Question

Prove that

(i) the three points A(1, 4, 2), B(3, 2, 4) and C(5, 0, 6) are collinear,

(ii) the four points P(2, ;1, 1), Q(1, 3, ;2), R((2, 1, ;3) and S(3, 2, 0) are coplanar.

Answer 1

(i)

$A(1, 4, 2), B(3, 2, 4) and C(5, 0, 6)$

The three point A,B,C are collinear if the direction ratio of AB,BC,AC are proportional.

Now,

$A(1, 4, 2), B(3, 2, 4)$

Direction Ratio $=3-1,2-4,4-2$

$= (2,-2,2)$

$so, a_1=2,b_1=-2,c_1=2$

Now,

$B(3, 2, 4) , C(5, 0, 6)$

Direction Ratio$=$ $( 5-3,0-2,6-4)$

$=(2,-2,2)$

So,

$a_2=2,b_2=-2,c_2=2$

ratios,

$a_2/a_1=2/2=1$

$b_2/b_1=-2/-2=1$

$c_2/c_1=2/2=1$

Therefore A,B,C are collinear

Hence proved.

(ii)

$P(2, 1, 1), Q(1, 3, 2), R(2, 1, 3) and S(3, 2, 0)$

So, let

$\overrightarrow{OP}=2{\hat{i} +\hat{j}}+\hat{k}$

$\overrightarrow{OQ}={\hat{i} +3\hat{j}}+2\hat{k}$

$\overrightarrow{OR}=2{\hat{i} +\hat{j}}+3\hat{k}$

$\overrightarrow{OS}={\hat{i} +2\hat{j}}$

now,

$\overrightarrow{PQ}=OQ-OP={\hat{i} +2\hat{j}}+\hat{k}$

$\overrightarrow{PR}=OE-OP=2\hat{k}$

$\overrightarrow{PS}={\hat{i} +\hat{j}}-\hat{k}$

Solving aboove three equations we get

$|PQ,PR,PS|=0$

Hence the given four points are coplanar.

Hence proved .