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Gelneren [198K]
2 years ago
12

Which of the following is an equation for the plane through the point (1, 1, −1) and parallel

Mathematics
1 answer:
Bas_tet [7]2 years ago
6 0

The given plane, x-y+z=3, has normal vector \vec n = \langle1,-1,1\rangle. Any plane parallel to this one has the same normal vector.

Let (x,y,z) be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is

\langle x,y,z\rangle - \langle 1,1,-1\rangle = \langle x-1, y-1, z+1 \rangle

and this is perpendicular to \vec n.

So the equation of the plane is

\langle x-1, y-1, z+1 \rangle \cdot \vec n = 0 \implies (x-1) - (y-1) + (z+1) = 0 \implies \boxed{x - y + z = -1}

or equivalently,

\boxed{-x + y - z = 1}

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2 years ago
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Mandarinka [93]

Answer:

The answer is

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Step-by-step explanation:

3+ – 5(4+ – 3v) can be written as

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Expand and simplify

That's

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Hope this helps you

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3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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