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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(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 $4$.

Similar Questions

The position vectors of points $A, B, C$ are given by $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} - 3\hat{j} - 5\hat{k}$,and $a\hat{i} - 3\hat{j} + \hat{k}$ respectively. If these points form a right-angled triangle with $\angle C = \pi/2$,find the value of $a$.

The position vectors of the vertices of a $\Delta ABC$ are $4\hat{i}-2\hat{j}$,$\hat{i}+4\hat{j}-3\hat{k}$,and $-\hat{i}+5\hat{j}+\hat{k}$ respectively. Then,$\angle ABC$ is equal to:

Let $a$ and $b$ be two unit vectors and $\theta$ is the angle between them. Then,$a+b$ is a unit vector,if

If $a = i + 2j - 3k$ and $b = 3i - j + 2k,$ then the angle between the vectors $a + b$ and $a - b$ is ............... $^o$

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module