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A plane in R³ contains the point (-r, 0, 1) and has the vector (1,4,3r) as a normal, where r is a real number. (a) Determine the equation of the plane in terms of r. [3] 1 The equation of the plane is x. n = p.n. where x = (x, y, z), p = (-1, 0, 1) and n = (1,4,3) So the equation of the plane is (x, z) · (1,4,5) = (·<, 0, 1) · (1,4.3.) A 1x + ty- 31₂ = x + 1 +0x x + 1 x 3 r Hy 1x + 4y - 3 r₂ = -8 +0 + 3 x = 2+ x+4y-3c₂ So x + 4y - 3r₂ = 2r 4

A plane in R³ contains the point (-r, 0, 1) and has the vector (1,4,3r) as a normal, where r is a real number. (a) Determine the equation of the plane in terms of r. [3] 1 The equation of the plane is x. n = p.n. where x = (x, y, z), p = (-1, 0, 1) and n = (1,4,3) So the equation of the plane is (x, z) · (1,4,5) = (·<, 0, 1) · (1,4.3.) A 1x + ty- 31₂ = x + 1 +0x x + 1 x 3 r Hy 1x + 4y - 3 r₂ = -8 +0 + 3 x = 2+ x+4y-3c₂ So x + 4y - 3r₂ = 2r 4

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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Just wondering if my answer is correct.  Any explantion of why it isnt or any critique if it is would be helpful thanks .

A plane in R³ contains the point (-r, 0, 1) and has the vector (1,4,3r) as a normal, where r is a real number. (a) Determine the equation of the plane in terms of r. [3]
Transcribed Image Text:A plane in R³ contains the point (-r, 0, 1) and has the vector (1,4,3r) as a normal, where r is a real number. (a) Determine the equation of the plane in terms of r. [3]
1 The equation of the plane is x. n = p.n. where x = (x, y, z), p = (-1, 0, 1) and n = (1,4,3) So the equation of the plane is (x, z) · (1,4,5) = (·<, 0, 1) · (1,4.3.) A 1x + ty- 31₂ = x + 1 +0x x + 1 x 3 r Hy 1x + 4y - 3 r₂ = -8 +0 + 3 x = 2+ x+4y-3c₂ So x + 4y - 3r₂ = 2r 4
Transcribed Image Text:1 The equation of the plane is x. n = p.n. where x = (x, y, z), p = (-1, 0, 1) and n = (1,4,3) So the equation of the plane is (x, z) · (1,4,5) = (·<, 0, 1) · (1,4.3.) A 1x + ty- 31₂ = x + 1 +0x x + 1 x 3 r Hy 1x + 4y - 3 r₂ = -8 +0 + 3 x = 2+ x+4y-3c₂ So x + 4y - 3r₂ = 2r 4
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