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. $\frac{1}{\sqrt{5}}\left(i+j+k\right)$
• B. $\frac{1}{\sqrt{3}}\left(i+6j+k\right)$
• C. $\frac{1}{\sqrt{3}}\left(i+j+k\right)$
• D. $\frac{1}{5}\left(7i+3j+4k\right)$
• E. None of these

Answer 1

The correct option is D

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

Determine the position vector $P$

Given the points $A(1,-1,2)$ and $B(2,3,-1)$

$P$ divides $AB$internally in the ratio of $2:3$, hence $m:n=2:3$

Let the coordinates of a point $P$ be $\left(\alpha ,\beta ,\gamma \right)$ therefore,

$$\begin{gathered} \alpha =\frac{m{x}_2+n{x}_1}{m+n} \\ =\frac{2\times 2+3\times 1}{2+3} \\ =\frac{4+3}{5} \\ =\frac{7}{5} \end{gathered}$$

Similarly,

$$\begin{gathered} \beta =\frac{m{y}_2+n{y}_1}{m+n} \\ =\frac{2\times 3+3\times -1}{2+3} \\ =\frac{6-3}{5} \\ =\frac{3}{5} \end{gathered}$$

And the third coordinate,

$$\begin{gathered} y=\frac{m{z}_2+n{z}_1}{m+n} \\ =\frac{2\times -1+3\times 2}{2+3} \\ =\frac{-2+6}{5} \\ =\frac{4}{5} \end{gathered}$$

Therefore the position vector of $P$ is $\left(\frac{7}{5}i+\frac{3}{5}j+\frac{4}{5}k\right)\text{or}\frac{1}{5}\left(7i+3j+4k\right)$

Hence, option D is the correct answer.