show-that-the-plane-x-2y-3z-d-0-is-perpendiculr-to-each-of-the-plane-is-2x-5y-4z-1-0-and-4x-7y-3z-2-0-

Question Number 8150 by trapti rathaur@ gmail.com last updated on 02/Oct/16

s h o w t h a t t h e p l a n e x + 2 y - 3 z + d = 0 i s p e r p e n d i c u l r t o e a c h o f

t h e p l a n e i s 2 x + 5 y + 4 z + 1 = 0 a n d 4 x + 7 y + 3 z + 2 = 0 .

Commented by 123456 last updated on 03/Oct/16

α : x + 2 y - 3 z + d = 0

β : 2 x + 5 y + 4 z + 1 = 0

γ : 4 x + 7 y + 3 z + 2 = 0

lets \vec{r}, \vec{s}, \vec{t} be the normal vector of

α, β, γ we have

\vec{r} = (1, 2, - 3)

\vec{s} = (2, 5, 4)

\vec{t} = (4, 7, 3)

if two plan are perp, their normal vector

are perp too

two vector are perp if their dot product

equal to 0

\vec{r} \cdot \vec{s} = (1, 2, - 3) \cdot (2, 5, 4)

= 1 \times 2 + 2 \times 5 - 3 \times 4

= 2 + 10 - 12

= 12 - 12

= 0

\vec{r} \cdot \vec{t} = (1, 2, - 3) \cdot (4, 7, 3)

= 1 \times 4 + 2 \times 7 - 3 \times 3

= 4 + 14 - 9

= 18 - 9

= 9 \neq 0

\vec{s} \cdot \vec{t} = (2, 5, 4) \cdot (4, 7, 3)

= 2 \times 4 + 5 \times 7 + 4 \times 3

= 8 + 35 + 12

= 55 \neq 0

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