Introduction to Three Dimensional Geometry

Let height of the wall = h than A(0, 0, 0) G (10, 10, h)

$\overrightarrow{AG}=10\widehat{i}+10\widehat{j}+h\widehat{k}$              

B(10, 0, 0) A (0, 10, h)

$\overrightarrow{BH}=-10\widehat{i}+10\widehat{j}+h\widehat{k}$              

  $cos\theta =\frac{1}{5}=\frac{\overrightarrow{BH}.\overrightarrow{AG}}{\left|\overrightarrow{BH}\right|\left|\overrightarrow{AG}\right|}$             

$=\frac{-100+100+h^2}{\left(200+h^2\right)}$              

$5h^2=200+h^2\Rightarrow h^2=50$              

$h=5\sqrt[]{2}$

A(2, 3, 9)

B(5, 2, 1)

C(1, $\lambda$ , 8)

D( $\lambda$ ,2, 3)

$\Delta =\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -8\hfill \\ \hfill -4\hfill & \hfill \lambda -2\hfill & \hfill 7\hfill \\ \hfill \lambda -2\hfill & \hfill -1\hfill & \hfill -6\hfill \end{array}\right|$

$\overrightarrow{AB}=\left(3,-1,-8\right)$

$\overrightarrow{AC}=\left(-4,\lambda -2,7\right)$

$\overrightarrow{AD}=\left(\lambda -2,-1,-6\right)$

$\Delta =3\left[-6\lambda +12+7\right]-1\left(7\lambda -14-24\right)-8\left(4-{\left(\lambda -2\right)}^2\right)$

$=57-18\lambda \Rightarrow +38-32+8\left({\lambda }^2-4\lambda +4\right)$

$=95-57\lambda +8{\lambda }^2$

$\Delta =0\Rightarrow {\lambda }_1{\lambda }_2=\frac{95}{8}$

x + 2y + z = 14

${\perp }^{r}linePQ:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{1}=t$                

Q (1 + t, 2 + 2t, 3 + t)

 x + 2y + z = 14 ⇒ 1 + t + 4 + 4t + 3 + t = 14 ⇒ 6t = 6

t = 1

⇒ Q (2, 4, 4)

PQ = $\sqrt[]{1+4+1}=\sqrt[]{6}$  

$tan60°=\frac{PQ}{QR}\Rightarrow QR=\frac{PQ}{\sqrt[]{3}}=\sqrt[]{2}$

ar (PQR) = $\frac{1}{2}\sqrt[]{6}\cdot \sqrt[]{2}=\sqrt[]{3}$  

A(2, 3, 9)

B(5, 2, 1)

C(1, $\lambda$ , 8)

D( $\lambda$ ,2, 3)

$\Delta =\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -8\hfill \\ \hfill -4\hfill & \hfill \lambda -2\hfill & \hfill 7\hfill \\ \hfill \lambda -2\hfill & \hfill -1\hfill & \hfill -6\hfill \end{array}\right|$           

    $\overrightarrow{AB}=\left(3,-1,-8\right)$

$\overrightarrow{AC}=\left(-4,\lambda -2,7\right)$            

$\overrightarrow{AD}=\left(\lambda -2,-1,-6\right)$

$\Delta =3\left[-6\lambda +12+7\right]-1\left(7\lambda -14-24\right)-8\left(4-{\left(\lambda -2\right)}^2\right)$

$=57-18\lambda \Rightarrow +38-32+8\left({\lambda }^2-4\lambda +4\right)$            

$=95-57\lambda +8{\lambda }^2$

$\Delta =0\Rightarrow {\lambda }_1{\lambda }_2=\frac{95}{8}$                       

x + 2y + z = 14

${\perp }^{r}linePQ:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{1}=t$              

Q (1 + t, 2 + 2t, 3 + t)

x + 2y + z = 14 -> 1 + t + 4 + 4t + 3 + t = 14 Þ 6t = 6

t = 1

-> Q (2, 4, 4)

PQ = $\sqrt[]{1+4+1}=\sqrt[]{6}$  

$tan60°=\frac{PQ}{QR}\Rightarrow QR=\frac{PQ}{\sqrt[]{3}}=\sqrt[]{2}$

ar (PQR) = $\frac{1}{2}\sqrt[]{6}\cdot \sqrt[]{2}=\sqrt[]{3}$  

$\Delta =4\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill \alpha \hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill \alpha \hfill & \hfill 1\hfill \end{array}\right|=4$

$\Rightarrow \alpha =\pm 8$

$\left(\alpha ,-\alpha \right),\left(-\alpha ,\alpha \right)and\left({\alpha }^2,\beta \right)$ are collinear.

$\therefore \left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill -\alpha \hfill & \hfill 1\hfill \\ \hfill -\alpha \hfill & \hfill \alpha \hfill & \hfill 1\hfill \\ \hfill {\alpha }^2\hfill & \hfill \beta \hfill & \hfill 1\hfill \end{array}\right|=0\Rightarrow \beta =-64$

$\begin{array}{cc}\hfill 2x-3y=\gamma +5\hfill & \hfill \alpha x+5y=\left(\beta +1\right)\hfill \end{array}\}infinitelymanysolution$

$\therefore \frac{2}{\alpha }=\frac{-3}{5}=\left(\frac{\gamma +5}{\beta +1}\right)$

(i) $\frac{2}{\alpha }=\frac{-3}{5}=\frac{\gamma +5}{\beta +1}$

$\alpha =\frac{5*2}{-3}$ 5x + 25 = -3β - 3

^{$5\gamma +3\beta =-28$}

^{$\left|9\alpha +5\gamma +3\beta \right|=\left|9*\frac{-10}{3}-28\right|$}

= ^{$\left|-30-28\right|=58$}

Sum of all elements of $A\cap B=2$  [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

$=2\left[\frac{100*101}{2}-3\left(\frac{33*34}{2}\right)-5\left(\frac{20*21}{2}\right)+15\left(\frac{6*7}{2}\right)\right]$

= 10100 – 3366 – 2100 + 630

= 5264

Kindly consider the following figure

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Lettheverticesofthe\Delta ABCbe\\ A\left(x_1,y_1,z_1\right),B\left(x_2,y_2,z_2\right)andC\left(x_3,y_3,z_3\right)\\ Disthemid-pointofAB\\ \therefore 1=\frac{x_1+x_2}{2}\Rightarrow x_1+x_2=2\dots \left(i\right)\\ 2=\frac{y_1+y_2}{2}\Rightarrow y_1+y_2=4\dots \left(ii\right)\\ and-3=\frac{z_1+z_2}{2}\Rightarrow z_1+z_2=-6\dots \left(iii\right)\\ Eisthemid-pointofBC\\ \therefore 3=\frac{x_2+x_3}{2}\Rightarrow x_2+x_3=6\dots \left(iv\right)\\ 0=\frac{y_2+y_3}{2}\Rightarrow y_2+y_3=0\dots \left(v\right)\\ and1=\frac{z_2+z_3}{2}\Rightarrow z_2+z_3=2\dots \left(vi\right)\\ Fisthemid-pointofAC\\ \therefore -1=\frac{x_1+x_3}{2}\Rightarrow x_1+x_3=-2\dots \left(vii\right)\\ 1=\frac{y_1+y_3}{2}\Rightarrow y_1+y_3=2\dots \left(viii\right)\\ and-4=\frac{z_1+z_3}{2}\Rightarrow z_1+z_3=-8\dots \left(ix\right)\\ Addingeq.\left(i\right),\left(iv\right)and\left(vii\right)weget\\ 2\left(x_1+x_2+x_3\right)=2+6-2\\ \therefore x_1+x_2+x_3=3\dots \left(x\right)\\ Addingeq.\left(ii\right),\left(v\right)and\left(viii\right)weget\\ 2\left(y_1+y_2+y_3\right)=4+0+2\\ \therefore y_1+y_2+y_3=3\dots \left(xi\right)\\ Addingeq.\left(iii\right),\left(vi\right)and\left(ix\right)weget\\ 2\left(z_1+z_2+z_3\right)=-6+2-8\\ \therefore z_1+z_2+z_3=-6\dots \left(xii\right)\\ Subtracting\left(i\right)fromeq.\left(x\right)weget\\ x_3=3-2=1\Rightarrow x_3=1\\ Subtracting\left(iv\right)fromeq.\left(x\right)weget\\ x_1=3-6=-3\Rightarrow x_1=-3\\ Subtracting\left(vii\right)fromeq.\left(x\right)weget\\ x_2=3-\left(-2\right)=5\Rightarrow x_2=5\end{array}$