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Let $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=2\hat{i}+2\hat{j}+\hat{k},\vec{c}=5\hat{i}+\hat{j}-\hat{k}$

be three vectors. Then the area of the regin formed by the

position vector $\vec{r}$ satisfy the equation $\hat{r}\cdot \vec{a}=5$

and $|\vec{r}-\vec{b}|+|\vec{r}-\vec{c}|=4$ is closest to the integer, is

What i try

Let $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}.$ Then $x+y+z=5\cdots \cdots (1)$

and $\displaystyle \sqrt{(x-2)^2+(y-2)^2+(z-1)^2}+\sqrt{(x-5)^2+(y-1)^2+(z+1)^2}=4$

$\displaystyle \sqrt{(x-2)^2+(y-2)^2+(4-x-y)^2}+\sqrt{(x-5)^2+(y-1)^2+(6-x-y)^2}=4\cdots (2)$

some help me to solve it please

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jacky
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2 Answers2

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To expand on the previous answer by @mathsdiscussion.com.

The equation $r.a=5$ determines a plane. Furthermore, it makes the problem much easier to solve if one notes that $b$ and $c$ lie on this plane and are distance $\sqrt 14$ apart.

A point $r$ on the plane satisfying the other equation is such that the sum of its distance from $a$ and its distance from $b$ is $4$. It therefore lies on an ellipse.

For the standard equation of the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

The distance between foci is $2\sqrt{(a^2-b^2)}$.

The sum of distances' is $2a$.

The area is $\pi ab$.

So now we need to solve $2a=4$ and $2\sqrt{(a^2-b^2)}=\sqrt 14$. Then $a=2, b=\frac{1}{\sqrt 2}$.

The area of the ellipse is $\pi ab=\pi \sqrt 2$.

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The intersection of two will be ellipse with length of major axis as 4 and minor axis as $ \sqrt{2}$ Hence area will be $\\$. $ \sqrt{2}π $.

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