Question

attempt 1: 0 out of 2 parts have been answered correctly. (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 $\mathbf{r}(t) = \langle x(t), y(t), z(t) \
angle$. $\mathbf{r}(t) = \langle 1 + 10t, 1 - 12t, 1 - 14t \
angle$

Explanation:

Part (b)

Step 1: Check Plane 1 ($-2x - 3y + 2z = -3$)

Substitute \(x = 1\), \(y = 1\), \(z = 1\) into the plane equation:
Left - hand side (LHS) \(=-2(1)-3(1)+2(1)=-2 - 3 + 2=-3\).
Right - hand side (RHS) \(=-3\).
Since \(LHS = RHS\), \(P_0\) lies on Plane 1.

Step 2: Check Plane 2 ($-x + 5y - 5z = -1$)

Substitute \(x = 1\), \(y = 1\), \(z = 1\) into the plane equation:
LHS \(=-1(1)+5(1)-5(1)=-1 + 5 - 5=-1\).
RHS \(=-1\).
Since \(LHS = RHS\), \(P_0\) lies on Plane 2.

Step 3: Verify Parametric Equations

The parametric equations are \(x(t)=1 + 10t\), \(y(t)=1-12t\), \(z(t)=1 - 14t\).

• When \(t = 0\), \(x(0)=1+10(0)=1\), \(y(0)=1-12(0)=1\), \(z(0)=1-14(0)=1\), which is the point \(P_0(1,1,1)\).
• To check if the line lies on both planes, substitute \(x(t)\), \(y(t)\), \(z(t)\) into Plane 1:

\(-2(1 + 10t)-3(1-12t)+2(1 - 14t)=-2-20t-3 + 36t+2 - 28t=-3\) (simplifies to RHS).

• Substitute into Plane 2:

\(-(1 + 10t)+5(1-12t)-5(1 - 14t)=-1-10t + 5-60t-5 + 70t=-1\) (simplifies to RHS).

Answer:

The point \(P_0(1,1,1)\) lies on both planes (verified by substitution) and the parametric equations \(x = 1 + 10t\), \(y = 1-12t\), \(z = 1 - 14t\) (or vector \(\mathbf{r}(t)=\langle1 + 10t,1-12t,1 - 14t
angle\)) represent the line of intersection.