Jee Advanced
Vector Algebra
JEE Advanced 2023 Paper 1 Online
INTEGER+4 / -02023
Let $P$ be the plane $\sqrt{3} x+2 y+3 z=16$ and let
$S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1\right.$ and the distance of $(\alpha, \beta, \gamma)$ from the plane $P$ is $\left.\frac{7}{2}\right\}$.
Let $\vec{u}, \vec{v}$ and $\vec{w}$ be three distinct vectors in $S$ such that $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\vec{u}, \vec{v}$ and $\vec{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ is :
$S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1\right.$ and the distance of $(\alpha, \beta, \gamma)$ from the plane $P$ is $\left.\frac{7}{2}\right\}$.
Let $\vec{u}, \vec{v}$ and $\vec{w}$ be three distinct vectors in $S$ such that $|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\vec{u}, \vec{v}$ and $\vec{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ is :