Question

Using vectors, find the value of $\lambda$ such that the points $(\lambda ,-10,3),(1,-1,3)$ and $(3,5,3)$ are collinear.

Answer 1

Let the given points be

$\begin{array}{c}A\left(\lambda ,-10,3\right),B\left(1,-1,3\right)andC\left(3,5,3\right)\\ \overrightarrow{AB}=\left(\hat{i}-\hat{j}+3\hat{k}\right)-\left(\lambda \hat{i}-10\hat{j}+3\hat{k}\right)=\left(1-\lambda \right)\hat{i}+9\hat{j}\\ \overrightarrow{AC}=\left(3\hat{i}+5\hat{j}+3\hat{k}\right)-\left(\lambda \hat{i}-10\hat{j}+3\hat{k}\right)=\left(3-\lambda \right)\hat{i}+15\hat{j}\\ IfthepointsA,B,Carecollinear,then\overrightarrow{AB}=K\overrightarrow{AC}forsomescalarK\\ \left(1-\lambda \right)\hat{i}+9\hat{j}=K\left[\left(3-\lambda \right)\hat{i}+15\hat{j}\right]\\ 1-\lambda =K\left(3-\lambda \right)and9=15K\\ 1-\lambda =\frac{3}{5}\left(3-\lambda \right)\\ 5-5\lambda =9-3\lambda \\ 2\lambda =-4\\ \lambda =-2\end{array}$