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equation of the the point (5-5/3) which plane 4y - 1+² 32 = 1 8+ X Find the The an

equation of the the point (5-5/3) which plane 4y - 1+² 32 = 1 8+ X Find the The an

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Find the equation of the plane through the point (5, -5, 3), which is parallel to the plane 4y - 6x + 3z = 1.
Transcribed Image Text:Find the equation of the plane through the point (5, -5, 3), which is parallel to the plane 4y - 6x + 3z = 1.
**Problem Statement:** Find the equation of the plane through the point \((5, -5, 3)\) which is parallel to the plane \(9y - 6t + 3z = 1\). **Explanation:** To find the equation of a plane parallel to a given one, you need to use the same normal vector. Therefore, the equation of the new plane will have the same coefficients for \(y\), \(t\), and \(z\) as the given plane. The given plane is in the form \(9y - 6t + 3z = 1\). This plane has the normal vector \( \vec{n} = (0, 9, -6, 3) \). Considering the new plane must pass through the point \((5, -5, 3)\), plug these coordinates into the plane equation: \(9(-5) - 6t + 3(3) = D\). Calculate \(D\) to find the constant for the new plane's equation. Then, substitute back into the general form to get the new equation of the plane.
Transcribed Image Text:**Problem Statement:** Find the equation of the plane through the point \((5, -5, 3)\) which is parallel to the plane \(9y - 6t + 3z = 1\). **Explanation:** To find the equation of a plane parallel to a given one, you need to use the same normal vector. Therefore, the equation of the new plane will have the same coefficients for \(y\), \(t\), and \(z\) as the given plane. The given plane is in the form \(9y - 6t + 3z = 1\). This plane has the normal vector \( \vec{n} = (0, 9, -6, 3) \). Considering the new plane must pass through the point \((5, -5, 3)\), plug these coordinates into the plane equation: \(9(-5) - 6t + 3(3) = D\). Calculate \(D\) to find the constant for the new plane's equation. Then, substitute back into the general form to get the new equation of the plane.
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