The question asks where the line passing through the points $(-3, 1, 0)$ and $(-1, 3, 4)$ intersects the plane given by the equation $2x + y - z = -2$.

Step-by-step solution:

Step 1: Find the parametric equation of the line

The line passing through two points can be represented parametrically. If the points are $(-3, 1, 0)$ and $(-1, 3, 4)$, we can write the parametric equations as follows.

Let the parameter $t$ vary between the two points:

$x = -3 + t(-1 - (-3)) = -3 + 2t$ $y = 1 + t(3 - 1) = 1 + 2t$ $z = 0 + t(4 - 0) = 4t$

Thus, the parametric equation of the line is: $(x, y, z) = (-3 + 2t, 1 + 2t, 4t)$

Step 2: Find the intersection with the plane

We substitute these parametric equations into the plane equation $2x + y - z = -2$:

$2(-3 + 2t) + (1 + 2t) - 4t = -2$

Simplifying this equation:

$-6 + 4t + 1 + 2t - 4t = -2$ $-5 + 2t = -2$ $2t = 3$ $t = \frac{3}{2}$

Step 3: Find the coordinates of the intersection

Now, substitute $t = \frac{3}{2}$ into the parametric equations for $x$, $y$, and $z$:

$x = -3 + 2\left(\frac{3}{2}\right) = -3 + 3 = 0$ $y = 1 + 2\left(\frac{3}{2}\right) = 1 + 3 = 4$ $z = 4\left(\frac{3}{2}\right) = 6$

Thus, the intersection point is $(0, 4, 6)$.

Final Answer:

The line intersects the plane at the point $(0, 4, 6)$.

Would you like further details or have any questions about the steps?

Here are 5 related questions:

1. How do we find the direction vector for a line passing through two points?
2. How can we derive the parametric equations of a line from two points?
3. How do you verify that a point lies on a given plane?
4. What is the geometric interpretation of a plane equation like $2x + y - z = -2$?
5. How can we use systems of equations to find the intersection of two planes?

Tip: The parametric form of a line is often very useful for finding intersections with planes, lines, and other geometric objects!