A line passes through A(4, -6, -2) and B(16, -2, 4). The point P(a, b, c) where a, b, c are non-nega

Q: Why did we simplify the direction ratios from $(12, 4, 6)$ to $(6, 2, 3)$? Does it change the line?

Great question! No, it does not change the line at all. Direction ratios define the *orientation* of the line, not its specific position. Think of it like a slope in 2D; whether you say the slope is $2/1$ or $4/2$, it's the same steepness. Simplifying makes the algebra much cleaner and reduces the chance of calculation errors.

Q: Why do we get two values for $\lambda$ ($\pm 3$)? Which one do I pick?

The parameter $\lambda$ represents the 'distance' along the line from point $A$. Since a line extends infinitely in both directions, there are two points on the line that are exactly $21$ units away from $A$—one in the direction of $B$ and one in the opposite direction. The problem gives us a constraint: $a, b, c$ must be non-negative. We test both values of $\lambda$ to see which one satisfies this condition.

Q: Can I use the distance formula directly without the parametric form?

You could, but it would be much harder! The parametric form $P = A + \lambda \vec{v}$ is a powerful tool because it reduces a 3D coordinate problem into a single-variable algebraic equation. It effectively 'flattens' the 3D line into a 1D number line, making the distance constraint trivial to solve.

Question

A line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$ where $a, b, c$ are non-negative integers, on the line $AB$ lies at a distance of 21 units, from the point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have two fixed points, $A(4, -6, -2)$ and $B(16, -2, 4)$, acting as anchors for a straight line stretching through the air.

To navigate this line, we first need to understand its orientation. We calculate the direction ratios by finding the difference between the coordinates of $B$ and $A$: $(16 - 4, -2 - (-6), 4 - (-2))$, which gives us $(12, 4, 6)$.

We can simplify these ratios to $(6, 2, 3)$ to make our future calculations elegant and manageable.

### The Parametric Bridge: Unlocking the Line

Now, how do we find any point $P$ on this line? We use the parametric form.

By setting the symmetric equation of the line equal to a parameter $\lambda$, we create a bridge:

$$ \frac{x - 4}{6} = \frac{y + 6}{2} = \frac{z + 2}{3} = \lambda $$

Every point $P(a, b, c)$ on this line can now be described as:

$$ a = 4 + 6\lambda, \quad b = -6 + 2\lambda, \quad c = -2 + 3\lambda $$

This $\lambda$ is our magic key; it tells us exactly how far we have traveled from point $A$ along the direction vector.

### The Constraint Hunt: Solving for $\lambda$

The problem states that point $P$ is exactly $21$ units away from $A$. Using the distance formula, we know that:

$$ AP = \sqrt{(6\lambda)^2 + (2\lambda)^2 + (3\lambda)^2} $$

Simplifying this, we get:

$$ \sqrt{36\lambda^2 + 4\lambda^2 + 9\lambda^2} = \sqrt{49\lambda^2} = 7|\lambda| $$

Setting $7|\lambda| = 21$, we find that $|\lambda| = 3$, meaning $\lambda$ can be $3$ or $-3$.

If $\lambda = 3$, then $P = (4 + 18, -6 + 6, -2 + 9) = (22, 0, 7)$.

If $\lambda = -3$, then $P = (4 - 18, -6 - 6, -2 - 9) = (-14, -12, -11)$.

The problem explicitly states that $a, b, c$ must be non-negative integers. Thus, we reject the second option and embrace **$P(22, 0, 7)$**.

### The Final Encounter: Calculating the Distance

With $P(22, 0, 7)$ secured, we introduce the final point $Q(4, -12, 3)$. Our mission is to find the distance $PQ$.

We deploy the distance formula one last time:

$$ PQ = \sqrt{(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2} $$

This simplifies to:

$$ PQ = \sqrt{18^2 + 12^2 + 4^2} = \sqrt{324 + 144 + 16} = \sqrt{484} $$

The square root of $484$ is $22$. The final distance is **$22$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Direction Ratios of Line $A B$

Calculating Direction Ratios

Equation of Line $A B$

Parametric Coordinates of $P$

Distance Constraint $A P = 21$

Solving for $λ$

Identifying Point $P$

Introducing Point $Q$

Setting up Distance $P Q$

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Bridge

The Constraint Hunt

The Final Encounter

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

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

Calculating Direction Ratios

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

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

Distance Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AP=21

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

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

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

Setting up Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ