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Show that the four points having position vectors \(6\hat i-7\hat j,\) \(16\hat i-19j-4\hat k\), \(2\hat i-5\hat j+10\hat k\)  are coplanar..

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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.

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