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 (a_{1}, a_{2}, a_{3}), 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 = (a_{1}, a_{2}, a_{3}), B = (b_{1}, b_{2}, b_{3}), C = (c_{1}, c_{2}, c_{3}) 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 (a_{1}, a_{2}, a_{3}), 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 = (a_{1}, a_{2}, a_{3}), B = (b_{1}, b_{2}, b_{3}), C = (c_{1}, c_{2}, c_{3}) 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 (a_{1}, a_{2}, a_{3}), 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 = (a_{1}, a_{2}, a_{3}), B = (b_{1}, b_{2}, b_{3}), C = (c_{1}, c_{2}, c_{3}) 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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