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.

Let the four points be denoted be P, Q, R and S for \(6\hat i-7\hat j\), \(16\hat i-19\hat j-4\hat k\), and \(2\hat i-5\hat j+10\hat k\) respectively such that we can say,

Let us find \(\vec{PQ},\vec{PR}\) and \(\vec {PS}.\)

**So,**

**\(\vec{PQ}\) **= Position vector of Q - Position vector of P

Also,

\(\vec {PR}\) = Position vector of R - Position vector of P

And,

\(\vec {PS}\) = Position vector of S - Position vector of P

Now, we need to show a relation between **\(\vec{PQ}\)**, **\(\vec{PR}\) **and \(\vec {PS}\)

**So,**

Comparing coefficients of \(\hat i,\hat j\) and \(\vec k\)

**We get**

–6x – 4y = 10 …**(i)**

10x + 2y = –12 …**(ii)**

–6x + 10y = –4 …**(iii)**

For solving equation **(i)** and **(ii)** for x and y, multiply equation **(ii)** by 2.

10x + 2y = –12 [× 2

⇒ 20x + 4y = –24 …**(iv)**

Solving equations **(iv) **and **(i)**,

**We get**

Put x = –1 in equation (i), we get

–6(–1) – 4y = 10

⇒ 6 – 4y = 10

⇒ –4y = 10 – 6

⇒ –4y = 4

⇒ y = –1

Substitute x = –1 and y = –1 in equation **(iii)**, we get

–6x + 10y = –4

⇒ –6(–1) + 10(–1) = –4

⇒ 6 – 10 = –4

⇒ –4 = –4

**∵, L.H.S = R.H.S**

⇒ The value of x and y satisfy equation **(iii)**

.**Thus, ** \(6\hat i-7\hat j\), \(16\hat i-19\hat j-4\hat k\), and \(2\hat i-5\hat j+10\hat k\) are coplanar.