The distance of the point (1, -2, 3) from the plane x - y + z = 5 measured parallel to a line, whose

Q: Wait, why can't we just use the standard perpendicular distance formula $d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$?

That is a fantastic question! That formula is specifically designed for the shortest distance, which is always perpendicular to the plane. However, this problem adds a constraint: the distance must be measured parallel to a specific line. Because the line is not necessarily normal to the plane, the path is slanted, not perpendicular. We must find the intersection point first.

Q: Is there a shortcut formula for this?

Yes! If you have a point $(x_0, y_0, z_0)$ and a plane $ax + by + cz = d$, the distance measured parallel to a line with direction ratios $(l, m, n)$ is given by $d = \frac{|ax_0 + by_0 + cz_0 - d|}{|al + bm + cn|}$. In our case, this would be $\frac{|1(1) - 1(-2) + 1(3) - 5|}{|1(2) - 1(3) + 1(-6)|} = \frac{|1 + 2 + 3 - 5|}{|2 - 3 - 6|} = \frac{1}{|-7|} = \frac{1}{7}$. Wait, that gives the value of $\lambda$. The actual distance is $\lambda \sqrt{l^2 + m^2 + n^2} = \frac{1}{7} \sqrt{2^2 + 3^2 + (-6)^2} = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{1}{7} \sqrt{49} = 1$. Both methods lead to the same beautiful result!

Q: What if the line is parallel to the plane?

If the line is parallel to the plane, the direction vector of the line $(l, m, n)$ will be perpendicular to the normal of the plane $(a, b, c)$. This means the dot product $al + bm + cn$ will be zero. In our formula, the denominator would become zero, implying the distance is undefined or infinite because the line never intersects the plane. Always check for this if you get a zero denominator!

Question

The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to a line, whose direction ratios are $2, 3, -6$ is :

Options:

(A) $3$

(B) $5$

(C) $2$

(D) $1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at a point $A(1, -2, 3)$ in a vast three-dimensional space. Below you lies a flat, infinite surface—a plane defined by the equation $x - y + z = 5$.

Usually, when we talk about the distance from a point to a plane, we instinctively think of the shortest path, the perpendicular drop. But today, the problem asks for something different. It asks for the distance measured parallel to a specific line with direction ratios $(2, 3, -6)$.

We are essentially firing a projectile from $A$ in the direction of the vector $\vec{v} = 2\hat{i} + 3\hat{j} - 6\hat{k}$ and asking how far it travels before it pierces the plane.

### The Parametric Bridge

To solve this, we need to map out the path of our projectile. Since we know the starting point $A(1, -2, 3)$ and the direction of travel $(2, 3, -6)$, we can write the equation of the line representing our path.

In its symmetric form, this is:

$$ \frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{-6} = \lambda $$

Here, $\lambda$ is our parameter—a variable that tracks our progress along the line. Any point $P$ on this line can be expressed as $(2\lambda + 1, 3\lambda - 2, -6\lambda + 3)$.

Think of $\lambda$ as a scout. As $\lambda$ changes, the scout moves along the line. We want to find the exact value of $\lambda$ where the scout steps onto the plane.

### The Moment of Intersection

The plane is defined by $x - y + z = 5$. For our scout to be on the plane, their coordinates must satisfy this equation.

We substitute our parametric expressions for $x, y,$ and $z$ into the plane equation:

$$ (2\lambda + 1) - (3\lambda - 2) + (-6\lambda + 3) = 5 $$

Expanding the terms, we get $2\lambda + 1 - 3\lambda + 2 - 6\lambda + 3 = 5$. Grouping the $\lambda$ terms together, we have $(2 - 3 - 6)\lambda$, which is $-7\lambda$.

Summing the constants, $1 + 2 + 3 = 6$. So, the equation simplifies to:

$$ -7\lambda + 6 = 5 $$

Solving for $\lambda$, we find $-7\lambda = -1$, which gives us $\lambda = \frac{1}{7}$.

### The Final Calculation

We have found our magic number! At $\lambda = \frac{1}{7}$, our line intersects the plane.

To find the distance, we calculate the length of the segment $AP$. The displacement vector from $A$ to $P$ is $(2\lambda, 3\lambda, -6\lambda)$. The distance $d$ is the magnitude of this vector:

$$ d = \sqrt{(2\lambda)^2 + (3\lambda)^2 + (-6\lambda)^2} = \sqrt{4\lambda^2 + 9\lambda^2 + 36\lambda^2} = \sqrt{49\lambda^2} = 7|\lambda| $$

Substituting our value $\lambda = \frac{1}{7}$, we get:

$$ d = 7 	imes \frac{1}{7} = 1 $$

The distance is exactly **1 unit**. It is elegant, precise, and deeply satisfying. We didn't just calculate a number; we traced a path through space and found the exact point of contact.

Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to a line, whose direction ratios are $2, 3, - 6$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

Equation of the Line

General Point on the Line

Intersection with the Plane

Solving for $λ$

Coordinates of Point $P$

Applying Distance Formula

Final Calculation

Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Parametric Bridge

The Moment of Intersection

The Final Calculation

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