If the line, (x-1)/2 = (y+1)/3 = (z-2)/4 meets the plane, x + 2y + 3z = 15 at a point P, then the di

Q: Why do we equate the line to $\lambda$? Can't we just solve for $x$ and $y$ directly?

Equating to $\lambda$ is the 'Parametric Form'. It is the most powerful tool in 3D geometry because it reduces a complex 3D problem into a simple 1D problem. Without $\lambda$, you would be juggling multiple variables and trying to solve a system of equations, which is a recipe for algebraic disaster. $\lambda$ acts as a 'position marker' along the line.

Q: What if the line never hits the plane? How would the math tell me that?

That is a brilliant question! If the line is parallel to the plane, the $\lambda$ terms would cancel out entirely during your substitution, leaving you with a mathematical contradiction like $0 = 5$. That is the universe's way of telling you that the line and plane never meet. In this problem, the math works out perfectly, which confirms an intersection exists.

Q: Why is the distance formula just $\sqrt{x^2 + y^2 + z^2}$?

Because the origin is $(0,0,0)$. The general distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. When one of those points is the origin, the subtraction becomes trivial, leaving you with the square root of the sum of the squares of the coordinates. It is the 3D extension of the Pythagorean theorem!

Question

If the line, $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4}$ meets the plane, $x + 2y + 3z = 15$ at a point $P$, then the distance of $P$ from the origin is

Options:

(A) $9/2$

(B) $2\sqrt{5}$

(C) $\sqrt{5}/2$

(D) $7/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through three-dimensional space. In the JEE Advanced landscape, 3D geometry is not about memorizing formulas—it is about visualization.

When you see a line and a plane, do not just see symbols on a page. Imagine a laser beam (the line) piercing through a sheet of paper (the plane). The point where that beam hits the paper is our target, point $P$. Our mission is to find the distance of this point from the origin.

### The Parametric Bridge

We are given the line: 
$$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4}$$

This symmetric form is elegant, but it is static. To find the intersection, we need to make this line 'move'. We introduce a parameter, $\lambda$. By setting the entire expression equal to $\lambda$, we create a bridge between the geometry of the line and the algebra of the plane.

This allows us to define any point on the line as a function of $\lambda$:
$$x = 2\lambda + 1$$
$$y = 3\lambda - 1$$
$$z = 4\lambda + 2$$

Think of $\lambda$ as a 'time' variable. As $\lambda$ changes, you are essentially walking along the line. For every value of $\lambda$, you are at a specific coordinate $(x, y, z)$. This is the most powerful technique in your 3D geometry toolkit.

### The Intersection Encounter

Now, we bring in the plane: $x + 2y + 3z = 15$. This plane is a constraint and a boundary. The point $P$ is special because it is the only point that satisfies both the line's path and the plane's boundary.

This means the coordinates we just derived for the line must satisfy the equation of the plane. We substitute our parametric expressions into the plane equation:
$$(2\lambda + 1) + 2(3\lambda - 1) + 3(4\lambda + 2) = 15$$

Let us expand it carefully, term by term:
$$2\lambda + 1 + 6\lambda - 2 + 12\lambda + 6 = 15$$

Now, we group the like terms. We collect all the $\lambda$ terms together and all the constants together:
$$(2 + 6 + 12)\lambda + (1 - 2 + 6) = 15$$

This simplifies beautifully to:
$$20\lambda + 5 = 15$$

Solving for $\lambda$ is now trivial:
$$20\lambda = 10$$
$$\lambda = \frac{1}{2}$$

### The Final Destination

Now that we have $\lambda$, we can find the exact coordinates of point $P$. We plug $\lambda = \frac{1}{2}$ back into our parametric equations:
$$x = 2\left(\frac{1}{2}\right) + 1 = 2$$
$$y = 3\left(\frac{1}{2}\right) - 1 = \frac{1}{2}$$
$$z = 4\left(\frac{1}{2}\right) + 2 = 4$$

So, our point $P$ is $(2, \frac{1}{2}, 4)$. The final step is to calculate the distance of this point from the origin $(0,0,0)$ using the 3D distance formula:
$$OP = \sqrt{x^2 + y^2 + z^2}$$

Substituting our coordinates:
$$OP = \sqrt{2^2 + \left(\frac{1}{2}\right)^2 + 4^2}$$
$$OP = \sqrt{4 + \frac{1}{4} + 16} = \sqrt{20 + \frac{1}{4}}$$

To add these, we find a common denominator:
$$OP = \sqrt{\frac{80}{4} + \frac{1}{4}} = \sqrt{\frac{81}{4}}$$

And finally, the square root of $\frac{81}{4}$ is **$\frac{9}{2}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is

Select Answer:

Visualized Solution

Visualizing the Setup

Parametric Form of the Line

Coordinates of a General Point

Intersection Condition

Substituting into Plane Equation

Expanding the Equation

Combining Like Terms

Solving for $λ$

Coordinates of Point $P$

Distance from Origin

Substituting Coordinates

Final Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Parametric Bridge

The Intersection Encounter

The Final Destination

Similar Questions

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersects at the point $P$ , then the distance of the point $P$ from the plane $z = a$ is :

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 :

The distance of the point $P (3, 4, 4)$ from the point of intersection of the line joining the points $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ and the plane $2 x + y + z = 7$ , is equal to ___

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ is:

Let the plane $P : 8 x + α_{1} y + α_{2} z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ . If the intercept of $P$ on the $y$ -axis is 1, then the distance between $P$ and $L$ is :

The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ be $P$ . Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$

Let $\frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z + 3}{- 1}$ lie on the plane $p x - q y + z = 5$ , for some $p, q \in R$ . The shortest distance of the plane from the origin is: