MHB Finding Points Parallel to a Plane

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The vector P = (1,2,-1) is parallel to the plane defined by the equation 7x + 2y + kz = 5, leading to the determination of the value of k. To find k, the normal vector of the plane, represented as <7, 2, k>, is used in the equation for points parallel to the plane. The calculation shows that substituting the point (1,2,-1) into the plane equation yields 11 - k = 5. Solving this equation results in k being equal to 6. The discussion emphasizes the relationship between a point and a plane's normal vector in determining parallelism.
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If the vector $P = (1,2,-1)$ is parallel to the plane $7x+2y+kz = 5$, then what's the value of $k$?
 
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Do you know how to find the general formula for points parallel to a plane?
 
Deveno said:
Do you know how to find the general formula for points parallel to a plane?

So we find another plane parallel to this in which the point $(1,2,-1)$ lies?

I think $\mathbf{n} \cdot (\mathbf{x-x_{0}}) = 0$ so

$<7,2,k><x-1,y-2,z+1> = 0 \\
\implies 7x+2y+kz+k = 11 \\

\therefore 11-k = 5 \\

\implies k = 6.$
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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