The points with position vectors 60i + 3j, 40i - 8j, ai - 52j are collinear if (a) a = -40 (b) a = 4

Q: Why can't we just use the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$? It seems faster.

You absolutely can! The slope method is essentially the same as the vector ratio method. However, in JEE Advanced, you will often encounter 3D problems where 'slope' isn't defined in the same way. Mastering the vector approach now prepares you for the 3D geometry that will appear later in your syllabus. Think of the vector method as a universal tool that works in 2D, 3D, and beyond.

Q: Does it matter if I use $\vec{AB}$ and $\vec{BC}$ instead of $\vec{AB}$ and $\vec{AC}$?

Great question! No, it does not matter. Any two vectors formed by these three points must be parallel. You could use $\vec{AB}$ and $\vec{BC}$, or $\vec{AC}$ and $\vec{BC}$. The result for $a$ will be identical. Choose the pair that makes the arithmetic easiest for you. The geometry remains invariant regardless of which vectors you choose.

Q: What if the denominator becomes zero? Is the ratio method still valid?

That is a sharp observation. If a component is zero, the ratio $\frac{u_x}{v_x}$ becomes undefined. In such cases, the cross-product method is your safest bet: $\vec{AB} \times \vec{AC} = 0$. This ensures that the area of the triangle formed by the three points is zero, which is the ultimate definition of collinearity. Always keep the cross-product in your back pocket for those tricky edge cases!

Question

The points with position vectors $60\hat{i} + 3\hat{j}, 40\hat{i} - 8\hat{j}, a\hat{i} - 52\hat{j}$ are collinear if

Options:

(A) $a = -40$

(B) $a = 40$

(C) $a = 20$

(D) none of these

Sigma Percentile · Accepted Answer

### The Geometry of Alignment: A Journey into Collinearity

Welcome, my dear student. Today, we are not just solving a problem about coordinates; we are exploring the profound concept of alignment.

In the world of JEE Advanced, collinearity is more than just points on a line—it is a test of your ability to translate geometric intuition into the rigorous language of vectors. Let us peel back the layers of this problem and see how the math reveals the truth.

### Phase 1: Defining the Landscape

Imagine you are standing on a coordinate plane. You have three markers: point $A$, point $B$, and point $C$. We are given their position vectors:

$$A = 60\hat{i} + 3\hat{j}$$

$$B = 40\hat{i} - 8\hat{j}$$

$$C = a\hat{i} - 52\hat{j}$$

Our mission is to find the value of $a$ that forces these three points to sit perfectly on a single, unbroken straight line. When we say 'collinear,' we are saying that the path from $A$ to $B$ must be the same path that continues to $C$.

There is no deviation, no turning, no bending. It is a singular, unified direction.

### Phase 2: The Vector Bridge

How do we translate this visual 'straightness' into algebra? We use vectors. If $A$, $B$, and $C$ are collinear, then the vector $\vec{AB}$ (the displacement from $A$ to $B$) and the vector $\vec{AC}$ (the displacement from $A$ to $C$) must be parallel.

They are essentially the same vector, just scaled by some constant factor. Mathematically, this means $\vec{AC} = k \vec{AB}$ for some scalar $k$.

Let us calculate these vectors. To find $\vec{AB}$, we subtract the position of $A$ from $B$:

$$\vec{AB} = \vec{B} - \vec{A} = (40 - 60)\hat{i} + (-8 - 3)\hat{j} = -20\hat{i} - 11\hat{j}$$

Now, let us find $\vec{AC}$:

$$\vec{AC} = \vec{C} - \vec{A} = (a - 60)\hat{i} + (-52 - 3)\hat{j} = (a - 60)\hat{i} - 55\hat{j}$$

Look at these two results. We have $\vec{AB} = -20\hat{i} - 11\hat{j}$ and $\vec{AC} = (a - 60)\hat{i} - 55\hat{j}$. For these to be parallel, their components must be proportional.

This is the 'Golden Ratio' of collinearity:

$$\frac{a - 60}{-20} = \frac{-55}{-11}$$

### Phase 3: The Algebra of Truth

I know that seeing fractions can sometimes make us anxious, but look at the right side of our equation. It is a gift! The negative signs cancel out, and $55$ divided by $11$ is a clean, beautiful $5$.

Our equation simplifies instantly:

$$\frac{a - 60}{-20} = 5$$

Now, we are in the home stretch. We need to isolate $a$. We multiply both sides by $-20$:

$$a - 60 = 5 	imes (-20)$$

$$a - 60 = -100$$

Finally, we add $60$ to both sides to reveal the value of $a$:

$$a = -100 + 60$$

**$$a = -40$$**

### The Conclusion: Why This Matters

There it is. When $a = -40$, the points align perfectly. But I want you to take a moment to appreciate what you just did.

You didn't just solve for a variable; you used the power of vector proportionality to enforce a geometric constraint. This is the essence of JEE Advanced physics and mathematics.

Whenever you face a problem involving collinearity in the future, remember this journey. Don't rush to the formula. Visualize the vectors.

Ask yourself: 'Are these two displacements pointing in the same direction?' If they are, their components must be in harmony. Keep practicing this, and you will find that even the most complex problems become a simple conversation between you and the geometry.

Animated Solution for Mathematics - Vector Algebra: The points with position vectors $60 \hat{i} + 3 \hat{j}, 40 \hat{i} - 8 \hat{j}, a \hat{i} - 52 \hat{j}$ are collinear if

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Visualized Solution

$D e f inin g t h e P o in t s A, B, C$

$T h e C o n d i t i o n f or C o l l in e a r i t y$

$C a l c u l a t in g V ec t or A B$

$C a l c u l a t in g V ec t or A C$

$S e tt in g u p t h e P r o p or t i o n$

$S im pl i f y in g t h e P r o p or t i o n$

$S o l v in g f or a$

$F ina l R es u l t$

The Sigma Insight: Components of a Vector

The Geometry of Alignment

Phase 1

Phase 2

Phase 3

The Conclusion

Similar Questions

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The points with position vectors 60i^+3j^​,40i^−8j^​,ai^−52j^​ are collinear if

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Defining the Points A,B,C

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

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

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

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

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

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

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