Let a = hat{i} + hat{j} + hat{k},b = 2hat{i} | vedclass

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(A) The equation $\vec{r} \cdot \vec{a} = 5$ represents a plane $x + y + z = 5$. The equation $|\vec{r} - \vec{b}| + |\vec{r} - \vec{c}| = 4$ represen

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$4$

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(A) The equation $\vec{r} \cdot \vec{a} = 5$ represents a plane $x + y + z = 5$. The equation $|\vec{r} - \vec{b}| + |\vec{r} - \vec{c}| = 4$ represents an ellipse in space with foci at $\vec{b}$ and $\vec{c}$,provided the plane contains the foci. First,check if $\vec{b}$ and $\vec{c}$ lie on the plane $x + y + z = 5$: For $\vec{b}(2, 2, 1)$,$2 + 2 + 1 = 5$ (True). For $\vec{c}(5, 1, -1)$,$5 + 1 - 1 = 5$ (True). Since both foci lie on the plane,the intersection is an ellipse in the plane. The length of the major axis is $2a = 4$,so $a = 2$. The distance between the foci is $2ae = |\vec{b} - \vec{c}| = |(2-5)\hat{i} + (2-1)\hat{j} + (1-(-1))\hat{k}| = |-3\hat{i} + \hat{j} + 2\hat{k}| = \sqrt{(-3)^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$. Thus,$2ae = \sqrt{14} \Rightarrow 4e = \sqrt{14} \Rightarrow e = \frac{\sqrt{14}}{4}$. Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 4(1 - \frac{14}{16}) = 4(\frac{2}{16}) = \frac{8}{16} = \frac{1}{2}$,so $b = \frac{1}{\sqrt{2}}$. The area of the ellipse is $\pi ab = \pi(2)(\frac{1}{\sqrt{2}}) = \sqrt{2}\pi \approx 1.414 	imes 3.14159 \approx 4.44$. The closest integer is $4$.

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