NCERT Solutions
Class 12 Maths
Three Dimensional Geometry

Ex.11.3 Q.10
Find the vector equation of the plane passing through the intersection of the planes
r.(2i + 2j – 3k) = 7, r.(2i + 5j + 3k) = 9 and through the point (2, 1, 3).
Given, equations of plane are
r.(2i + 2j – 3k) = 7 ……….1
r.(2i + 5j + 3k) = 9 ……….2
The equation of any plane through the intersection of the planes given in equations 1 and 2 is given by,
[r.(2i + 2j – 3k) - 7] + λ[r.(2i + 5j + 3k) - 9] = 0, where λ є R
=> r.[(2i + 2j – 3k) + λ[r.(2i + 5j + 3k)] = 9λ + 7
=> r.[(2 + 2λ)i + (2 + 5λ)j + (3λ - 3)k] = 9λ + 7 ……….3
The plane passes through the point (2, 1, 3).
Therefore, its position vector is given by,
r = 2i + j + 3k
Substitute the value of r in the equation 3, we get
=> (2i + j + 3k).[(2 + 2λ)i + (2 + 5λ)j + (3λ - 3)k] = 9λ + 7
=> 2(2 + 2λ) + (2 + 5λ) + 3(3λ - 3) = 9λ + 7
=> 4 + 4λ + 2 + 5λ + 9λ - 9 = 9λ + 7
=> 9λ = 10
=> λ =
Substituting λ = in equation 3, we obtain
=>
=>
=>
=> r.(38i + 68j + 3k) = 153
This is the vector equation of the required plane.