Finding Points Parallel to a Plane

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SUMMARY

The discussion centers on determining the value of \( k \) for the plane equation \( 7x + 2y + kz = 5 \) such that the vector \( P = (1, 2, -1) \) is parallel to the plane. The solution involves using the normal vector \( \mathbf{n} = (7, 2, k) \) and applying the formula \( \mathbf{n} \cdot (\mathbf{x - x_0}) = 0 \). Through calculations, it is established that \( k = 6 \) to maintain parallelism with the given point.

PREREQUISITES
  • Understanding of vector mathematics and normal vectors
  • Familiarity with plane equations in three-dimensional space
  • Knowledge of the dot product and its geometric interpretation
  • Basic algebra for solving equations
NEXT STEPS
  • Study the properties of normal vectors in vector geometry
  • Learn how to derive equations of planes given points and normal vectors
  • Explore the concept of parallelism in three-dimensional geometry
  • Investigate applications of planes and vectors in physics and engineering
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Students and professionals in mathematics, physics, and engineering who are working with three-dimensional geometry and vector analysis will benefit from this discussion.

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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.$
 

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