Let P be the plane sqrt{3} x+2 y+3 z=16 and l | vedclass

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

Question

(B) Given $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$,so $	riangle UVW$ is an equilateral triangle.Let $O$ be the origin. The vectors $\v

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Given $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$,so $	riangle UVW$ is an equilateral triangle.Let $O$ be the origin. The vectors $\vec{u}, \vec{v}, \vec{w}$ are unit vectors,so they lie on a sphere of radius $1$ centered at $O$.The distance of the plane $P: \sqrt{3}x + 2y + 3z = 16$ from the origin $O(0,0,0)$ is $OQ = \frac{|0+0+0-16|}{\sqrt{(\sqrt{3})^2 + 2^2 + 3^2}} = \frac{16}{\sqrt{3+4+9}} = \frac{16}{4} = 4$.The distance of any point $(\alpha, \beta, \gamma) \in S$ from the plane $P$ is given as $\frac{7}{2}$. Let $Q$ be the projection of $O$ on $P$. Since $U, V, W$ are at a constant distance from $P$,they lie on a circle which is the intersection of the sphere and a plane parallel to $P$. Let $P'$ be this plane. The distance of $P'$ from $O$ is $OP = OQ - PQ = 4 - \frac{7}{2} = \frac{1}{2}$.The radius of the circle formed by the intersection of the sphere and plane $P'$ is $R = \sqrt{1^2 - (1/2)^2} = \sqrt{3/4} = \frac{\sqrt{3}}{2}$.Since $	riangle UVW$ is equilateral and inscribed in this circle of radius $R$,its side length $a = 2R \cos(30^\circ) = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2}$.The area of $	riangle UVW = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \cdot \frac{9}{4} = \frac{9\sqrt{3}}{16}$.The volume of the tetrahedron with vertices $O, U, V, W$ is $\frac{1}{3} 	imes 	ext{Area}(	riangle UVW) 	imes OP = \frac{1}{3} 	imes \frac{9\sqrt{3}}{16} 	imes \frac{1}{2} = \frac{3\sqrt{3}}{32}$.The volume $V$ of the parallelepiped determined by $\vec{u}, \vec{v}, \vec{w}$ is $6 	imes 	ext{Volume of tetrahedron} = 6 	imes \frac{3\sqrt{3}}{32} = \frac{9\sqrt{3}}{16}$.Therefore,$\frac{80}{\sqrt{3}} V = \frac{80}{\sqrt{3}} 	imes \frac{9\sqrt{3}}{16} = 5 	imes 9 = 45$.

Similar Questions

If $P$,$Q$,and $R$ are the feet of the perpendiculars drawn from the point $A(1, 1, 1)$ to the planes $P_1: x + 2y + 2z = 2$,$P_2: 2x - 2y + z = -8$,and to the line of intersection of $P_1$ and $P_2$ respectively,then the area of $\Delta PQR$ is:

The equation of the line passing through the intersection of the plane $x+2y+3z=4$ and the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}$ and parallel to the vector $(2\hat{i}-3\hat{j}) \times (\hat{i}+2\hat{j}-\hat{k})$ is

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module