Two-system-of-rectangular-axes-have-the-same-origin-If-a-plane-cuts-them-at-distance-a-b-c-and-p-q-r-respectively-then-prove-with-the-help-of-an-appropriate-diagram-that-1-a-2-1-b-2-

Question Number 80146 by Khyati last updated on 31/Jan/20

T w o s y s t e m o f r e c t a n g u l a r a x e s h a v e

t h e s a m e o r i g i n . I f a p l a n e c u t s t h e m

a t d i s t a n c e a, b, c a n d p, q, r

r e s p e c t i v e l y, t h e n p r o v e w i t h t h e h e l p

o f a n a p p r o p r i a t e d i a g r a m t h a t :

\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}}

Answered by mr W last updated on 31/Jan/20

Commented by mr W last updated on 31/Jan/20

s a y t h e o r i g i n i s p o i n t O a n d t h e p l a n e

i s P .

t h e d i s t a n c e f r o m O t o t h e p l a n e i s h,

w h i c h i s c o n s t a n t .

{l e t}^{'} s l o o k a t a n a r b i t r a r y r e c t a n g u l a r

a x i s s y s t e m w i t h o r i g i n a t O a n d t h e

a x e s i n t e r s e c t t h e p l a n e a t A, B, C

w i t h d i s t a n c e s a, b, c t o t h e o r i g i n .

O, A, B, C b u i l d a t e t r a h e d r o n .

A B = \sqrt{a^{2} + b^{2}}

B C = \sqrt{b^{2} + c^{2}}

C A = \sqrt{c^{2} + a^{2}}

t h e v o l u m e o f t h e t e t r a h e d r o n i s V .

V = \frac{1}{3} \times \frac{a b}{2} \times c = \frac{a b c}{6}

o r

V = \frac{1}{3} \times Δ_{A B C} \times h w i t h Δ_{A B C} = a r e a A B C

Δ_{A B C} = \frac{1}{2} \times A B \times A C \times \sin ∠ B A C

Δ_{A B C} = \frac{\sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})}}{2} \times \sin ∠ B A C

{B C}^{2} = {A B}^{2} + {A C}^{2} - 2 \times A B \times A C \times \cos ∠ B A C

b^{2} + c^{2} = a^{2} + b^{2} + c^{2} + a^{2} - 2 \sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})} \times \cos ∠ B A C

a^{2} = \sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})} \times \cos ∠ B A C

\Rightarrow \cos ∠ B A C = \frac{a^{2}}{\sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})}}

\Rightarrow \sin ∠ B A C = \frac{\sqrt{(a^{2} + b^{2}) (c^{2} + a^{2}) - a^{4}}}{\sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})}}

\Rightarrow \sin ∠ B A C = \frac{\sqrt{b^{2} c^{2} + a^{2} (b^{2} + c^{2})}}{\sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})}}

Δ_{A B C} = \frac{\sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})}}{2} \times \frac{\sqrt{b^{2} c^{2} + a^{2} (b^{2} + c^{2})}}{\sqrt{(a^{2} + b^{2}) (c^{2} + a^{2})}}

\Rightarrow Δ_{A B C} = \frac{\sqrt{b^{2} c^{2} + a^{2} (b^{2} + c^{2})}}{2}

V = \frac{1}{3} \times \frac{\sqrt{b^{2} c^{2} + a^{2} (b^{2} + c^{2})}}{2} \times h = \frac{h \sqrt{b^{2} c^{2} + a^{2} (b^{2} + c^{2})}}{6}

\Rightarrow \frac{h \sqrt{b^{2} c^{2} + a^{2} (b^{2} + c^{2})}}{6} = \frac{a b c}{6}

\Rightarrow h^{2} [b^{2} c^{2} + a^{2} (b^{2} + c^{2})] = a^{2} b^{2} c^{2}

\Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{h^{2}}

s i n c e h i s c o n s t a n t, w e g e t a l s o f o r

a n o t h e r s y s t e m :

\Rightarrow \frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}} = \frac{1}{h^{2}}

t h e r e f o r e :

\Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}}

Commented by mr W last updated on 31/Jan/20

A n o t h e r w a y :

e q n . o f p l a n e i n s y s t e m 1 :

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0

d i s t a n c e f r o m o r i g i n t o p l a n e i s

h = \frac{∣ \frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1 ∣}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}}

s i m i l a r l y i n s y s t e m 2 t h e

d i s t a n c e f r o m o r i g i n t o p l a n e i s

h = \frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{p^{2}} + \frac{1}{r^{2}}}}

s i n c e h i s i n b o t h s y s t e m s t h e s a m e,

\frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = \frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{p^{2}} + \frac{1}{r^{2}}}}

\Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}} + \frac{1}{p^{2}} + \frac{1}{r^{2}}

Commented by TawaTawa last updated on 31/Jan/20

Weldon sir, God bless you