Show that the vectors given
by and are non-coplanar. Express vector as a linear combination of the vectors and

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.