Let the system of linear equations– x + 2y – 9z = 7– x + 3y + 7z = 9– 2x + y + 5z = 8– 3x + y + 13z = λhas a unique solutions x = α, y = β, z = γ. Then the distance of the point (α, β, γ) from the plane 2x

Engineering

Mathematics

Solving System of Linear Equation Cramers Rule

Distance from a Plane

Question

Let the system of linear equations

– x + 2y – 9z = 7
– x + 3y + 7z = 9
– 2x + y + 5z = 8
– 3x + y + 13z = λ
has a unique solutions x = α, y = β, z = γ. Then the distance of the point (α, β, γ) from the plane 2x – 2y + z = λ is

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– x + 2y – 9z = 7 ……….(i)
– x + 3y + 7z = 9 ………(ii)
– 2x + y + 5z = 8 ……….(iii)
Solve (i), (ii) & (iii) for x, y, z
x = – 3, y = 2, z = 0
x = – 3, β = 2, z = 0
⇒ λ = 11
Now, Distance of point (– 3, 2, 0) from plane:

$= \frac{| - 6 - 4 - 11 |}{\sqrt{4 + 4 + 1}} = \frac{21}{3} = 7$