Let L₁: (x-1)/3 = (y-1)/-1 = (z+1)/0 and L₂: (x-2)/2 = y/0 = (z+4)/α, α ∈ R, be two lines, which int

Q: Wait, why do we reject the negative root here? Or do we even have one?

In this specific problem, the linear equations for the intersection point yield a unique solution for $\lambda$ and $\mu$. There are no quadratic equations involved, so we don't need to worry about rejecting roots. The geometry is clean and deterministic!

Q: Why can't we just use the distance formula between $A$ and $L_{2}$?

The distance formula between a point and a line gives the perpendicular distance $AP$. However, the problem asks for $PB$, which is the distance from the foot of the perpendicular $P$ to the intersection point $B$. These are two different segments of a right-angled triangle $\triangle APB$.

Q: Is the projection formula always valid for finding $PB$?

Yes, absolutely! Since $P$ is the foot of the perpendicular from $A$ to $L_{2}$, the vector $\vec{AP}$ is perpendicular to the line $L_{2}$. The vector $\vec{AB}$ can be decomposed into $\vec{AP} + \vec{PB}$. Since $\vec{AP}$ is perpendicular to the line, the length $PB$ is indeed the magnitude of the projection of $\vec{AB}$ onto the direction vector of $L_{2}$.

Question

Let $L_{1}:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_{2}:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}$, $\alpha\in R$, be two lines, which intersect at the point B. If P is the foot of perpendicular from the point $A(1,1,-1)$ on $L_{2}$ then the value of $26\alpha(PB)^{2}$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space, looking at two infinite lines, $L_{1}$ and $L_{2}$. They are not parallel; they are destined to meet at a single, precise location we call $B$.

To find $B$, we must translate the geometric description into the language of algebra. We represent any point on $L_{1}$ as $(3\lambda + 1, -\lambda + 1, -1)$ and any point on $L_{2}$ as $(2\mu + 2, 0, \alpha\mu - 4)$.

Because these lines intersect at $B$, the coordinates must be identical at that point.

### The Intersection Point

We begin by equating the $y$-coordinates: $-\lambda + 1 = 0$. This immediately reveals that $\lambda = 1$.

With $\lambda$ in hand, we turn to the $x$-coordinates: $3\lambda + 1 = 2\mu + 2$. Substituting $\lambda = 1$, we get $3(1) + 1 = 2\mu + 2$, which simplifies to $4 = 2\mu + 2$, giving us $\mu = 1$.

Now, we have the key to the $z$-coordinate. Equating the $z$-components: $-1 = \alpha\mu - 4$. Substituting $\mu = 1$, we find $-1 = \alpha - 4$, which leads us to the value $\alpha = 3$.

The intersection point $B$ is found by plugging $\lambda = 1$ into the parametric form of $L_{1}$, yielding $B = (4, 0, -1)$.

### The Vector $\vec{AB}$ and the Foot of the Perpendicular

Now, consider the point $A(1, 1, -1)$. We draw a vector $\vec{AB}$ from $A$ to $B$.

The components are $\vec{AB} = (4-1, 0-1, -1-(-1)) = (3, -1, 0)$. We are looking for the foot of the perpendicular $P$ from $A$ onto $L_{2}$.

The segment $PB$ is the projection of $\vec{AB}$ onto the line $L_{2}$. The direction vector of $L_{2}$ is $\vec{d} = (2, 0, \alpha)$. Since we found $\alpha = 3$, our direction vector is $\vec{d} = (2, 0, 3)$.

### The Final Calculation

The length $PB$ is the scalar projection of $\vec{AB}$ onto $\vec{d}$, given by:

$$PB = \frac{|\vec{AB} \cdot \vec{d}|}{|\vec{d}|}$$

The dot product $\vec{AB} \cdot \vec{d} = (3)(2) + (-1)(0) + (0)(3) = 6$. The magnitude $|\vec{d}|$ is:

$$|\vec{d}| = \sqrt{2^2 + 0^2 + 3^2} = \sqrt{4 + 0 + 9} = \sqrt{13}$$

Thus, $PB = \frac{6}{\sqrt{13}}$. We need the value of $26\alpha(PB)^{2}$.

Squaring $PB$, we get $(PB)^{2} = \frac{36}{13}$. Finally, the calculation proceeds as:

$$26 	imes 3 	imes \frac{36}{13} = 2 	imes 3 	imes 36 = 216$$

We have navigated the 3D space, solved for the unknown parameter, and calculated the projection with precision. The answer, **216**, is the elegant conclusion to our geometric adventure.

Enter Numerical Value:

Visualized Solution

Visualizing the Lines $L_{1}$ and $L_{2}$

Parametric Form of $L_{1}$

Parametric Form of $L_{2}$

Equating $y$ -coordinates

Solving for $λ$ and $μ$

Equating $z$ -coordinates and solving for $α$

Finding the Intersection Point $B$

Identifying Point $A$ and Vector $A B$

Concept of Foot of Perpendicular $P$

Direction Vector of $L_{2}$

Calculating the Projection $P B$

Finding $P B^{2}$

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Intersection Point

The Vector $A B$ and the Foot of the Perpendicular

The Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Parametric Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Parametric Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinates

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z-coordinates and solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Identifying Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Concept of Foot of Perpendicular .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Direction Vector of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Calculating the Projection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PB

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PB2