To find the equation of the line where the two planes $Q$ and $R$ intersect, we can follow these steps:

1. Find the direction vector of the line of intersection.

The direction vector of the line will be parallel to the cross product of the normal vectors of the planes $Q$ and $R$.

• The normal vector of plane $Q: -2x + y - 4z = 1$ is $\mathbf{n}_Q = \langle -2, 1, -4 \rangle$.
• The normal vector of plane $R: x + y + z = -1$ is $\mathbf{n}_R = \langle 1, 1, 1 \rangle$.

Now, we calculate the cross product $\mathbf{n}_Q \times \mathbf{n}_R$:

$\mathbf{n}_Q \times \mathbf{n}_R = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & -4 \\ 1 & 1 & 1 \end{vmatrix}$

This gives: $\mathbf{n}_Q \times \mathbf{n}_R = \mathbf{i} \left( 1 \times 1 - (-4) \times 1 \right) - \mathbf{j} \left( -2 \times 1 - (-4) \times 1 \right) + \mathbf{k} \left( -2 \times 1 - 1 \times 1 \right)$ $= \mathbf{i}(1 + 4) - \mathbf{j}(-2 + 4) + \mathbf{k}(-2 - 1)$ $= 5\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}$

Thus, the direction vector $\mathbf{v}$ of the line is $\langle 5, -2, -3 \rangle$.

2. Find a point on the line of intersection.

To find a point on the line, we can solve the system of equations given by the two planes. One way to do this is to set one variable to an arbitrary value and solve for the other two.

Let’s set $z = 0$, and solve for $x$ and $y$.

From plane $Q: -2x + y = 1$, we get: $-2x + y = 1 \quad \text{(equation 1)}$ From plane $R: x + y = -1$, we get: $x + y = -1 \quad \text{(equation 2)}$

Solve this system: From equation (2), solve for $y$: $y = -1 - x$ Substitute this into equation (1): $-2x + (-1 - x) = 1$ $-2x - 1 - x = 1$ $-3x - 1 = 1$ $-3x = 2$ $x = -\frac{2}{3}$ Now substitute $x = -\frac{2}{3}$ into $y = -1 - x$: $y = -1 - \left( -\frac{2}{3} \right) = -1 + \frac{2}{3} = -\frac{1}{3}$

Thus, a point on the line of intersection is $\left( -\frac{2}{3}, -\frac{1}{3}, 0 \right)$.

3. Parametric equation of the line.

The parametric equation of the line is given by: $\mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{v}$ where $\mathbf{r}_0$ is the point on the line $\left( -\frac{2}{3}, -\frac{1}{3}, 0 \right)$ and $\mathbf{v} = \langle 5, -2, -3 \rangle$ is the direction vector.

Thus, the parametric equations for the line are: $x = -\frac{2}{3} + 5t$ $y = -\frac{1}{3} - 2t$ $z = -3t$

Final Answer:

The parametric equations of the line where the planes intersect are: $x = -\frac{2}{3} + 5t, \quad y = -\frac{1}{3} - 2t, \quad z = -3t$

Would you like a deeper explanation of any step? Here are some related questions:

1. How can you convert parametric equations to symmetric form?
2. What is the geometric interpretation of the cross product of two vectors?
3. How do you verify if a point lies on both planes?
4. Can two planes intersect at more than one line?
5. How do you find the angle between two planes?

Tip: Always check that your direction vector satisfies both plane equations to confirm accuracy!