Locus of points in 3D

Question

A rigid body rotates about an axis through origin with an angular velocity $10(3^{1/2})$ rad/s. If $w$ points in the direction of $i+j+k$ , then the equation of locus of the points having tangential speed $20 m/s$ is

9th question :

Answer 1

Let's consider a generic point $P$ on the rigid body:

$$P(x,y,z)$$

Since the planes orthogonal to the rotation axis have equation:

$$x+y+z=k$$

the corresponding point $Q$ on rotating axis such that $PQ$ is orthogonal to the rotation axis is:

$$Q=(t,t,t)=\left(\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3}\right)$$

NOTE

It easy to check that $$OQ \perp PQ$$ infact

$OQ=(t,t,t)$, $PQ=(x-t,y-t,z-t)$

$$OQ \cdot PQ=t(x+y+z)-3t^2=t\cdot3t-3t^2=0$$

Thus

$$|v_P|=PQ\cdot 10 \sqrt 3$$

$$PQ = \sqrt{(x-t)^2+(y-t)^2+(2t-x-y)^2}$$

The condition is equivalent to:

$$|v_P|=PQ\cdot 10 \sqrt 3= 10 \sqrt 3\sqrt{(x-t)^2+(y-t)^2+(z-t)^2}=20$$

and finally

$$(x-t)^2+(y-t)^2+(z-t)^2=\frac43$$

with

$$t=\frac{x+y+z}{3}$$