Determine properties of a point whose movement in space is determined by a vector parametric equation

Consider a point moving in space with position at time $t$ given by

$X(t) = (1-t) \mathbf{i} + (2-3t) \mathbf{j} + (2t-1) \mathbf{k}.$

Prove that the motion of the point is along a line.
Find a vector parallel to this line.
Find the time $t$ at which the point intersects the plane with Cartesian equation $2x+3y+2z+1 = 0$ .
What is the Cartesian equation for the plane parallel to the plane in part (c) which contains the point $X(3)$ ?
Let $M$ be the plane perpendicular to $L$ containing the point $X(2)$ . Find a Cartesian equation for $M$ .

Proof. We use the formula for the motion of the particle to compute
$\begin{align*} X(t) &= (1-t) \mathbf{i} + (2-3t) \mathbf{j} + (2t-1) \mathbf{k} \\ &= \mathbf{i} + 2 \mathbf{j} - \mathbf{k} + t(-\mathbf{i} - 3\mathbf{j} + 2 \mathbf{k}) \\ &= (1,2,-1) + t(-1,-3,2). \end{align*}$

This is the parametric equation for the line through $(1,2,-1)$ parallel to the vector $(-1,-3,2). \qquad \blacksquare$
From part (a) we have a vector $N$ parallel to $L$ given by $N = (1,3,-2)$ .
First, the line on which the point moves is the set of points
$L = \{ (1,2,-1) + t(-1,-3,2) \} = \{ (1-t, 2-3t, -1+2t) \}.$

So, to find the intersection with the plane $2x + 3y + 2z + 1 = 0$ we compute

$\begin{align*} && 2(1-t) + 3(2-3t) + 2(-1+2t) + 1 &= 0 \\ \implies && 2 - 2t + 6 - 9t -2 + 4t + 1 &= 0 \\ \implies && 7 -7t &= 0 \\ \implies && t &= 1. \end{align*}$
First, we have
$X(3) = (1,2,-1) + 3(-1,-3,2) = (1,2,-1) + (-3,-9,6) = (-2,-7,5).$

Since we know the plane is parallel to the one in part (c) it has a Cartesian equation of the form

$2x + 3y + 2z = d.$

We compute $d = 2(-2) + 3(-7) + 2(5) = -15$ . Hence, the plane has Cartesian equation

$2x + 3y + 2z + 15 = 0.$
Since the plane is perpendicular to the line $L$ we know that it has a normal vector in the same direction as $L$ , so $N = (1,3,-2)$ (from part (b)). Thus, we have a Cartesian equation of the form
$x + 3y - 2z = d.$

Since the point

$X(2) = (1,2,-1) + 2(-1,-3,2) = (1,2,-1) + (-2, -6, 4) = (-1, -4, 3)$

is on the plane we have $d = -1 + 3(-4) - 2(3) = -19$ . Therefore, the plane is given by

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