# 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.

Toolbox:
• The position vector of a point which divides the line in the ratio $m:n$ internally is given by $\overrightarrow r=\large\frac{m\overrightarrow a+n\overrightarrow b}{m+n}$
Step 1:
Let the position vectors $\overrightarrow{OA}=\hat i-2\hat j-8\hat k$ and $\overrightarrow {OC}=11\hat i+3\hat j+7\hat k$ respectively.
Now let the point B divides the line AC in the ratio $\lambda:1$
Hence the position vector of $B$ is $\overrightarrow {OB}=\large\frac{\lambda(11\hat i+3\hat j+7\hat k)+1(\hat i-2\hat j-8\hat k)}{\lambda+1}$
Step 2:
But we know $\overrightarrow{OB}=5\hat i-2\hat k$
Therefore $5\hat i-2\hat k=\large\frac{\lambda(11\hat i+3\hat j+7\hat k)+1(\hat i-2\hat j-8\hat k)}{\lambda+1}$
$\Rightarrow 5\hat i-2\hat k=\hat i\large\frac{(11\lambda+1)}{\lambda+1}$$+\hat j\large\frac{(3\lambda-2)}{\lambda+1}$$+\large\frac{(7\lambda-8)}{\lambda+1}$
Step 3:
Now equating the coefficients of $\hat j$ on both sides we get
$0=\large\frac{3\lambda-2}{\lambda+1}$
$3\lambda=2$
Therefore $\lambda=\large\frac{2}{3}$
Step 4:
Now let us check if $A,B$ and $C$ are collinear.
$\large\frac{11\lambda+1}{\lambda+1}=\large\frac{11\big(\Large\frac{2}{3}\big)+1}{\Large\frac{2}{3}+1}$
$\qquad\quad=\large\frac{22+3}{2+3}=\frac{25}{5}$$=5$
Similarly $\large\frac{7\lambda-8}{\lambda+1}=\frac{7\times\Large\frac{2}{3}-8}{\Large\frac{2}{3}+1}$
$\qquad\qquad\qquad=\large\frac{14-24}{2+3}$
$\qquad\qquad\qquad=\large\frac{-10}{5}$
$\qquad\qquad\qquad=-2$
Step 5:
This proves $A,B$ and $C$ are collinear and B divides it in the ratio $2:3$