Question

Let $A(1,-1,2)$ and $B(2,3,-1)$ be two points. If a point $P$ divides $AB$ internally in the ratio $2:3,$ then the position vector of $P$ is

• A. $\frac{1}{\sqrt{5}}(\hat{i}+\hat{j}+\hat{k})$
• B. $\frac{1}{\sqrt{3}}(\hat{i}+6\hat{j}+\hat{k})$
• C. $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
• D. $\frac{1}{5}(7\hat{i}+3\hat{j}+4\hat{k})$

Answer 1

Answer: (D)

$\frac{1}{5}(7\hat{i}+3\hat{j}+4\hat{k})$

Answer 2

The position vector of P
$=\frac{3A+2B}{3+2}$
$=\frac{3(1,-1,2)+2(2,3,-1)}{5}$
$=\frac{(7,3,4)}{5}$
$=\frac{1}{5}(7\hat{i}+3\hat{j}+4\hat{k})$