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Question Number 45809 by rahul 19 last updated on 17/Oct/18

Commented by rahul 19 last updated on 17/Oct/18

$C a l c u l a t i o n e r r o r . . . .$
Commented by MJS last updated on 17/Oct/18

$how you got P ? {it}^{'} s not in the plain at all! ?$ $5 x - 4 y - z = 1$ $5 \times 4 - 4 \times 11 - 7 = - 31$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18
	$The eqn of st line QR is ((x−2)/(1−2))=((y−3)/(−1−3))=((z−5)/(4−5)) ((x−2)/1)=((y−3)/4)=((z−5)/1)=r (r+2,4r+3,r+5) is the point P...interdection of st line and plane... 5(r+2)−4(4r+3)−1(r+5)=1 5r+10−16r−12−r−5−1=0 −12r−8=0 r=((−2)/3) point P(((−2)/3)+2 ,((−8)/3)+3 ,((−2)/3)+5) point P((4/3),(1/3),((13)/3)) let point S(α,β,γ) poinS lies on QR so ((α−2)/1)=((β−3)/4)=((γ−5)/1)=k (α,β,γ)=(k+2,4k+3,k+5) direction ratio of TS that means betwedn (k+2,4k+3,k+5)and (2,1,4) is (k,4k+2,k+1) TS⊥QR atS so 1(k)+4(4k+2)+1(k+1)=0 18k+8+1=0 k=((−1)/2) so point S is{((−1)/2)+2,4(((−1)/2))+3,((−1)/2)+5} ={(3/2),1,(9/2)}←it is point S {(4/3),(1/3),((13)/3)}←it is point P PS=(√(((4/3)−(3/2))^2 +((1/3)−1)^2 +(((13)/3)−(9/2))^2 )) =(√((((−1)/6))^2 +(((−2)/3))^2 +(((−1)/6))^2 )) =(√((1/(36))+(4/9)+(1/(36)))) =(√((1+16+1)/(36))) =(√((18)/(36))) =(1/((√2) ))ans$
	$T h e e q n o f s t l i n e Q R i s$ $\frac{x - 2}{1 - 2} = \frac{y - 3}{- 1 - 3} = \frac{z - 5}{4 - 5}$ $\frac{x - 2}{1} = \frac{y - 3}{4} = \frac{z - 5}{1} = r$ $(r + 2, 4 r + 3, r + 5) i s t h e p o i n t P . . . i n t e r d e c t i o n$ $o f s t l i n e a n d p l a n e . . .$ $5 (r + 2) - 4 (4 r + 3) - 1 (r + 5) = 1$ $5 r + 10 - 16 r - 12 - r - 5 - 1 = 0$ $- 12 r - 8 = 0 r = \frac{- 2}{3}$ $p o i n t P (\frac{- 2}{3} + 2, \frac{- 8}{3} + 3, \frac{- 2}{3} + 5)$ $p o i n t P (\frac{4}{3}, \frac{1}{3}, \frac{13}{3})$ $l e t p o i n t S (α, β, γ)$ $p o i n S l i e s o n Q R s o \frac{α - 2}{1} = \frac{β - 3}{4} = \frac{γ - 5}{1} = k$ $(α, β, γ) = (k + 2, 4 k + 3, k + 5)$ $d i r e c t i o n r a t i o o f T S t h a t m e a n s b e t w e d n$ $(k + 2, 4 k + 3, k + 5) a n d (2, 1, 4) i s$ $(k, 4 k + 2, k + 1)$ $T S ⊥ Q R a t S s o$ $1 (k) + 4 (4 k + 2) + 1 (k + 1) = 0$ $18 k + 8 + 1 = 0$ $k = \frac{- 1}{2}$ $s o p o i n t S i s {\frac{- 1}{2} + 2, 4 (\frac{- 1}{2}) + 3, \frac{- 1}{2} + 5}$ $= {\frac{3}{2}, 1, \frac{9}{2}} \leftarrow i t i s p o i n t S$ ${\frac{4}{3}, \frac{1}{3}, \frac{13}{3}} \leftarrow i t i s p o i n t P$ $P S = \sqrt{{(\frac{4}{3} - \frac{3}{2})}^{2} + {(\frac{1}{3} - 1)}^{2} + {(\frac{13}{3} - \frac{9}{2})}^{2}}$ $= \sqrt{{(\frac{- 1}{6})}^{2} + {(\frac{- 2}{3})}^{2} + {(\frac{- 1}{6})}^{2}}$ $= \sqrt{\frac{1}{36} + \frac{4}{9} + \frac{1}{36}}$ $= \sqrt{\frac{1 + 16 + 1}{36}} = \sqrt{\frac{18}{36}} = \frac{1}{\sqrt{2}} a n s$
	Commented by rahul 19 last updated on 17/Oct/18
	thanks sir!��☺️
	Answered by MJS last updated on 17/Oct/18

	$line Q R = l_{1} : X = Q + t (R - Q)$ $l_{1} : (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 2 \\ 3 \\ 5 \end{matrix}) - t (\begin{matrix} 1 \\ 4 \\ 1 \end{matrix})$ $plane p_{1} : 5 x - 4 y - z = 1$ $P = l_{1} \cap p_{1}$ $5 (2 - t) - 4 (3 - 4 t) - (5 - t) = 1$ $t = \frac{2}{3} \Rightarrow P = (\begin{matrix} \frac{4}{3} \\ \frac{1}{3} \\ \frac{13}{3} \end{matrix})$ $plane ⊥ l_{1} = p_{2} : x + 4 y + z = c$ $[the direction of l_{1} is the perpendicular of p_{2}]$ $T \in p_{2} \Rightarrow c = 10$ $p_{2} : x + 4 y + z = 10$ $S = l_{1} \cap p_{2}$ $(2 - t) + 4 (3 - 4 t) + (5 - t) = 10$ $t = \frac{1}{2} \Rightarrow S = (\begin{matrix} \frac{3}{2} \\ 1 \\ \frac{9}{2} \end{matrix})$ $∣ P S ∣= \sqrt{{(\frac{3}{2} - \frac{4}{3})}^{2} + {(1 - \frac{1}{3})}^{2} + {(\frac{9}{2} - \frac{13}{3})}^{2}} = \frac{\sqrt{2}}{2}$
	Commented by rahul 19 last updated on 17/Oct/18
	thanks sir!��☺️
	Commented by MJS last updated on 17/Oct/18

	$\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
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