Question

Find the equation of the plane through (3,4,5)and parallael to the plane 2x+3y-z=0

Answer 1

Our plane is parallel to the plane,

$\longrightarrow2x+3y-z=0$

to which it is clear that the vector $\left<2,\ 3,\ -1\right>$ is perpendicular. So this vector is also perpendicular to our plane, since the planes are parallel.

Our plane passes through $(3,\ 4,\ 5).$ Let $(x,\ y,\ z)$ be a point on our plane so that the vector $\left<x-3,\ y-4,\ z-5\right>$ lie on our plane.

Now the two vectors are perpendicular to each other, so their dot product is zero, i.e.,

$\longrightarrow\left<x-3,\ y-4,\ z-5\right>\cdot\left<2,\ 3,\ -1\right>=0$

$\longrightarrow2(x-3)+3(y-4)-(z-5)=0$

$\longrightarrow\underline{\underline{2x+3y-z=13}}$

This is the equation of our plane.

Answer 2

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Our plane is parallel to the plane,

⟶2x+3y−z=0

to which it is clear that the vector ⟨2, 3, −1⟩ is perpendicular. So this vector is also perpendicular to our plane, since the planes are parallel.

Our plane passes through (3, 4, 5). Let (x, y, z) be a point on our plane so that the vector ⟨x−3, y−4, z−5⟩ lie on our plane.

Now the two vectors are perpendicular to each other, so their dot product is zero, i.e.,

⟶⟨x−3, y−4, z−5⟩⋅⟨2, 3, −1⟩=0

⟶2(x−3)+3(y−4)−(z−5)=0

2x+3y−z=13

This is the equation of our plane.