Question

(a) find a vector v parallel to the line of intersection of the planes $-2x - 3y + 2z = -3$ and $-x + 5y - 5z = -1$. $\mathbf{v} = \langle 10, -12, -14 \
angle$ (b) show that the point $p_0(1, 1, 1)$ lies on both planes and hence on the line of intersection. then find parametric equations $x = x(t)$, $y = y(t)$, $z = z(t)$ for the line of intersection. enter your answer as the vector $\mathbf{r}(t) = \langle x(t), y(t), z(t) \
angle$. submit answer

Explanation:

Part (b)

Step 1: Verify Point in First Plane

The first plane is \(-2x - 3y + 2z = -3\). Substitute \(x = 1\), \(y = 1\), \(z = 1\):
\[
-2(1) - 3(1) + 2(1) = -2 - 3 + 2 = -3
\]
This matches the plane’s equation, so \(P_0\) is on the first plane.

Step 2: Verify Point in Second Plane

The second plane is \(-x + 5y - 5z = -1\). Substitute \(x = 1\), \(y = 1\), \(z = 1\):
\[
-1(1) + 5(1) - 5(1) = -1 + 5 - 5 = -1
\]
This matches the plane’s equation, so \(P_0\) is on the second plane. Thus, \(P_0\) lies on the line of intersection.

Step 3: Parametric Equations of the Line

A line in 3D is given by \(\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}\), where \(\mathbf{r}_0 = \langle 1, 1, 1
angle\) (position vector of \(P_0\)) and \(\mathbf{v} = \langle 10, -12, -14
angle\) (direction vector from part (a)).

Break into components:

• \(x(t) = 1 + 10t\)
• \(y(t) = 1 - 12t\)
• \(z(t) = 1 - 14t\)

Thus, the vector \(\mathbf{r}(t)\) is:
\[
\mathbf{r}(t) = \langle 1 + 10t, 1 - 12t, 1 - 14t
angle
\]

Final Answer (Part b)

\[
\mathbf{r}(t) = \boxed{\langle 1 + 10t, 1 - 12t, 1 - 14t
angle}
\]

Answer:

Part (b)

Step 1: Verify Point in First Plane

The first plane is \(-2x - 3y + 2z = -3\). Substitute \(x = 1\), \(y = 1\), \(z = 1\):
\[
-2(1) - 3(1) + 2(1) = -2 - 3 + 2 = -3
\]
This matches the plane’s equation, so \(P_0\) is on the first plane.

Step 2: Verify Point in Second Plane

The second plane is \(-x + 5y - 5z = -1\). Substitute \(x = 1\), \(y = 1\), \(z = 1\):
\[
-1(1) + 5(1) - 5(1) = -1 + 5 - 5 = -1
\]
This matches the plane’s equation, so \(P_0\) is on the second plane. Thus, \(P_0\) lies on the line of intersection.

Step 3: Parametric Equations of the Line

A line in 3D is given by \(\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}\), where \(\mathbf{r}_0 = \langle 1, 1, 1
angle\) (position vector of \(P_0\)) and \(\mathbf{v} = \langle 10, -12, -14
angle\) (direction vector from part (a)).

Break into components:

• \(x(t) = 1 + 10t\)
• \(y(t) = 1 - 12t\)
• \(z(t) = 1 - 14t\)

Thus, the vector \(\mathbf{r}(t)\) is:
\[
\mathbf{r}(t) = \langle 1 + 10t, 1 - 12t, 1 - 14t
angle
\]

Final Answer (Part b)

\[
\mathbf{r}(t) = \boxed{\langle 1 + 10t, 1 - 12t, 1 - 14t
angle}
\]