Answer :

(i) A(2, 2, -1), B(3, 4, 2) and C(7, 0, 6)


Given Points :


A = (2, 2, -1)


B = (3, 4, 2)


C = (7, 0, 6)


To Find : Equation of plane passing through points A, B & C


Formulae :


1) Position vectors :


If A is a point having co-ordinates (a1, a2, a3), then its position vector is given by,



2) Vector :


If A and B be two points with position vectors respectively, where




then,




3) Cross Product :


If are two vectors




then,



4) Dot Product :


If are two vectors




then,



5) Equation of Plane :


If A = (a1, a2, a3), B = (b1, b2, b3), C = (c1, c2, c3) are three non-collinear points,


Then, the vector equation of the plane passing through these points is



Where,



For given points,


A = (2, 2, -1)


B = (3, 4, 2)


C = (7, 0, 6)


Position vectors are given by,





Now, vectors are








Therefore,






Now,



= 40 + 16 + 12


= 68


………eq(1)


And



= 20x + 8y – 12z


………eq(2)


Vector equation of the plane passing through points A, B & C is



From eq(1) and eq(2)


20x + 8y – 12z = 68


This is 5x + 2y – 3z = 17 vector equation of required plane.


(ii) Given Points :


A = (0, -1, -1)


B = (4, 5, 1)


C = (3, 9, 4)


To Find : Equation of plane passing through points A, B & C


Formulae :


1) Position vectors :


If A is a point having co-ordinates (a1, a2, a3), then its position vector is given by,



2) Vector :


If A and B be two points with position vectors respectively, where




then,




3) Cross Product :


If are two vectors




then,



4) Dot Product :


If are two vectors




then,



5) Equation of Plane :


If A = (a1, a2, a3), B = (b1, b2, b3), C = (c1, c2, c3) are three non-collinear points,


Then, vector equation of the plane passing through these points is



Where,



For given points,


A = (0, -1, -1)


B = (4, 5, 1)


C = (3, 9, 4)


Position vectors are given by,





Now, vectors are








Therefore,






Now,



= 0 + 14 – 22


= - 8


………eq(1)


And



= 10x - 14y + 22z


………eq(2)


Vector equation of plane passing through points A, B & C is



From eq(1) and eq(2)


10x - 14y + 22z = - 8


This is 5x - 7y + 11z = - 4 vector equation of required plane


(iii) Given Points :


A = (-2, 6, -6)


B = (-3, 10, 9)


C = (-5, 0, -6)


To Find : Equation of plane passing through points A, B & C


Formulae :


1) Position vectors :


If A is a point having co-ordinates (a1, a2, a3), then its position vector is given by,



2) Vector :


If A and B be two points with position vectors respectively, where




then,




3) Cross Product :


If are two vectors




then,



4) Dot Product :


If are two vectors




then,



5) Equation of Plane :


If A = (a1, a2, a3), B = (b1, b2, b3), C = (c1, c2, c3) are three non-collinear points,


Then, vector equation of the plane passing through these points is



Where,



For given points,


A = (-2, 6, -6)


B = (-3, 10, 9)


C = (-5, 0, -6)


Position vectors are given by,





Now, vectors are








Therefore,






Now,



= - 180 - 270 – 108


= - 558


………eq(1)


And



= 90x - 45y + 18z


………eq(2)


Vector equation of plane passing through points A, B & C is



From eq(1) and eq(2)


90x - 45y + 18z = - 558


This is 10x - 5y + 2z = - 62 vector equation of required plane


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