Question

A rigid body rotates about an axis through the origin with an angular velocity $10\sqrt{3}$ radians/sec.If $\overline{\omega }$ points in the direction of $\hat{i}+\hat{j}+\hat{k}$ then the equation to the locus of the points having tangential speed 20 m/sec is :

• A. $x^2+y^2+z^2-xy-yz-zx=0$
• B. $x^2+y^2+z^2-2xy-2yz-2zx-1=0$
• C. $x^2+y^2+z^2-xy-yz-zx-2=0$
• D. $x^2+y^2+z^2-2xy-2yz-2zx-2=0$

Answer 1

The correct option is C $x^2+y^2+z^2-xy-yz-zx-2=0$

As we know, $\overrightarrow{v}=\overrightarrow{w}\times \overrightarrow{r}$

here, $\overrightarrow{w}$ = 10$\sqrt{3}\left(\frac{\overrightarrow{i+}\overrightarrow{j}+\overrightarrow{k}}{\sqrt{3}}\right)$

and, let $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

then $\overrightarrow{v}=10[(z-y)\overrightarrow{i}-(z-x)\overrightarrow{j}+(y-x)\overrightarrow{k}]$

and it is given that $\left|\overrightarrow{v}\right|$ = 20

10$\sqrt{(z-y{)}^2+(z-x{)}^2+(y-x{)}^2}$ = 20

$x^2{+y}^2+z^2-xy-yz-zx-2=0$