Finding locus of feet of perpendicular Given curve  $r = (a \cos t , a \sin t , at )$.  Show that locus of feet of perpendicular from origin to the tangent is a curve that completely lies on hyperboloid $x^2+y^2-z^2=a^2$
Tangent vector at any point $t$ is $(-a \sin t, a \cos t,1 )$. How do I proceed ? Thanks
 A: 
As you have find the tangent vector at the point, the equation of the tangent line would be $$ \vec r = (a \cos t, a \sin t, a t) + \lambda (-a \sin t, a \cos t,1) $$
Let $P( a \cos d + \lambda -a \sin d ,a \sin d +  \lambda a \cos d, a d + \lambda)$ be some point on this tangent line, $O$ be origin and $\vec d$ be direction ratio of the lines.
Then for $\vec {OP} $ to be perpendicular to $ \vec d$, $ \vec d \cdot \vec {OP} = 0$ , which on solving gives $\lambda = \frac{-a d}{a^2 + 1} $
On placing this value on in the coordinates of point $P$ , the foot of perpendicular can be obtained, which satisfies the equation of the hyperboloid.
Note: Relevant 3D graph can be found here .
A: For the tangent line at some point of the curve with $t=t_0$ we get this equation:
$r:\vec{OX}=(a\cos t_0-\Lambda a \sin t_0, a \sin t_0+\Lambda a \cos t_0,at_0+a\Lambda );\;\Lambda\in\mathbb R$
The vectors direction of $r$, $\vec d_r=(-a \sin t_0, a \cos t_0,t_0 )$ (there is a mistake in your caculation) and position for the foot (that lies on the tangent to the curve) are perpendicular:
$\vec{OX_0}\cdot\vec{d_r}=0;\;\vec d_s=\vec{OX_0}$
$0=-a^2\sin t_0\,\cos t_0+\Lambda_0 a^2\sin^2 t_0+a^2 \cos t_0\sin t_0+\Lambda_0 a^2\cos^2 t_0+a^2t_0+a^2\Lambda_0$
$0=a^2(t_0+2\Lambda)\implies\Lambda_0=-t_0/2$
Then, the feet all are into the curve:
$\vec{OX_0}=a(\cos t_0-\dfrac{t_0}{2}\sin t_0, \sin t_0+\dfrac{t_0}2\cos t_0,\dfrac{t_0}2)=(x_0,y_0,z_0);\;t_0\in\mathbb R$
Now
$x_0^2+y_0^2=a^2(\cos^2 t_0-t_0\sin t_0\cos t_0+\dfrac{t_0^2}4\cos^2 t_0+\sin^2t_0+t_0\sin t_0\cos t_0+\dfrac{t_0^2}4\cos^2t_0)=$
$=a^2(1+\dfrac{t_0^2}4)$
$z_0^2=a^2\dfrac{t_0^2}4$
So is, $x_0^2+y_0^2-z_0^2=a^2$
A: Working in homogeneous coordinates, $\mathbf r = (a\cos t,a\sin t,a t,1)^T$, $\mathbf r'=(-a\sin t,a\cos t,a,0)^T$. The latter is also the vector that represents the plane through the origin that’s perpendicular to the tangent at $\mathbf r$. The foot of the perpendicular from the origin is the intersection of this plane with the tangent line at $\mathbf r$. Using the line’s Plücker matrix to compute the intersection, this point is $$\mathbf p = \left(\mathbf r\mathbf r'^T-\mathbf r'\mathbf r^T\right)\mathbf r' = a^2\left(a(2\cos t+t\sin t),a(2\sin t-t\cos t),at,2\right)^T.$$ The equation of the hyperboloid is, in homogeneous matrix form, $\mathbf x^T\operatorname{diag}(1,1,-1,-a^2)\mathbf x=0$, so, removing the common factor of $a^2$ we compute $$\begin{align} (\mathbf p/a^2)^T \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-a^2\end{bmatrix} (\mathbf p/a^2) &= a^2(2\cos t+t\sin t)^2+a^2(2\sin t - t\cos t)^2-a^2t^2-4a^2 \\
&= 4a^2+a^2t^2-a^2t^2-4a^2 = 0. \end{align}$$
