Question

Show that the following poionts are collinear
(i) $A(3,2,-4),B(9,8,-10),C(-2,-3,1)$
(ii) $P(4,5,2),Q(3,2,4),R(5,8,0)$

Answer 1

$\left(i\right)A(3,2,-4),B(9,8,-10),C(-2,-3,1)$

$\therefore$ For $A,B,C$ to be collinear,

$\lambda \overrightarrow{a}+\mu \overrightarrow{b}+\delta \overrightarrow{c}=0$ where $\lambda ,\mu ,\delta \ne 0$

$\therefore (3\hat{i}+2\hat{j}-4\hat{k})\lambda +(9\hat{i}+8\hat{j}-10\hat{k})\mu +(-2\hat{i}-3\hat{j}+\hat{k})\delta =0(3\lambda +9\mu -2\delta )\hat{i}+(2\lambda +8\mu -3\delta )\hat{j}+(-4\lambda -10\mu +1\delta )\hat{k}=0\therefore 3\lambda +9\mu -2\delta =02\lambda +8\mu -3\delta =0-4\lambda -10\mu +1\delta =0$

Solving by cramer's Rule,

$\left|\begin{array}{ccc}3& 9& -2\\ 2& 8& -3\\ -4& -10& 1\end{array}\right|=0$

$\therefore$ unique solutions for $\mu ,\lambda ,\delta$ exist and scalars are non zero.

$\therefore$ $A,B,C$ are collinear.

$\left(ii\right)P(4,5,2),Q(3,2,4),R(5,8,0)$

Similarly,

$\left|\begin{array}{ccc}4& 5& 2\\ 3& 2& 4\\ 5& 8& 0\end{array}\right|=0$

unique solutions exists and all the scalars are non zero.

$\therefore$ $P,Q,R$ are collinear.