A cube of unit edge length is held before a plane. Prove that the sum of the squares of the projected lengths of edges of the cube on the plane (irrespective of the orientation of the cube) is 8.


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Question Number 62121 by ajfour last updated on 16/Jun/19

$A c u b e o f u n i t e d g e l e n g t h$ $i s h e l d b e f o r e a p l a n e . P r o v e t h a t$ $t h e s u m o f t h e s q u a r e s o f t h e$ $p r o j e c t e d l e n g t h s o f e d g e s o f t h e c u b e$ $o n t h e p l a n e (i r r e s p e c t i v e o f$ $t h e o r i e n t a t i o n o f t h e c u b e) i s 8 .$
	Answered by mr W last updated on 16/Jun/19

	Commented by mr W last updated on 16/Jun/19

	$s a y O 1234567 i s t h e u n i t c u b e .$ $s a y A B C i s t h e p l a n e w i t h$ $A (a, 0, 0), B (0, b, 0) a n d C (0, 0, c)$ $e q n . o f t h e p l a n e i s$ $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ $\vec{n} = (\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$ $∣ n ∣= \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}$ $f o r t w o p o i n t s P (x_{1}, y_{1}, z_{1}) a n d Q (x_{2}, y_{2}, z_{2})$ $\vec{P Q} = (Δ x, Δ y, Δ z) w i t h$ $Δ x = x_{2} - x_{1}, Δ y = y_{2} - y_{1}, Δ z = z_{2} - z_{1}$ $∣ P Q ∣= \sqrt{{(Δ x)}^{2} + {(Δ y)}^{2} + {(Δ z)}^{2}}$ $\cos θ = \frac{\vec{P Q} . \vec{n}}{∣ P Q ∣ \times ∣ n ∣} = \frac{\frac{Δ x}{a} + \frac{Δ y}{b} + \frac{Δ z}{c}}{\sqrt{{(Δ x)}^{2} + {(Δ y)}^{2} + {(Δ z)}^{2}} \times \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}}$ $l e n g t h o f p r o j e c t i o n o f P Q o n p l a n e i s$ $∣ P^{'} Q^{'} ∣=∣ P Q ∣ \sin θ$ $\Rightarrow∣ P^{'} Q^{'} ∣^{2} =∣ P Q ∣^{2} \sin^{2} θ =∣ P Q ∣^{2} (1 - \cos^{2} θ)$ $\Rightarrow∣ P^{'} Q^{'} ∣^{2} = {(Δ x)}^{2} + {(Δ y)}^{2} + {(Δ z)}^{2} - \frac{{(\frac{Δ x}{a} + \frac{Δ y}{b} + \frac{Δ z}{c})}^{2}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}$ $n o w w e a p p l y t h i s f o r a l l 12 e d g e s o f t h e$ $u n i t c u b e .$ $e d g e O 1 = 32 = 45 = 76 : (4 e d g e s)$ $Δ x = 1, Δ y = 0, Δ z = 0$ $∣ O^{'} 1^{'} ∣^{2} = 1 - \frac{\frac{1}{a^{2}}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}$ $e d g e 12 = O 3 = 74 = 65 : (4 e d g e s)$ $Δ x = 0, Δ y = 1, Δ z = 0$ $∣ 1^{'} 2^{'} ∣^{2} = 1 - \frac{\frac{1}{b^{2}}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}$ $e d g e 25 = 34 = O 7 = 15 : (4 e d g e s)$ $Δ x = 0, Δ y = 0, Δ z = 1$ $∣ 2^{'} 5^{'} ∣^{2} = 1 - \frac{\frac{1}{c^{2}}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}$ $s u m f o r a l l e d g e s :$ $4 (1 - \frac{\frac{1}{a^{2}}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} + 1 - \frac{\frac{1}{b^{2}}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} + 1 - \frac{\frac{1}{c^{2}}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}})$ $= 4 (3 - \frac{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}})$ $= 4 (3 - 1)$ $= 8$ $t h e r e s u l t i s i n d e p e n d e n t f r o m t h e$ $o r i e n t a t i o n o f t h e p l a n e .$
	Commented by ajfour last updated on 16/Jun/19

	$T h a n k s S i r, v e r y b r a v e!$
	Answered by ajfour last updated on 16/Jun/19

	Commented by ajfour last updated on 16/Jun/19
	$let a^� =cos θi^� +sin θk^� b^� =cos φcos ψi^� +cos φsin ψj^� +sin φk^� as c^� = a^� ×b^� c^� =−cos φsin ψsin θi^� + (−sin φcos θ+cos φcos ψsin θ)j^� +cos φsin ψcos θk^� and as a^� .b^� =0 ⇒ cos θcos φcos ψ+sin θsin φ=0 ........(i) let required sum be S. S=4{cos^2 θ+cos^2 φcos^2 ψ+ cos^2 φsin^2 ψ+cos^2 φsin^2 ψsin^2 θ +(−sin φcos θ+cos φcos ψsin θ)^2 } ⇒ (S/4)=cos^2 θ+cos^2 φ+sin^2 θcos^2 φ +sin^2 φcos^2 θ −2sin φcos φsin θcos θcos ψ Now using (i) (S/4)=cos^2 θ+cos^2 φ+sin^2 θcos^2 φ +sin^2 φcos^2 θ+2sin^2 θsin^2 φ (S/4) = cos^2 θ+cos^2 φ +sin^2 θ−sin^2 θsin^2 φ sin^2 φ−sin^2 φsin^2 θ +2sin^2 θsin^2 φ ⇒ (S/4) = 2 or S=8 .$
	$l e t \bar{a} = \cos θ \hat{i} + \sin θ \hat{k}$ $\bar{b} = \cos ϕ \cos ψ \hat{i} + \cos ϕ \sin ψ \hat{j}$ $+ \sin ϕ \hat{k}$ $a s \bar{c} = \bar{a} \times \bar{b}$ $\bar{c} = - \cos ϕ \sin ψ \sin θ \hat{i} +$ $(- \sin ϕ \cos θ + \cos ϕ \cos ψ \sin θ) \hat{j}$ $+ \cos ϕ \sin ψ \cos θ \hat{k}$ $a n d a s \bar{a} . \bar{b} = 0$ $\Rightarrow \cos θ \cos ϕ \cos ψ + \sin θ \sin ϕ = 0$ $. . . . . . . . (i)$ $l e t r e q u i r e d s u m b e S .$ $S = 4 {\cos^{2} θ + \cos^{2} ϕ \cos^{2} ψ +$ $\cos^{2} ϕ \sin^{2} ψ + \cos^{2} ϕ \sin^{2} ψ \sin^{2} θ$ $+ {(- \sin ϕ \cos θ + \cos ϕ \cos ψ \sin θ)}^{2}}$ $\Rightarrow \frac{S}{4} = \cos^{2} θ + \cos^{2} ϕ + \sin^{2} θ \cos^{2} ϕ$ $+ \sin^{2} ϕ \cos^{2} θ$ $- 2 \sin ϕ \cos ϕ \sin θ \cos θ \cos ψ$ $N o w u s i n g (i)$ $\frac{S}{4} = \cos^{2} θ + \cos^{2} ϕ + \sin^{2} θ \cos^{2} ϕ$ $+ \sin^{2} ϕ \cos^{2} θ + 2 \sin^{2} θ \sin^{2} ϕ$ $\frac{S}{4} = \cos^{2} θ + \cos^{2} ϕ$ $+ \sin^{2} θ - \sin^{2} θ \sin^{2} ϕ$ $\sin^{2} ϕ - \sin^{2} ϕ \sin^{2} θ$ $+ 2 \sin^{2} θ \sin^{2} ϕ$ $\Rightarrow \frac{S}{4} = 2 o r S = 8 .$
	Commented by mr W last updated on 16/Jun/19

	$t h i s i s t h e w a y t o p r o v e i w i s h e d b u t$ $i c o u l d n o t c a r r y o u t . t h a n k s s i r!$
	Commented by ajfour last updated on 16/Jun/19

	$i^{'} m g l a d y o u l i k e i t, S i r . I t$ $n a t u r a l l y t u r n e d e a s i e r t h a n I$ $e x p e c t e d .$
	Commented by mr W last updated on 16/Jun/19

	$t h i s i s m y t r y u s i n g y o u r m e t h o d$ $w i t h o u t a s s u m i n g t h a t \vec{a} i s i n t h e$ $p l a n e x z .$ $\vec{a} = (a_{1}, a_{2}, a_{3}) w i t h a_{1}^{2} + a_{2}^{2} + a_{3}^{3} = 1$ $\vec{b} = (b_{1}, b_{2}, b_{3}) w i t h b_{1}^{2} + b_{2}^{2} + b_{3}^{3} = 1$ $\vec{a} . \vec{b} = 0$ $\Rightarrow a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = 0$ $\Rightarrow a_{1} b_{1} + a_{2} b_{2} = - a_{3} b_{3}$ $\vec{c} = \vec{a} \times \vec{b} = (a_{2} b_{3} - a_{3} b_{2}, a_{1} b_{3} - a_{3} b_{1}, a_{1} b_{2} - a_{2} b_{1})$ $\frac{S}{4} = a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} + c_{1}^{2} + c_{2}^{2}$ $= a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} + {(a_{2} b_{3} - a_{3} b_{2})}^{2} + {(a_{1} b_{3} - a_{3} b_{1})}^{2}$ $= a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} + a_{2}^{2} b_{3}^{2} - 2 a_{2} a_{3} b_{2} b_{3} + a_{3}^{2} b_{2}^{2} + a_{1}^{2} b_{3}^{2} - 2 a_{1} a_{3} b_{1} b_{3} + a_{3}^{2} b_{1}^{2}$ $= a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} + (a_{1}^{2} + a_{2}^{2}) b_{3}^{2} - 2 (a_{1} b_{1} + a_{2} b_{2}) a_{3} b_{3} + a_{3}^{2} (b_{1}^{2} + b_{2}^{2})$ $= a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} + (a_{1}^{2} + a_{2}^{2}) b_{3}^{2} + 2 a_{3}^{2} b_{3}^{2} + a_{3}^{2} (b_{1}^{2} + b_{2}^{2})$ $= a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} + (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) b_{3}^{2} + a_{3}^{2} (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$ $= a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + b_{1}^{2} + b_{2}^{2} + b_{3}^{2}$ $= 1 + 1$ $= 2$ $\Rightarrow S = 2 \times 4 = 8$
	Commented by ajfour last updated on 16/Jun/19

	$I t m u s t h a v e b e e n p o s s i b l e, t h i s$ $g e n e r a l w a y, & i t s n o t t h a t l e n g t h y$ $t o o, N i c e S i r .$
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