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 points are $A(\lambda ,-10,3),B(1,-1,3),C(3,5,3)$

$\therefore \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$=(\hat{i}-\hat{j}+3\hat{k})-(k\hat{i}-10\hat{j}+3\hat{k})=(1-k)\hat{i}+9\hat{j}$

and $\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$

$=(3\hat{i}+5\hat{j}+3\hat{k})-(\hat{i}-\hat{j}+3\hat{k})=2\hat{i}+6\hat{j}$

If point $A,B,C$ are collinear then there exists some scalar $\lambda$ such that

$\Rightarrow \overrightarrow{AB}=\lambda \overrightarrow{BC}$

$\Rightarrow (1-k)\hat{i}+9\hat{j}=\lambda (2\hat{i}+6\hat{j})$

Comparing the coefficients we get

$1-k=2\lambda$ and $9=6\lambda$ or $\lambda =\frac{9}{6}=\frac{3}{2}$

$\Rightarrow 1-k=2\times \frac{3}{2}$

$\therefore 1-k=3$

$\therefore k=-2$