus, the

41. Kindly consider the following

We know that through three collinear points $A,B,C$ i.e., through a straight line, we can pass an infinite number of planes.

(a) The given points are $A\left(1,1,-1\right),B\left(6,4,-5\right),andC\left(-4,-2,3\right).$

$\begin{array}{l}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 6\hfill & \hfill 4\hfill & \hfill -5\hfill \\ \hfill -4\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right|=(12-10)-(18-20)-(-12+16)\\ \end{array}$

$\begin{array}{l}=2+2-4\\ =0\end{array}$

Since $A,B,C$ are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are $A\left(1,1,0\right),B\left(1,2,1\right),andC\left(-2,2,-1\right).$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=(-2-2)-(2+2)=-8\ne 0$

T

40. Kindly go through the solution

(a) The position vector of point $\left(1,0,-2\right)$ is  $\overrightarrow{a}=\widehat{i}-2\widehat{k}$

The normal vector N perpendicular to the plane is  $\overrightarrow{N}=\widehat{i}+\widehat{j}-\widehat{k}$

The vector equation of the plane is given by,  $\left(\overrightarrow{r}-\overrightarrow{a}\right).\overrightarrow{N}=0$

$\Rightarrow \left[\overrightarrow{r}-\left(\widehat{i}-2\widehat{k}\right)\right].\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=0.........(1)$

$\overrightarrow{r}$ is the position vector of any point $\left(x, y, z\right)$ in the plane.

$\therefore \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$

Therefore, equation (1) becomes

$\begin{array}{l}\left[\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)-\left(\widehat{i}-2\widehat{k}\right)\right].\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=0\\ \Rightarrow \left[(x-1)\widehat{i}+y\widehat{j}+(z+2)\widehat{k}\right].\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=0\\ \Rightarrow (x-1)+y-(z+2)=0\\ \Rightarrow x+y-z-3=0\\ \Rightarrow x+y-z=3\end{array}$

This is the Cartesian

39. Let f (x) = cos (x³) sin² (x⁵).

f' (x) = cos (x³) $\frac{d}{dx}$  sin² (x⁵) + sin² (x⁵) $\frac{d}{dx}$ cos (x³)

= cos (x³) 2sin (x⁵) $\frac{d}{dx}$  sin (x⁵) + sin² (x⁵) [sin (x³)] $\frac{d}{dx}$ x³.

= 2 cos (x³) sin (x⁵). cos (x⁵) $\frac{d}{dx}$   (x⁵) - sin² (x⁵) sin (x³). 3x²

= 2. cos (x³) sin (x⁵) cos (x⁵). 5 - 3x²sin² (x⁵) sin (x³)

= x² sin (x⁵