If the line (x-2)/3 = (y-1)/2 = (z-1)/-1 intersects the plane 2x + 3y - z + 13 = 0 at a point P and

Q: Wait, why do we use $\lambda$ instead of just solving for $x, y, z$ directly?

Great question! A line in 3D space has infinitely many points, but they are all constrained by the same ratio. By introducing the parameter $\lambda$, we collapse the three variables $x, y, z$ into a single variable. This makes the intersection with a plane a simple algebraic problem rather than a complex system of equations.

Q: What if the line is parallel to the plane? How would the math look?

If the line were parallel to the plane, the direction vector of the line would be perpendicular to the normal vector of the plane. In your calculation, the $\lambda$ terms would cancel out entirely, leaving you with a contradiction (like $0 = 5$). If that happens, it means there is no intersection point!

Q: Is there a faster way to find the distance without finding the coordinates of $P$ and $Q$?

While you could technically use vector projections, finding the coordinates is the most robust and least error-prone method for this type of problem. It builds a solid geometric foundation and ensures you don't miss any subtle details. Stick to the parametric method—it's elegant and reliable!

Question

If the line $\frac{x-2}{3} = \frac{y-1}{2} = \frac{z-1}{-1}$ intersects the plane $2x + 3y - z + 13 = 0$ at a point P and the plane $3x + y + 4z = 16$ at a point Q, then PQ is equal to :

Options:

(A) $2\sqrt{14}$

(B) $\sqrt{14}$

(C) $2\sqrt{7}$

(D) $14$

Sigma Percentile · Accepted Answer

### The Geometry of a Laser Beam

Imagine you are standing in a vast, empty room. In front of you, there are two large, invisible sheets of glass—these are our planes.

A laser beam, our line $L$, cuts through the air, piercing both sheets. Our goal is to find the distance between the two points where the laser hits the glass. This is a journey through 3D space.

### The Parametric Bridge

To navigate this 3D space, we need a way to describe any point on our laser beam. We use the symmetric form of the line:

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

By setting this equal to a parameter, $\lambda$, we create a bridge. Every point on this line can now be written as a function of $\lambda$:

$$
x = 3\lambda + 2, \quad y = 2\lambda - 1, \quad z = -\lambda + 1
$$

This is our master key. Whenever we need to know where the line is, we just ask $\lambda$.

### The First Impact: Point $P$

Now, let's find where the laser hits the first plane, $2x + 3y - z + 13 = 0$. Since point $P$ lies on both the line and the plane, its coordinates must satisfy the plane's equation.

We substitute our parametric expressions into the plane equation:

$$
2(3\lambda + 2) + 3(2\lambda - 1) - (-\lambda + 1) + 13 = 0
$$

Expanding this, we get $6\lambda + 4 + 6\lambda - 3 + \lambda - 1 + 13 = 0$. Combining the terms, we find $13\lambda + 13 = 0$, which gives us $\lambda = -1$.

Plugging this back into our parametric equations, we find $P = (-1, -3, 2)$. The first impact is confirmed.

### The Second Impact: Point $Q$

We repeat the process for the second plane, $3x + y + 4z = 16$. Substituting our parametric coordinates:

$$
3(3\lambda + 2) + (2\lambda - 1) + 4(-\lambda + 1) = 16
$$

Expanding gives $9\lambda + 6 + 2\lambda - 1 - 4\lambda + 4 = 16$. Simplifying, we get $7\lambda + 9 = 16$, which leads to $7\lambda = 7$, or $\lambda = 1$.

Substituting $\lambda = 1$ back into our parametric equations, we find $Q = (5, 1, 0)$.

### The Final Distance

We have our two points: $P(-1, -3, 2)$ and $Q(5, 1, 0)$. The distance $PQ$ is the final piece of the puzzle.

Using the 3D distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$, we calculate:

$$
PQ = \sqrt{(5 - (-1))^2 + (1 - (-3))^2 + (0 - 2)^2}
$$

This simplifies to:

$$
PQ = \sqrt{6^2 + 4^2 + (-2)^2} = \sqrt{36 + 16 + 4} = \sqrt{56}
$$

Simplifying the radical, $\sqrt{56} = \sqrt{4 	imes 14} = 2\sqrt{14}$.

We have arrived at our destination. The distance between the two points is **$2\sqrt{14}$**. You have successfully navigated the 3D space!

Animated Solution for Mathematics - Three Dimensional Geometry: If the line $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z - 1}{- 1}$ intersects the plane $2 x + 3 y - z + 13 = 0$ at a point P and the plane $3 x + y + 4 z = 16$ at a point Q, then PQ is equal to :

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of the Line

Finding Intersection Point $P$

Solving for $λ$ at Point $P$

Coordinates of Point $P$

Finding Intersection Point $Q$

Solving for $λ$ at Point $Q$

Coordinates of Point $Q$

Calculating Distance $P Q$

Substituting Coordinates into Distance Formula

Evaluating the Squares

Final Answer

The Sigma Insight: Intersection of a Line and a Plane

The Geometry of a Laser Beam

The Parametric Bridge

The First Impact

The Second Impact

The Final Distance

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Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of the Line

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculating Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ