Vectors parallel to the same plane, or lie on the same plane are called coplanar vectors

The three vectors are coplanar if one of them is expressible as a linear combination of the other two.

Given that

Let

Comparing the coefficients of , and , we get

1 = 2x + y …(1)

2 = x + y …(2)

3 = 3x + y …(3)

Solving equation (1) and (2), we get

⇒ x = –1

Substitute x = –1 in equation (2), we get

2 = x + y

⇒ 2 = –1 + y

⇒ y = 2 + 1

⇒ y = 3

Put x = –1 and y = 3 in equation (3), we get

3 = 3x + y

⇒ 3 = 3(–1) + 3

⇒ 3 = –3 + 3

⇒ 3 ≠ 0

∴ L.H.S ≠ R.H.S

⇒ The value of x and y doesn’t satisfy equation (3).

Thus, , and are not coplanar.

Let be depicted as,

…(*)

Substitute the value of , , and .

Comparing the coefficients in , and , we get

2 = x + 2y + z …(1)

–1 = 2x + y + z …(2)

–3 = 3x + 3y + z …(3)

From equation (1),

2 = x + 2y + z

⇒ z = 2 – x – 2y …(4)

Putting the value of z from equation (4) in equations (2) & (3), we get

From equation (2),

–1 = 2x + y + z

⇒ –1 = 2x + y + (2 – x – 2y)

⇒ –1 = 2x + y + 2 – x – 2y

⇒ 2x – x + y – 2y = –1 – 2

⇒ x – y = –3 …(5)

From equation (3),

–3 = 3x + 3y + z

⇒ –3 = 3x + 3y + (2 – x – 2y)

⇒ –3 = 3x + 3y + 2 – x – 2y

⇒ 3x – x + 3y – 2y = –3 – 2

⇒ 2x + y = –5 …(6)

Solving equation (5) and (6), we have

⇒ 3x = –8

Substituting in equation (5), we get

x – y = –3

⇒ –8 – 3y = –3 × 3

⇒ –8 – 3y = –9

⇒ 3y = 9 – 8

⇒ 3y = 1

Now, substitute and in z = 2 – x – 2y, we get

⇒ z = 4

We have got , and z = 4.

Put these values in equation (*), we get

Thus, we have found the relation.