Question

Show that the points A (1, −2, −8), B (5, 0, −2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer 1

Given points $A\left(1,-2,-8\right),B\left(5,0,-2\right),C\left(11,3,7\right)$.
Therefore, $\overrightarrow{AB}=5\hat{i}+0\hat{j}-2\hat{k}-\hat{i}+2\hat{j}+8\hat{k}=4\hat{i}+2\hat{j}+6\hat{k}$
$\overrightarrow{BC}=11\hat{i}+3\hat{j}+7\hat{k}-5\hat{i}+2\hat{k}=6\hat{i}+3\hat{j}+9\hat{k}$
and, $\overrightarrow{AC}=11\hat{i}+3\hat{j}+7\hat{k}-\hat{i}+2\hat{j}+8\hat{k}=10\hat{i}+5\hat{j}+15\hat{k}$
Clearly, $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$
Hence $A,B,C$ are collinear.
Suppose $B$ divides in the ratio AC in the ratio $\lambda :1$. Then the position vector $B$ is
$\left(\frac{11\lambda +1}{\lambda +1}\right)\hat{i}+\left(\frac{3\lambda -2}{\lambda +1}\right)\hat{j}+\left(\frac{7\lambda -8}{\lambda +1}\right)\hat{k}$
But the position vector of $B$ is $5\hat{i}+0\hat{j}-2\hat{k}.$

$$\begin{gathered} \frac{11\lambda +1}{\lambda +1}=5,\frac{3\lambda -2}{\lambda +1}=0,\frac{7\lambda -8}{\lambda +1}=-2 \\ \Rightarrow 11\lambda +1=5\lambda +5,3\lambda -2=0,7\lambda -8=-2\lambda -2 \\ \Rightarrow 6\lambda =4,3\lambda =2,9\lambda =6 \\ \Rightarrow \lambda =\frac{2}{3},\lambda =\frac{2}{3},\lambda =\frac{2}{3} \end{gathered}$$