A line with direction ratios 2, 1, 2 meets the lines x = y + 2 = z and x + 2 = 2y = 2z respectively

Q: Wait, why can't we use the same parameter $t$ for both lines? That would make the algebra so much easier!

I understand the temptation! But think about the geometry: if you use the same parameter $t$ for both lines, you are forcing the points $P$ and $Q$ to be at the exact same 'distance' from the origin or reference point along their respective lines. That is a massive, incorrect assumption. $P$ and $Q$ are independent entities; they need independent parameters to roam freely on their respective lines. Always use distinct parameters for distinct lines.

Q: Why do we equate the ratios of direction vectors? Is that always allowed?

Absolutely! Two vectors are parallel if and only if one is a scalar multiple of the other. If vector $\vec{v_1} = (a_1, b_1, c_1)$ and $\vec{v_2} = (a_2, b_2, c_2)$ are parallel, then $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. This is the algebraic key to unlocking the coordinates. Since the line $PQ$ is defined by the direction ratios $(2, 1, 2)$, any vector lying on that line must be proportional to $(2, 1, 2)$.

Q: Is it a coincidence that the foot of the perpendicular $F$ is the same as point $P$? I feel like I did something wrong.

It is a beautiful geometric coincidence! In this specific problem, the geometry of the lines and the position of point $A$ align perfectly such that the shortest path from $A$ to the line $PQ$ lands exactly on the intersection point $P$. Don't worry, you didn't do anything wrong—you actually discovered a hidden elegance in the problem. Always trust your math; if the calculation leads you there, it is correct!

Question

A line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ respectively at the point $P$ and $Q$. if the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$, then $l^2$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are navigating the architecture of 3D space. Imagine you are standing in a vast, empty room where two lines, $L_1$ and $L_2$, are suspended in the air.

A third line, our line of interest $PQ$, pierces through both of them. Our goal is to find the exact coordinates of the points $P$ and $Q$ where this piercing occurs, and then measure the distance from a specific point $A(1, 2, 12)$ to this line $PQ$.

### The Parametric Dance

To conquer 3D geometry, we must master the art of parameterization. We cannot simply guess where $P$ and $Q$ are; we must let them 'roam' along their lines.

For $L_1$, defined by $x = y + 2 = z$, we introduce a parameter $t$. This gives us the general point $P(t, t-2, t)$.

For $L_2$, defined by $x + 2 = 2y = 2z$, we rewrite the equation as $\frac{x+2}{2} = y = z$. Introducing a different parameter $s$, we obtain the general point $Q(2s-2, s, s)$. We use different parameters because $P$ and $Q$ are independent.

### The Bridge of Proportionality

We calculate the direction ratios of the line segment $PQ$ by subtracting the coordinates of $Q$ from $P$. This yields the vector $\vec{PQ} = (t - 2s + 2, t - s - 2, t - s)$.

The problem explicitly states that the direction ratios of $PQ$ are $(2, 1, 2)$. Since our calculated vector must be parallel to this given direction, their components must be proportional:

$$\frac{t - 2s + 2}{2} = \frac{t - s - 2}{1} = \frac{t - s}{2}$$

By equating the first and third parts, the $t$ terms vanish, leading to $s = 2$. Substituting this back into the second and third parts, we solve for $t$ and find $t = 6$. The abstract points are now concrete: $P(6, 4, 6)$ and $Q(2, 2, 2)$.

### The Perpendicular Drop

We have a point $A(1, 2, 12)$ and we need the perpendicular distance to line $PQ$. We define a general point $F$ on $PQ$ using a new parameter $\lambda$.

Since the line $PQ$ passes through $Q(2, 2, 2)$ with direction $(2, 1, 2)$, we define $F = (2\lambda + 2, \lambda + 2, 2\lambda + 2)$. We construct the vector $\vec{AF} = F - A = (2\lambda + 1, \lambda, 2\lambda - 10)$.

Because $AF$ is perpendicular to $PQ$, their dot product must be zero:

$$\vec{AF} \cdot \vec{PQ} = 0 \Rightarrow 2(2\lambda + 1) + 1(\lambda) + 2(2\lambda - 10) = 0$$

Solving this yields $9\lambda = 18$, so $\lambda = 2$. Substituting $\lambda = 2$ back into $F$, we find the foot of the perpendicular is $F(6, 4, 6)$, which coincides with point $P$.

### Final Calculation

Finally, we calculate the distance $l$ between $A(1, 2, 12)$ and $F(6, 4, 6)$. The square of the distance is:

$$l^2 = (6-1)^2 + (4-2)^2 + (6-12)^2 = 25 + 4 + 36 = 65$$

Thus, the distance is **$l = \sqrt{65}$**. You have successfully navigated the 3D space and mastered this geometry challenge.

Enter Numerical Value:

Visualized Solution

Visualizing the 3D Geometry

General Point on $L_{1}$

General Point on $L_{2}$

Direction Ratios of $P Q$

Comparing Direction Ratios

Solving for $s$ and $t$

Coordinates of $P$ and $Q$

The Perpendicular from Point $A$

General Point $F$ on $P Q$

Perpendicularity Condition

Coordinates of Foot $F$

Calculating the Distance Squared ( $l^{2}$ )

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Dance

The Bridge of Proportionality

The Perpendicular Drop

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the 3D Geometry

General Point on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

General Point on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

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