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(A) Let $\vec{r}$ be the position vector of the point $C$. Since point $C$ divides the line segment joining the points with position vectors $\vec{a}

Vedclass · Accepted Answer

$\frac{7}{5}$

Vedclass · Answer

(A) Let $\vec{r}$ be the position vector of the point $C$. Since point $C$ divides the line segment joining the points with position vectors $\vec{a} = 2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $\vec{b} = 3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $m:n = 2:3$,the position vector $\vec{r}$ is given by the section formula: $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n} = \frac{2(3 \hat{i}-\hat{j}-2 \hat{k}) + 3(2 \hat{i}-3 \hat{j}+2 \hat{k})}{2+3}$ $= \frac{(6 \hat{i}-2 \hat{j}-4 \hat{k}) + (6 \hat{i}-9 \hat{j}+6 \hat{k})}{5} = \frac{12 \hat{i}-11 \hat{j}+2 \hat{k}}{5} = \frac{12}{5} \hat{i}-\frac{11}{5} \hat{j}+\frac{2}{5} \hat{k}$. Now,the distance of $C$ from the point $P$ with position vector $\vec{p} = 2 \hat{i}-\hat{j}+\hat{k}$ is given by $|\vec{r} - \vec{p}|$. $\vec{r} - \vec{p} = (\frac{12}{5}-2) \hat{i} + (-\frac{11}{5}+1) \hat{j} + (\frac{2}{5}-1) \hat{k} = \frac{2}{5} \hat{i} - \frac{6}{5} \hat{j} - \frac{3}{5} \hat{k}$. The distance $D = |\vec{r} - \vec{p}| = \sqrt{(\frac{2}{5})^2 + (-\frac{6}{5})^2 + (-\frac{3}{5})^2} = \sqrt{\frac{4}{25} + \frac{36}{25} + \frac{9}{25}} = \sqrt{\frac{49}{25}} = \frac{7}{5}$.

If a 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 the distance of $C$ from the point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ is

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