Let a line L pass through the point P(2, 3, 1) and be parallel to the line x + 3y - 2z - 2 = 0 = x -

Q: Why can't we just use the distance formula for a point to a plane? This is a 3D geometry problem, right?

That is a common trap! The distance formula for a point to a plane is for a point and a *plane*. Here, we are dealing with a *line*. A line in 3D space is not a plane; it is a one-dimensional object. We must find the foot of the perpendicular from the point to the line, which requires the parametric form of the line.

Q: Why do we use the cross product of the normals to find the direction of the line?

Think of the two planes as two sheets of paper intersecting. The line of intersection lies on both planes. Therefore, the line must be perpendicular to the normal vector of the first plane AND the normal vector of the second plane. The cross product of two vectors is the standard tool to find a third vector perpendicular to both.

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

Yes, you could use the formula for the distance from a point to a line: \frac{|\vec{PQ} \times \vec{d}|}{|\vec{d}|}. However, finding the foot of the perpendicular (the point R) is a very robust method that helps you visualize the geometry and is less prone to algebraic errors if you are careful with the dot product.

Question

Let a line $L$ pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of $L$ from the point $(5, 3, 8)$ is $\alpha$, then $3\alpha^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have a line $L$ that passes through a specific point $P(2, 3, 1)$, and this line is the intersection of two distinct planes.

Our mission is to find the perpendicular distance $\alpha$ from an external point $Q(5, 3, 8)$ to this line $L$. Once we have that distance, the final step is to calculate $3\alpha^2$.

### Phase 1: The Direction of the Line

The line $L$ is parallel to the intersection of two planes: $x + 3y - 2z - 2 = 0$ and $x - y + 2z = 0$. The normal vectors to these planes are $\vec{n_1} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{n_2} = \hat{i} - \hat{j} + 2\hat{k}$.

Because the line $L$ lies on both planes, its direction vector $\vec{d}$ must be perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. We calculate this using the cross product:

$$
\vec{d} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & -2 \ 1 & -1 & 2 \end{vmatrix}
$$

Expanding this determinant, we find:

$$
\vec{d} = \hat{i}(6 - 2) - \hat{j}(2 - (-2)) + \hat{k}(-1 - 3) = 4\hat{i} - 4\hat{j} - 4\hat{k}
$$

For simplicity, we scale this vector to $\vec{d} = (1, -1, -1)$. This represents the direction of our line.

### Phase 2: The Parametric Journey

Given point $P(2, 3, 1)$ and direction vector $\vec{d} = (1, -1, -1)$, the symmetric form of the line is:

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

Using the parameter $\lambda$, any point $R$ on the line can be expressed as:

$$
R \equiv (\lambda + 2, -\lambda + 3, -\lambda + 1)
$$

### Phase 3: The Perpendicular Quest

The shortest distance from $Q(5, 3, 8)$ to the line occurs when the vector $\vec{QR}$ is perpendicular to the line $L$. We define $\vec{QR}$ by subtracting the coordinates of $Q$ from $R$:

$$
\vec{QR} = ((\lambda+2)-5)\hat{i} + ((-\lambda+3)-3)\hat{j} + ((-\lambda+1)-8)\hat{k} = (\lambda-3)\hat{i} - \lambda\hat{j} - (\lambda+7)\hat{k}
$$

For $\vec{QR}$ to be perpendicular to the line, the dot product $\vec{QR} \cdot \vec{d}$ must be zero:

$$
(\lambda-3)(1) + (-\lambda)(-1) + (-(\lambda+7))(-1) = 0
$$

Simplifying the equation, we get $\lambda - 3 + \lambda + \lambda + 7 = 0$, which leads to $3\lambda + 4 = 0$. Thus, we find $\lambda = -\frac{4}{3}$.

### Phase 4: The Final Calculation

Substituting $\lambda = -\frac{4}{3}$ into our expression for $\vec{QR}$:

$$
\vec{QR} = \left(-\frac{4}{3}-3ight)\hat{i} - \left(-\frac{4}{3}ight)\hat{j} - \left(-\frac{4}{3}+7ight)\hat{k} = -\frac{13}{3}\hat{i} + \frac{4}{3}\hat{j} - \frac{17}{3}\hat{k}
$$

The square of the distance $\alpha$ is the squared magnitude of this vector:

$$
\alpha^2 = |\vec{QR}|^2 = \left(-\frac{13}{3}ight)^2 + \left(\frac{4}{3}ight)^2 + \left(-\frac{17}{3}ight)^2 = \frac{169 + 16 + 289}{9} = \frac{474}{9}
$$

Finally, we calculate $3\alpha^2$:

$$
3\alpha^2 = 3 	imes \frac{474}{9} = \frac{474}{3} = 158
$$

The final result is **158**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let a line $L$ pass through the point $P (2, 3, 1)$ and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z$ . If the distance of $L$ from the point $(5, 3, 8)$ is $α$ , then $3 α^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Visualizing the 3D Geometry

Finding the Direction Vector $d$

Setting up the Cross Product

Calculating the Direction Vector

Equation of Line $L$

General Point $R$ on Line $L$

Vector $QR$ and Perpendicularity

Applying the Dot Product Condition

Solving for $λ$

Finding the Distance Vector $QR$

Calculating $α^{2}$

Final Calculation of $3 α^{2}$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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