A point P moves so that overrightarrowOP=(1-t)vec…

A point $P$ moves so that $\overrightarrow{OP}=(1-t)\vec i+(2-3t)\vec j+(2t-1)\vec k$ $1^\circ$ Show that $P$ moves along a line $p$. $2^\circ$ Find the equation of $p$. $3^\circ$ When is $P$ in the plane $2x+3y+2z+1=0$? $4^\circ$ Find the plane normal to $p$ through the position at $t=2$.

Explanation of Method

A point whose position vector is linear in time moves along a line; where it crosses a plane, and the normal plane at a given moment.

Step 1: $1^\circ$, $2^\circ$ The line

\[ \overrightarrow{OP}=\{1,2,-1\}+t\{-1,-3,2\}. \]

This is a fixed point plus $t$ times a fixed vector, so $P$ moves along a line through $P_0(1,2,-1)$ with direction $\vec d=\{-1,-3,2\}$ (with constant velocity $\vec d$).

\[p:\ \frac{x-1}{-1}=\frac{y-2}{-3}=\frac{z+1}{2}\]

Step 2: $3^\circ$ Crossing the plane

\[ 2(1-t)+3(2-3t)+2(2t-1)+1=7-7t=0\ \Longrightarrow\ t=1,\quad P(0,-1,1). \]

\[t=1\]

Step 3: $4^\circ$ Normal plane at $t=2$

$P_2=(-1,-4,3)$, and the normal of the plane is $\vec d$: $-(x+1)-3(y+4)+2(z-3)=0$.

\[x+3y-2z+19=0\]

Check: $-1-12-6+19=0$. $\checkmark$

$P_0\,(t=0)$ $t=1$ $t=2$ $p$

$2x+3y+2z+1=0$
$x+3y-2z+19=0$

The path $p$ (red) crosses the blue plane at $t=1$, at $(0,-1,1)$. The gray plane is normal to $p$ at $P_2(-1,-4,3)$.

Result

\[\boxed{\,p:\ \frac{x-1}{-1}=\frac{y-2}{-3}=\frac{z+1}{2}\,}\]

\[\boxed{\,t=1\,}\]

\[\boxed{\,x+3y-2z+19=0\,}\]