# Show that the four points having position vectors 6i- 7j, 16i - 19j - 4k, 3j - 6k, 2i- 5j + 10k are coplanar..

247 views
in Vectors
closed

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..

+1 vote
by (29.5k points)
selected by

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.