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.
