If a point $\mathrm{C}$ divides the line segment joining the points with the position vectors $2 \hat{i}-3…
If a point $\mathrm{C}$ divides the line segment joining the points with the position vectors $2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $2: 3$, then the distance of $\mathrm{C}$ from the point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ is
$\frac{7}{5}$
$\frac{4}{5}$
$\frac{4}{5}$
$\frac{3}{5}$
Solution
Let $\vec{r}$ is the position vector of the point $C$.
Since, point $C$ divides the line segment joining the points with the position vectors $2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $2: 3$, then, we have
$\vec{r}=\frac{3(2 \hat{i}-3 \hat{j}+2 \hat{k})+2(3 \hat{i}-\hat{j}-2 \hat{k})}{3+2}$
$=\frac{12 \hat{i}-11 \hat{j}+2 \hat{k}}{5}$
$\Rightarrow \quad \vec{r}=\frac{1}{5}(12 \hat{i}-11 \hat{j}+2 \hat{k})=\frac{12}{5} \hat{i}-\frac{11}{5} \hat{j}+\frac{2}{5} \hat{k}$
Now, distance of $C$ from the point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ is
$\begin{aligned} & D=\left|\left(\frac{12}{5}-2\right) \hat{i}+\left(\frac{-11}{5}+1\right) \hat{j}+\left(\frac{2}{5}-1\right) \hat{k}\right| \\ & =\left|\frac{2}{5} \hat{i}-\frac{16}{5} \hat{j}-\frac{3}{5} \hat{k}\right| \\ & \therefore \quad D=\sqrt{\frac{4}{25}+\frac{36}{25}+\frac{9}{25}} \\ & \therefore \quad D=\frac{7}{5}\end{aligned}$