Ncert Solutions Maths class 12th

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_1, y_1, z_1).$

$\begin{array}{l}2x + 3y + 4z - 12 = 0\hfill \end{array}$

$\begin{array}{l}\Rightarrow 2x + 3y + 4z = 12 \dots \left(1\right)\hfill \end{array}$

(a) It is given that equation of the plane is

$\overrightarrow{r}.\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=2$

For any arbitrary point $P\left(x, y, z\right)$ on the plane, position vector  $\overrightarrow{r}$ I s given by,  $\overrightarrow{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Substituting the value of  $\overrightarrow{r}$ in equation (1), we obtain

$\left(x\widehat{i}+y\widehat{j}-z\widehat{k}\right).\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=2$

$\Rightarrow x+y-z=2$

This is the Cartesian equation of the plane.

(b)  $\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-4\widehat{k}\right)=1$

For any arbitrary po

38. Let f(x) = $\frac{sin(ax+b)}{cos(cx+d)}.$

f'(x) = $\frac{cos(cx+d)\frac{d}{dx}sin(ax+b)-sin(ax+b)\frac{d}{dx}cos(cx+d)}{co{s}^2(cx+d).}$

= $\frac{cos(cx+d)cos(ax+b)\frac{d}{dx}(ax+b)+sin(ax+b)sin(cx+d)\frac{d(x+d)}{dx}}{co{s}^2(cx+d)}$

= $=\frac{cos(cx+d)cos(ax+b)·a+sin(ax+b)sin(cx+d)·c}{co{s}^2(cx+d)}$

37. Kindly go through the solution

(a) It is given that equation of the plane is

$\overrightarrow{r}.\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=2$

For any arbitrary point $P\left(x, y, z\right)$ on the plane, position vector  $\overrightarrow{r}$ I s given by,  $\overrightarrow{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Substituting the value of  $\overrightarrow{r}$ in equation (1), we obtain

$\left(x\widehat{i}+y\widehat{j}-z\widehat{k}\right).\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=2$

$\Rightarrow x+y-z=2$

This is the Cartesian equation of the plane.

(b)  $\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-4\widehat{k}\right)=1$

For any arbitrary po

(a) The equation of the plane is  $z =2or0x +0y + z =2..........\left(1\right)$

The direction ratios of normal are $0,0,and1.$

$\therefore +02 +12 =1$

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

$0.x+0.y+1.z=2$

This is of the form  $lx + my + nz = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perp

36. Let f (x) = sin (ax + b)

f' (x) = $\frac{d}{dx}$ sin (ax + b)

= cos (ax + b) $\frac{d}{dx}$  (ax + b)

= a cos (ax + b).

35. Let f (x) = cos (sin x).

f' (x) $\frac{d}{dx}$ cos (sin x)

= - sin (sin x) $\frac{d}{dx}$ sin x

= - sin (sin x) cos x.

$\begin{array}{l}\overrightarrow{r}=\left(1-t\right)\widehat{i}+\left(t-2\right)\widehat{j}+\left(3-2t\right)\widehat{k}\\ \overrightarrow{r}=\widehat{i}-t\widehat{i}+t\widehat{j}-2\widehat{j}+3\widehat{k}-2t\widehat{k}\\ \overrightarrow{r}=\widehat{i}-2\widehat{j}+3\widehat{k}+t\left(-\widehat{i}+\widehat{j}-2\widehat{k}\right)-----(1)\end{array}$

$\begin{array}{l}\Rightarrow \overrightarrow{r}=\left(s+1\right)\widehat{i}+\left(2s-1\right)\widehat{j}-\left(2s+1\right)\widehat{k}\\ \overrightarrow{r}=s\widehat{i}+\widehat{i}+2s\widehat{j}-\widehat{j}-2s\widehat{k}-\widehat{k}\\ \overrightarrow{r}=\widehat{i}-\widehat{j}-\widehat{k}+s\left(\widehat{i}+2\widehat{j}-2\widehat{k}\right)-----(2)\end{array}$

Here,   

Hence, the shortage distance between the line is given by

$d=\left|\frac{\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right).\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)}{\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|}\right|=\frac{8}{}$

$\begin{array}{l}\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\lambda \left(\widehat{i}-3\widehat{j}+2\widehat{k}\right)and-----(1)\\ \overrightarrow{r}=4\widehat{i}+5\widehat{j}+6\widehat{k}+\mu \left(2\widehat{i}+3\widehat{j}+\widehat{k}\right)-----(2)\end{array}$

Here, comparing (1) and (2) with $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ , we have

$\begin{array}{l}{a}_1=\widehat{i}+2\widehat{j}+3\widehat{k},b_1=\widehat{i}-3\widehat{j}+2\widehat{k}\\ a_2=4\widehat{i}+5\widehat{j}+6\widehat{k},b_2=2\widehat{i}+3\widehat{j}+\widehat{k}\end{array}$

Therefore,

              $\begin{array}{l}{\overrightarrow{a}}_2-{\overrightarrow{a}}_1=3\widehat{i}+3\widehat{j}+3\widehat{k}\\ {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\\ =\widehat{i}\left(-3-6\right)-\widehat{j}\left(1-4\right)+\widehat{k}\left(3+6\right)\\ =-9\widehat{i}+3\widehat{j}+9\widehat{k}\\ \left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|==\\ =\\ =3\end{array}$