The distance of the point (-1, 2, -2) from the line of intersection of the planes 2x + 3y + 2z = 0 a

Q: Wait, why did you assume the line passes through the origin? Is that always true?

Great observation! It is not always true. However, look at the equations $2x + 3y + 2z = 0$ and $x - 2y + z = 0$. Since there is no constant term in either equation, the point $(0, 0, 0)$ satisfies both. If a point satisfies both plane equations, it must lie on their intersection. Always check for the origin first—it saves so much time!

Q: Why can't I just use the distance formula from a point to a plane? There are two planes here.

The distance from a point to a plane is a scalar value representing the shortest path to that surface. But the intersection of two planes is a line, not a point. You need the shortest distance to that specific line, which requires finding the foot of the perpendicular on that line, not just the distance to the individual planes.

Q: Does the order of the cross product matter when finding the direction vector?

Mathematically, $\vec{n}_1 \times \vec{n}_2$ and $\vec{n}_2 \times \vec{n}_1$ are just negatives of each other. Since a line's direction is defined by its ratio, the sign doesn't change the line itself. You will get the same result regardless of the order!

Q: Is there a faster way than finding the foot of the perpendicular?

You could use the formula for the distance from a point to a line: $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$, where $A$ is a point on the line and $\vec{v}$ is the direction vector. Both methods are elegant, but the foot-of-the-perpendicular method is often more intuitive for visualizing the geometry.

Q: I got $\lambda = 1/2$, but my distance calculation is messy. Did I make a mistake?

Don't panic! When you plug $\lambda = 1/2$ into $Q = (\lambda, 0, -\lambda)$, you get $(1/2, 0, -1/2)$. The distance formula involves squaring these fractions, which leads to $9/4 + 4 + 9/4 = 34/4$. Taking the root gives $\sqrt{34}/2$. Your math is likely perfect—it's just the nature of 3D geometry to involve these square roots!

Question

The distance of the point $(-1, 2, -2)$ from the line of intersection of the planes $2x + 3y + 2z = 0$ and $x - 2y + z = 0$ is :

Options:

(A) $\frac{1}{\sqrt{2}}$

(B) $\frac{5}{2}$

(C) $\frac{\sqrt{42}}{2}$

(D) $\frac{\sqrt{34}}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty room. In front of you, two massive, invisible sheets of glass—our planes—intersect. They meet at a single, razor-sharp line that stretches into infinity.

You are holding a point $P(-1, 2, -2)$ in space, and your goal is to find the shortest distance from your hand to that line of intersection. This is a quest to find the 'foot of the perpendicular'—the exact spot on that line that is closest to you.

### Finding the Path

To navigate this space, we first need to know the direction of the line. We are given two planes: $P_1: 2x + 3y + 2z = 0$ and $P_2: x - 2y + z = 0$.

Every plane has a 'normal' vector, a compass needle pointing straight out of its surface. For $P_1$, the normal is $\vec{n}_1 = (2, 3, 2)$, and for $P_2$, it is $\vec{n}_2 = (1, -2, 1)$.

If the line lies on both planes, it must be perpendicular to both normal vectors. This is the magic of the cross product. By calculating $\vec{v} = \vec{n}_1 	imes \vec{n}_2$, we find the direction of the line:

$$
\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 2 \ 1 & -2 & 1 \end{vmatrix} = 7\hat{i} - 0\hat{j} - 7\hat{k}
$$

We can simplify this direction ratio to $(1, 0, -1)$. This is the 'compass heading' of our line.

### Anchoring the Line

Now, where does this line live? We need a starting point. Notice that both plane equations have no constant term; they are equal to zero.

This is a gift! It means the origin $(0, 0, 0)$ satisfies both equations. Our line passes through the origin.

With a point $(0, 0, 0)$ and a direction $(1, 0, -1)$, the equation of our line is simply:

$$
\frac{x}{1} = \frac{y}{0} = \frac{z}{-1} = \lambda
$$

This means any point $Q$ on this line can be written as $Q = (\lambda, 0, -\lambda)$.

### The Perpendicular Connection

We are looking for a specific point $Q$ on the line such that the vector $\vec{PQ}$ is perfectly perpendicular to the line itself. Our point $P$ is $(-1, 2, -2)$.

The vector $\vec{PQ}$ is:

$$
\vec{PQ} = (\lambda - (-1), 0 - 2, -\lambda - (-2)) = (\lambda + 1, -2, -\lambda + 2)
$$

For $\vec{PQ}$ to be perpendicular to the line's direction $\vec{v} = (1, 0, -1)$, their dot product must vanish into nothingness:

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

Solving this, we get $\lambda + 1 + \lambda - 2 = 0$, which simplifies beautifully to $2\lambda = 1$, or $\lambda = \frac{1}{2}$.

### The Final Reveal

With $\lambda = \frac{1}{2}$, we find the coordinates of our target point $Q$ on the line: $Q = (\frac{1}{2}, 0, -\frac{1}{2})$. Now, the distance $PQ$ is just the magnitude of the vector we constructed earlier:

$$
PQ = \sqrt{(\frac{1}{2} + 1)^2 + (-2)^2 + (-\frac{1}{2} + 2)^2}
$$

$$
PQ = \sqrt{(\frac{3}{2})^2 + (-2)^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + 4 + \frac{9}{4}} = \sqrt{\frac{18}{4} + \frac{16}{4}} = \sqrt{\frac{34}{4}}
$$

Which gives us the final, elegant result: **$\frac{\sqrt{34}}{2}$**. You have successfully navigated the intersection of two planes, found the hidden line, and calculated the shortest path to it.

Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(- 1, 2, - 2)$ from the line of intersection of the planes $2 x + 3 y + 2 z = 0$ and $x - 2 y + z = 0$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

Finding Line Direction $v$

Calculating Cross Product

Identifying a Point on the Line

Equation of the Line $L$

Defining General Point $Q$

Vector $P Q$ Construction

The Perpendicularity Condition

Solving for $λ$

Coordinates of Foot $Q$

Applying Distance Formula

Final Calculation

Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Path

Anchoring the Line

The Perpendicular Connection

The Final Reveal

Similar Questions

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