Three Dimensional Geometry

ai+aj+ck,  i+k and ci+cj+bk are co-planar,
|a c; 1 0 1; c b| = 0
a (0-c) - a (b-c) + c (c-0) = 0
-ac - ab + AC + c² = 0
c² = ab
c = √ab

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

=> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21*\frac{34}{7}=102$

Now equation of line OA be

$\frac{x-1}{4}=\frac{y-3}{-5}=\frac{z-5}{2}=\lambda$

direction cosines of plane are 4, -5, 2

Equation of any point on OA be

$O\left(4\lambda +1,-5\lambda +3,2\lambda +5\right)$

Since O lies on given plane so

$4\left(4\lambda +1\right)-5\left(-5\lambda +3\right)+2\left(2\lambda +5\right)=8$

So, O (9/5,2,27/5). Hence by mid-point formula

B $\left(\frac{13}{5},1,\frac{29}{5}\right)\Rightarrow 5\left(\alpha +\beta +\gamma \right)=47$

Let $A\left(-2,-21,29\right),B\left(-1,-16,23\right),P\left(\lambda ,2,1\right),Q\left(4,-2,2\right)$

Given $\overrightarrow{AB}\perp \overrightarrow{PQ}$

$\therefore \overrightarrow{AB}.\overrightarrow{PQ}=0$

$\left(\widehat{i}+5\widehat{j}-6\widehat{k}\right).\left(\left(4-\lambda \right)\widehat{i}-4\widehat{j}+\widehat{k}\right)=0$

$4-\lambda -20-6=0$

$\lambda =-22$

$\therefore {\left(\frac{\lambda }{11}\right)}^2+\left(\frac{-4\lambda }{11}\right)-4=4+8-4=8$

$P_1:3x+15+21z=9$           

P₂ : x – 3y – z = 5

P₃ : 2x + 10y + 14z = 5

Ratio of the direction cosines of P₁ and P₂

$\frac{3}{2}=\frac{15}{10}=\frac{21}{14}$

Hence, P₁ and P₃ are parallel.

All the points A (1, 5, 35), B (7, 5,5),  $C(1,\lambda ,7),D(2\lambda ,1,2)$ are coplanar. Hence

$\overrightarrow{AB}*\overrightarrow{AC}.\overrightarrow{AD}=\left|\begin{array}{ccc}\hfill 6\hfill & \hfill 0\hfill & \hfill 30\hfill \\ \hfill 0\hfill & \hfill \lambda -5\hfill & \hfill 28\hfill \\ \hfill 2\lambda -1\hfill & \hfill -4\hfill & \hfill 33\hfill \end{array}\right|=0$

$\Rightarrow 5{\lambda }^2-44\lambda +95=0$

$\Rightarrow sumofroots=\frac{44}{5}$

${\mathcal{l}}_1:\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t$

${\mathcal{l}}_2:\frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s$

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=\left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}:\overrightarrow{r}=\overrightarrow{0}+\lambda \left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}\&{\mathcal{l}}_1\to$

Point of intersection $\mathcal{l}\&{\mathcal{l}}_1$ is P (2, 3, 2)

Point Q on ${\mathcal{l}}_2$ is (3 + 2s, 3 + 2s, 2 + s).

  $\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r     . (i)

b = -2 + 10r           . (ii)

a = 18r – 2              . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = -3

$\therefore$   a = 14 + 14 (-3) = -56 and b = -2 -30 = -32

$\left|a+b\right|\left|-56-32\right|=88$

$\overrightarrow{O}P=\left(2,-1,1\right)$

Normal vector to the plane

$=\overrightarrow{A}B*A\overrightarrow{C}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(3,-1,1\right)=\overrightarrow{n}$

Projection of $\overrightarrow{O}Pon\overrightarrow{n}is\left|\frac{\overrightarrow{O}P.\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\right|$

PN = $\frac{6+1+1}{\sqrt[]{11}}=\frac{8}{\text{exception 25:}}$

$\frac{\text{exception 25:}}{}$Projection of OP on plane = ON =  $\sqrt[]{O{P}^2-P{N}^2}=\sqrt[]{6-\frac{64}{11}}=\sqrt[]{\frac{2}{11}}$