Maths

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

It is known that if n1 and n2 are normal to the planes,  $\overrightarrow{r}.\overrightarrow{n_1}=d_1\&\overrightarrow{r}.\overrightarrow{n_2}=d_2$  then the angle between them, Q, is given by,

$cosQ=\left|\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}\right|..........(1)$

$\begin{array}{l}Here,\overrightarrow{n_1}=2\widehat{i}+2\widehat{j}-3\widehat{k}\&\overrightarrow{n_2}=3\widehat{i}-3\widehat{j}+5\widehat{k}\\ \therefore \overrightarrow{n_1}.\overrightarrow{n_2}=\left(2\widehat{i}+2\widehat{j}-3\widehat{k}\right)\left(3\widehat{i}-3\widehat{j}+5\widehat{k}\right)=2.3+2.(-3)+(-3).5=-15\\ \left|\overrightarrow{n_1}\right|==\\ \left|\overrightarrow{n_2}\right|==\end{array}$

Substituting the value of $\overrightarrow{n_1}.\overrightarrow{n_2}$ $\left|\overrightarrow{n_1}\right|\&\left|\overrightarrow{n_2}\right|$ in equation (1), we obtain

$\begin{array}{l}cosQ=\left|\frac{-15}{.}\right|\\ \Rightarrow cosQ=\frac{15}{}\\ \Rightarrow cos{Q}^{-1}=\left(\frac{15}{}\right)\end{array}$

The equation of the plane through the intersection of the planes, $x+y+z=1and2x+3y+4z=5$ , is

  $\left(x+y+z-1\right) +\lambda \left(2x+3y+4z-5\right)$

$\Rightarrow \left(2\lambda +1\right)x+\left(3\lambda +1\right)y+\left(4\lambda +1\right)z-\left(5\lambda +1\right)=0 ..........\left(1\right)$

The direction ratios,  $a_1, b_1, c_1,$ of this plane are $\left(2\lambda +1\right),\left(3\lambda +1\right),and\left(4\lambda +1\right).$

The plane in equation (1) is perpendicular to $x-y+z=0$

Its direction ratios,  $a_2, b_2, c_2,$ are $1,-1,and1$ .

Since the planes are perpendicular,

$\begin{array}{l}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ \Rightarrow \left(2\lambda +1\right)-\left(3\lambda +1\right)+\left(4\lambda +1\right)=0\\ \Rightarrow 3\lambda +1=0\\ \Rightarrow \lambda =-\frac{1}{3}\end{array}$

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

$\begin{array}{l}\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3}=0\\ \Rightarrow x-z+2=0\end{array}$

This is the required equation of the plane.

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 $\lambda =\frac{10}{9}$ in equation (3), we obtain

$\begin{array}{l}\overrightarrow{r}.\left(\frac{38}{9}\widehat{i}+\frac{68}{9}\widehat{j}+\frac{3}{9}\widehat{k}\right)=17\\ \Rightarrow \overrightarrow{r}.\left(38\widehat{i}+68\widehat{j}+3\widehat{k}\right)=153\end{array}$

This is the vector equation of the required plane.

26. $$L.H.S $L.H.S.=cos\left(\frac{3\pi }{2}+x\right)cos(2\pi +x)\left[cot\left(\frac{3\pi }{2}-x\right)+cot(2\pi +x)\right]$

Here,

$cos\left(\frac{3\pi }{2}+x\right)=cos\left(\frac{4\pi -\pi }{2}+x\right)=cos\left(2\pi -\frac{\pi }{2}+x\right)$

$=cos\left[2\pi -\left(\frac{\pi }{2}-x\right)\right];VI quadrant.$

$cos\left(\frac{\pi }{2}-x\right);i I^{st} quadrant.$

$=sinx$

$cos(2x+x)=cosx,as xlies in I^{st}quadrant.$

$cot(2\pi +x)=cotx;asxliesin{I}^{st}quadrant.$

$cot\left(\frac{3\pi }{2}-x\right)=cot\left[\frac{4\pi -\pi }{2}-x\right]$

$=cot\left[2\pi -\frac{\pi }{2}-x\right]$

$=cot\left[2x-\left(\frac{\pi }{2}+x\right)\right];V{I}^{th} quadrant.$

$=-cot\left(\frac{\pi }{2}+x\right);I{I}^{nd}quadrat.$

$=-(-tanx)$

$=tanx.$

So.L.H.S $L.H.S=sinxcosx[tanx+cotx]$

$=sinxcosx\left[\frac{sinx}{cosx}+\frac{cosx}{sinx}\right]$

$=sinxcosx\frac{\left(si{n}^{2}x+co{s}^{2}x\right)}{cosxsinx}$

$=si{n}^{2}x+co{s}^{2}x$

= 1

= R.H.S.

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).

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

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