On which of the following lines lies the point of intersection of the line, (x-4)/2 = (y-5)/2 = (z-3

Q: Wait, why do we use the parameter $\lambda$ instead of just solving the equations directly?

Great question! In 3D geometry, a line is not defined by a single equation but by a set of ratios. By setting these ratios equal to a parameter $\lambda$, we transform the line into a set of parametric equations: $x = 2\lambda + 4$, $y = 2\lambda + 5$, and $z = \lambda + 3$. This allows us to treat the line as a collection of points, making it incredibly easy to find where it 'collides' with a plane by simply substituting these expressions into the plane's equation.

Q: How do I know if the point I found actually lies on the line in the options?

For a point $(x_0, y_0, z_0)$ to lie on a line given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, you must substitute the coordinates into the equation. If the resulting ratios are all equal (e.g., $-1 = -1 = -1$), then the point satisfies the line equation. If they are not equal, the point is not on that line.

Q: Is there a faster way to check the options without substituting the point?

While you could check for parallel lines or coplanarity, substituting the point is the most robust and foolproof method. Since you have already calculated the intersection point $P(0, 1, 1)$, testing it is a quick algebraic check that guarantees accuracy. Don't look for shortcuts when a simple substitution provides absolute certainty!

Question

On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2} = \frac{y-5}{2} = \frac{z-3}{1}$ and the plane, $x + y + z = 2$ ?

Options:

(A) $\frac{x-2}{2} = \frac{y-3}{2} = \frac{z+3}{3}$

(B) $\frac{x-4}{1} = \frac{y-5}{1} = \frac{z-5}{-1}$

(C) $\frac{x-1}{1} = \frac{y-3}{2} = \frac{z+4}{-5}$

(D) $\frac{x+3}{3} = \frac{4-y}{3} = \frac{z+1}{-2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty 3D space. You have a flat, infinite sheet of paper—a plane—defined by the equation:

$$x + y + z = 2$$

Now, imagine a laser beam, a line defined by the symmetric equations:

$$\frac{x-4}{2} = \frac{y-5}{2} = \frac{z-3}{1}$$

Our mission is to find the exact point where this laser meets the paper. This is a fundamental problem of spatial intersection.

### The Parametric Bridge

To find this point, we need a way to describe every single location on that laser beam. We use the parametric form by setting the line's ratios equal to a parameter $\lambda$:

$$\frac{x-4}{2} = \frac{y-5}{2} = \frac{z-3}{1} = \lambda$$

This single variable $\lambda$ acts as a coordinate tracker. If we know $\lambda$, we know exactly where we are on the line.

Rearranging these, we get the coordinates of any point on the line in terms of $\lambda$:

$$x = 2\lambda + 4, \quad y = 2\lambda + 5, \quad z = \lambda + 3$$

This triplet $(2\lambda+4, 2\lambda+5, \lambda+3)$ is our key to the intersection.

### The Collision Point

The plane acts as a constraint; it demands that the sum of the coordinates of any point on it must be $2$. We take our general point from the line and force it to satisfy the plane's equation:

$$(2\lambda + 4) + (2\lambda + 5) + (\lambda + 3) = 2$$

Grouping the terms, we obtain:

$$(2\lambda + 2\lambda + \lambda) + (4 + 5 + 3) = 2$$

$$5\lambda + 12 = 2$$

Solving for $\lambda$ is straightforward:

$$5\lambda = -10 \quad \Rightarrow \quad \lambda = -2$$

This value, $\lambda = -2$, is the specific parameter at which our laser hits the plane. Substituting $\lambda = -2$ back into our parametric equations gives us the intersection point $P$:

$$x = 2(-2) + 4 = 0$$
$$y = 2(-2) + 5 = 1$$
$$z = (-2) + 3 = 1$$

Thus, our point of intersection is **$P(0, 1, 1)$**.

### The Final Verification

The problem asks us to identify which of the given lines passes through this point $P(0, 1, 1)$. We test the options by substituting $x=0, y=1, z=1$ into the provided line equations.

Consider the line:

$$\frac{x-1}{1} = \frac{y-3}{2} = \frac{z+4}{-5}$$

Substituting our point $P(0, 1, 1)$ into this equation:

$$\frac{0-1}{1} = -1, \quad \frac{1-3}{2} = -1, \quad \frac{1+4}{-5} = -1$$

Since all three ratios equal $-1$, the point **$P(0, 1, 1)$** lies perfectly on this line. We have successfully navigated the 3D landscape to confirm the intersection.

Animated Solution for Mathematics - Three Dimensional Geometry: On which of the following lines lies the point of intersection of the line, $\frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1}$ and the plane, $x + y + z = 2$ ?

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Parametric Form of the Line

General Point on the Line

Substitution in Plane Equation

Atomic Compute: Grouping Terms

Atomic Compute: Solving for $λ$

Finding Intersection Point $P$

Checking the Options

Atomic Compute: Verify Option 3

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Parametric Bridge

The Collision Point

The Final Verification

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