Questions: (a) Find parametric equations for the line through (4,2,4) that is perpendicular to the plane x-y+2z=2. (Use the parameter t.) (x(t), y(t), z(t))= (b) In what points does this line intersect the coordinate planes? xy-plane (x(t), y(t), z(t))= yz-plane (x(t), y(t), z(t))= xz-plane (x(t), y(t), z(t))=

(a) Find parametric equations for the line through (4,2,4) that is perpendicular to the plane x-y+2z=2. (Use the parameter t.)

(x(t), y(t), z(t))=

(b) In what points does this line intersect the coordinate planes?

xy-plane (x(t), y(t), z(t))=

yz-plane (x(t), y(t), z(t))=

xz-plane (x(t), y(t), z(t))=

Transcript text: (a) Find parametric equations for the line through $(4,2,4)$ that is perpendicular to the plane $x-y+2 z=2$. (Use the parameter $t$.) $\square$ \[ (x(t), y(t), z(t))=(\square) \] (b) In what points does this line intersect the coordinate planes? $\square$ $x y$-plane $\quad(x(t), y(t), z(t))=(\square)$ $\square$ $y z$-plane $\quad(x(t), y(t), z(t))=$ ) $x z$-plane $\quad(x(t), y(t), z(t))=$ $\square$ )

Solution

Solution Steps

Solution Approach

(a) To find the parametric equations for the line through the point (4, 2, 4) that is perpendicular to the plane $x - y + 2z = 2$, we need to determine the normal vector to the plane. The normal vector can be derived from the coefficients of $x$, $y$, and $z$ in the plane equation, which is $(1, -1, 2)$. Using this normal vector and the given point, we can write the parametric equations for the line.

(b) To find the points where this line intersects the coordinate planes, we set the corresponding coordinate to zero and solve for the parameter $t$. Specifically:

For the $xy$-plane, set $z = 0$.
For the $yz$-plane, set $x = 0$.
For the $xz$-plane, set $y = 0$.

Step 1: Parametric Equations of the Line

The line through the point $(4, 2, 4)$ that is perpendicular to the plane $x - y + 2z = 2$ has a normal vector given by $(1, -1, 2)$. The parametric equations for the line can be expressed as: \[ \begin{align_} x(t) &= t + 4, \\ y(t) &= 2 - t, \\ z(t) &= 2t + 4. \end{align_} \]

Step 2: Intersection with the $xy$-Plane

To find the intersection of the line with the $xy$-plane, we set $z = 0$: \[ 2t + 4 = 0 \implies t = -2. \] Substituting $t = -2$ into the parametric equations gives: \[ \begin{align_} x(-2) &= -2 + 4 = 2, \\ y(-2) &= 2 - (-2) = 4. \end{align_} \] Thus, the intersection point with the $xy$-plane is $(2, 4, 0)$.

Step 3: Intersection with the $yz$-Plane

To find the intersection of the line with the $yz$-plane, we set $x = 0$: \[ t + 4 = 0 \implies t = -4. \] Substituting $t = -4$ into the parametric equations gives: \[ \begin{align_} x(-4) &= 0, \\ y(-4) &= 2 - (-4) = 6, \\ z(-4) &= 2(-4) + 4 = -4. \end{align_} \] Thus, the intersection point with the $yz$-plane is $(0, 6, -4)$.

Step 4: Intersection with the $xz$-Plane

To find the intersection of the line with the $xz$-plane, we set $y = 0$: \[ 2 - t = 0 \implies t = 2. \] Substituting $t = 2$ into the parametric equations gives: \[ \begin{align_} x(2) &= 2 + 4 = 6, \\ y(2) &= 0, \\ z(2) &= 2(2) + 4 = 8. \end{align_} \] Thus, the intersection point with the $xz$-plane is $(6, 0, 8)$.