Question

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=2\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{c}=5\hat{i}+\hat{j}-\hat{k}$ be three vectors. The area of the region formed by the set of points whose position vector $\overrightarrow{r}$ satisfy the equations $\overrightarrow{r}.\overrightarrow{a}=5$ and $|\overrightarrow{r}-\overrightarrow{b}|+|\overrightarrow{r}-\overrightarrow{c}|=4$ is closest to the integer

• A. $4$
• B. $9$
• C. $14$
• D. $19$

Answer 1

The correct option is A $4$

Let $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
$\overrightarrow{r}.\overrightarrow{a}=5\Rightarrow x+y+z=5$
$\overrightarrow{b}=2\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{c}=5\hat{i}+\hat{j}-\hat{k}$ lies on the plane.
$\overrightarrow{b}-\overrightarrow{c}=-3\hat{i}+\hat{j}+2\hat{k}\Rightarrow |\overrightarrow{b}-\overrightarrow{c}|=\sqrt{14}$
$|\overrightarrow{r}-\overrightarrow{b}|+|\overrightarrow{r}-\overrightarrow{c}|=4$
This is a equation of ellipsoid
So the intersection of the plane and the ellipsoid we will get ellipse where
$2a=4\Rightarrow a=2$ and
$2ae=|\overrightarrow{b}-\overrightarrow{c}|=\sqrt{14}\Rightarrow 4e=\sqrt{14}\Rightarrow e=\frac{\sqrt{14}}{4}\Rightarrow e^2=\frac{14}{16}=\frac{7}{8}\Rightarrow 1-\frac{b^2}{a^2}=\frac{7}{8}\Rightarrow 1-\frac{b^2}{4}=\frac{7}{8}\Rightarrow b=\frac{1}{\sqrt{2}}$
So area of an ellipse is given by
$=\pi ab=\pi \times 2\times \frac{1}{\sqrt{2}}=\sqrt{2}\pi \approx 5$
Hence the nearest integer is $4$.