Let a plane P contain two lines r = i + λ(i + j), λ ∈ R and r = -j + μ(j - k), μ ∈ R. If Q(α, β, γ)

Q: Wait, why do we reject the negative root here? Or rather, does the sign of the normal vector matter?

Great question! The sign of the normal vector $\vec{n}$ does not matter at all. Whether you choose $\vec{n}$ or $-\vec{n}$, the equation of the plane $ax + by + cz + d = 0$ will simply be scaled by $-1$, which represents the exact same plane. The formula for the foot of the perpendicular is self-correcting; the ratio on the right side will adjust its sign accordingly to ensure the correct coordinates for $Q$.

Q: How do we know for sure that these two lines are coplanar? Could they be skew lines?

In this specific problem, the question explicitly states that the plane $P$ contains both lines. This is a gift! It guarantees that the lines are coplanar and intersect or are parallel. If the problem hadn't stated this, you would first need to check the condition for coplanarity: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = 0$. Since we are told they lie in the plane, we can proceed directly to finding the normal.

Q: Why can't we just use the distance formula to find the foot of the perpendicular?

The distance formula only gives you the length of the perpendicular segment, not the coordinates of the point where it hits the plane. To find the foot $Q$, you need the direction of the normal vector. The formula $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = k$ essentially tells us that the line segment $MQ$ is parallel to the normal vector $\vec{n}$, which is the geometric definition of a perpendicular drop.

Question

Let a plane $P$ contain two lines $\vec{r} = \hat{i} + \lambda(\hat{i} + \hat{j}), \lambda \in R$ and $\vec{r} = -\hat{j} + \mu(\hat{j} - \hat{k}), \mu \in R$. If $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1, 0, 1)$ to $P$, then $3(\alpha + \beta + \gamma)$ equals

Sigma Percentile · Accepted Answer

### The Architecture of a Plane: A 3D Odyssey

Imagine you are standing in a vast, empty 3D space. You are given two lines, $L_1$ and $L_2$, and told they define a flat, infinite surface—a plane $P$.

Your mission is to find the exact point $Q$ on this plane that acts as the 'shadow' of a point $M$ hovering above it. This is not just algebra; it is the art of 3D navigation.

### Phase 1: Defining the Foundation

Every plane is defined by its orientation, and its orientation is defined by its normal vector $\vec{n}$. Think of the normal vector as a flagpole sticking straight out of the surface.

If we know the direction of this flagpole, we know the tilt of the entire plane. We are given two lines:
$L_1: \vec{r} = \hat{i} + \lambda(\hat{i} + \hat{j})$
$L_2: \vec{r} = -\hat{j} + \mu(\hat{j} - \hat{k})$

From these, we extract the direction vectors: $\vec{d_1} = (1, 1, 0)$ and $\vec{d_2} = (0, 1, -1)$. Because these lines lie in the plane, their direction vectors are essentially 'flat' against the surface.

To find the normal vector $\vec{n}$, we need a vector perpendicular to both $\vec{d_1}$ and $\vec{d_2}$. The cross product is our tool of choice: $\vec{n} = \vec{d_1} 	imes \vec{d_2}$.

Calculating the determinant:
$$
\vec{n} = 
\begin{vmatrix} 
\hat{i} & \hat{j} & \hat{k} \ 
1 & 1 & 0 \ 
0 & 1 & -1 
\end{vmatrix} = -\hat{i} + \hat{j} + \hat{k}
$$

### Phase 2: Constructing the Plane

Now that we have the normal vector $\vec{n} = (-1, 1, 1)$, we need a single point to anchor our plane. Looking at $L_1$, the position vector $\hat{i}$ tells us that the point $A(1, 0, 0)$ is on the line, and thus on the plane.

Using the point-normal form, $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$, we substitute our values:
$$-1(x-1) + 1(y-0) + 1(z-0) = 0$$

Simplifying this, we arrive at the elegant equation of our plane:
$$x - y - z - 1 = 0$$
This equation is the DNA of our plane; every point on it must satisfy this relationship.

### Phase 3: The Perpendicular Drop

We are now at the climax of our journey. We have a point $M(1, 0, 1)$ and we want to drop a perpendicular to the plane $x - y - z - 1 = 0$.

The foot of this perpendicular, $Q(\alpha, \beta, \gamma)$, is the point where the line passing through $M$ and perpendicular to the plane intersects the plane itself. We use the standard formula:

$$
\frac{\alpha - 1}{1} = \frac{\beta - 0}{-1} = \frac{\gamma - 1}{-1} = -\frac{1(1) - 1(0) - 1(1) - 1}{1^2 + (-1)^2 + (-1)^2}
$$

The right side simplifies beautifully:
$$
-\frac{1 - 0 - 1 - 1}{1 + 1 + 1} = -\frac{-1}{3} = \frac{1}{3}
$$

### The Final Resolution

With the ratio equal to $\frac{1}{3}$, we can solve for each coordinate individually:

$\alpha - 1 = \frac{1}{3} \implies \alpha = \frac{4}{3}$

$-\beta = \frac{1}{3} \implies \beta = -\frac{1}{3}$

$\frac{\gamma - 1}{-1} = \frac{1}{3} \implies \gamma - 1 = -\frac{1}{3} \implies \gamma = \frac{2}{3}$

The sum $\alpha + \beta + \gamma = \frac{4}{3} - \frac{1}{3} + \frac{2}{3} = \frac{5}{3}$.

The question asks for $3(\alpha + \beta + \gamma)$, which is $3 	imes \frac{5}{3} = 5$.

**The final answer is 5.**

Enter Numerical Value:

Visualized Solution

Visualizing the Lines and Plane

Finding the Normal Vector $n$

Calculating the Cross Product

Establishing the Plane Equation

Simplifying the Plane Equation

The Point $M$ and its Foot $Q$

Formula for Foot of Perpendicular

Substituting Values into the Formula

Evaluating the Ratio

Solving for Coordinates of $Q$

Calculating the Final Sum

Final Answer

The Sigma Insight: Equation of a Plane

The Architecture of a Plane

Phase 1

Phase 2

Phase 3

The Final Resolution

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