Question

If three points $A,B,C$ whose position vectors are respectively $\hat{i}-2\hat{j}-8\hat{k},5\hat{j}-2\hat{k}$ and $11\hat{i}+3\hat{j}+7\hat{k}$ are collinear , then the ratio in which B, divides AC is

• A. 1 : 2
• B. 2 : 3
• C. 2:1
• D. none of these.

Answer 1

Answer: (B)

Let $B\left(5\hat{i}+2\hat{k}\right)$ divide $AC$ in the ratio $K:1$ at $P$
where $A$ is $\left(\hat{i}-2\hat{j}-8\hat{k}\right)$ and $C$ is $\left(11\hat{i}+3\hat{j}+3\hat{k}\right)$.
$\therefore$ position vector of $P$ is
$\frac{K\left(11\hat{i}+3\hat{j}+7\hat{k}\right)+1\left(\hat{i}-2\hat{j}-8\hat{k}\right)}{K+1}$
i.e., $\frac{\left(11K+1\right)}{K+1}\hat{i}+\frac{3\left(K-2\right)}{K+1}\hat{j}+\frac{\left(7K-8\right)}{K-1}\hat{k}$
If $P$ coincides with $B$, then
$\frac{11K+1}{K+1}=5,\frac{3K-2}{K+1}=0,\frac{7K-8}{K+1}=-1$
$\therefore 3K=2$ i.e, $K=\frac{2}{3}$
Hence the reqd. ratio is $2:3$.