If the points with position vectors αi + 10j + 13k, 6i + 11j + 11k, (9/2)i + βj - 8k are collinear,

Q: Why can't we just use the slope formula like we do in 2D coordinate geometry?

That is a brilliant question! In 2D, we have a single slope $m = \frac{y_2 - y_1}{x_2 - x_1}$. However, in 3D space, a line doesn't have a single 'slope' in the same way. Instead, we use direction ratios. While you *could* use the condition that the direction ratios of $AB$ and $BC$ are proportional, the Section Formula is often more elegant because it handles the coordinates directly without needing to worry about ratios of ratios. It keeps your algebra clean and linear.

Q: Do I really need to find the individual values of $\alpha$ and $\beta$?

Not necessarily! Notice the target expression: $(19\alpha - 6\beta)^2$. In many JEE Advanced problems, the expression is designed to be solved by finding specific linear combinations of variables rather than the variables themselves. If you find $19\alpha$ and $6\beta$ directly, you save time and reduce the risk of arithmetic errors. Always keep an eye on the finish line before you start running!

Q: What if I get a negative value for $k$? Does that mean I made a mistake?

Not at all! A positive $k$ implies internal division (the point $B$ lies between $A$ and $C$), while a negative $k$ implies external division (the point $B$ lies outside the segment $AC$). The math is robust; it will tell you the truth regardless of your assumptions. In this specific problem, the geometry dictates the result, so trust your algebraic derivation.

Question

If the points with position vectors $\alpha\hat{i} + 10\hat{j} + 13\hat{k}$, $6\hat{i} + 11\hat{j} + 11\hat{k}$, $\frac{9}{2}\hat{i} + \beta\hat{j} - 8\hat{k}$ are collinear, then $(19\alpha - 6\beta)^2$ is equal to

Options:

(A) 36

(B) 25

(C) 49

(D) 16

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant rigidity of 3D space. When we talk about collinear points, we are talking about the most fundamental structure in geometry: the straight line.

Imagine three points, $A$, $B$, and $C$, floating in a 3D coordinate system. If they are collinear, they are bound by a single, unbreakable rule: they must share the same direction.

We are given three points with position vectors:
$A = (\alpha, 10, 13)$
$B = (6, 11, 11)$
$C = (\frac{9}{2}, \beta, -8)$

Our mission is to find the value of $(19\alpha - 6\beta)^2$. Instead of brute-forcing vectors, we will use the **Section Formula**, the most powerful tool in our arsenal for collinearity problems.

### The Anchor: Why the $z$-coordinate Matters

When you look at the coordinates of $A$, $B$, and $C$, you might feel overwhelmed by the variables $\alpha$ and $\beta$. But look closer; the $z$-coordinates are $13$, $11$, and $-8$. They are completely free of variables.

We assume point $B$ divides the segment $AC$ in the ratio $k:1$. The Section Formula tells us that the $z$-coordinate of $B$ is given by:

$$z_B = \frac{k z_C + z_A}{k + 1}$$

Substituting our known values, we get:

$$11 = \frac{k(-8) + 13}{k + 1}$$

Multiplying both sides by $(k+1)$, we get $11(k+1) = -8k + 13$. Expanding this, we have $11k + 11 = -8k + 13$. Bringing the $k$ terms to one side, we find $19k = 2$, which gives us our ratio $k = \frac{2}{19}$.

### Unlocking the Unknowns

Now that we have $k = \frac{2}{19}$, we can find $\alpha$ and $\beta$ using the same logic. Let's tackle the $x$-coordinate first:

$$x_B = \frac{k x_C + x_A}{k + 1}$$

Substituting $x_B = 6$, $x_C = \frac{9}{2}$, and $x_A = \alpha$, we have:

$$6 = \frac{(\frac{2}{19})(\frac{9}{2}) + \alpha}{\frac{2}{19} + 1}$$

The denominator $\frac{2}{19} + 1$ is $\frac{21}{19}$, and the numerator simplifies to $\frac{9}{19} + \alpha$. Our equation becomes:

$$6 = \frac{\frac{9}{19} + \alpha}{\frac{21}{19}}$$

Multiplying both sides by $\frac{21}{19}$ and then by $19$ to clear the denominators, we get $126 = 9 + 19\alpha$. This leads us directly to $19\alpha = 117$.

### The Final Stretch

We repeat this exact process for the $y$-coordinate to find $6\beta$:

$$y_B = \frac{k y_C + y_A}{k + 1}$$

Substituting $y_B = 11$, $y_C = \beta$, and $y_A = 10$:

$$11 = \frac{(\frac{2}{19})\beta + 10}{\frac{21}{19}}$$

Multiplying by the denominator $\frac{21}{19}$ and then by $19$ to eliminate the fractions:

$$231 = 2\beta + 190$$

$$2\beta = 41$$

Since we need $6\beta$, we multiply by $3$ to obtain $6\beta = 123$.

### The Grand Finale

We have arrived at the destination. We need to calculate $(19\alpha - 6\beta)^2$. Substituting our derived values $19\alpha = 117$ and $6\beta = 123$:

$$(117 - 123)^2 = (-6)^2 = 36$$

The complexity of the 3D coordinates collapses into a simple, elegant integer. The final answer is **36**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the points with position vectors $α \hat{i} + 10 \hat{j} + 13 \hat{k}$ , $6 \hat{i} + 11 \hat{j} + 11 \hat{k}$ , $\frac{9}{2} \hat{i} + β \hat{j} - 8 \hat{k}$ are collinear, then $(19 α - 6 β)^{2}$ is equal to

Select Answer:

Visualized Solution

Identifying the Points

The Concept of Collinearity

Setting up the $z$ -coordinate

Solving for $k$

Setting up the $x$ -coordinate

Calculating $19 α$

Setting up the $y$ -coordinate

Calculating $6 β$

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Anchor

Unlocking the Unknowns

The Final Stretch

The Grand Finale

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