Let A(t) = f1(t)i + f2(t)j and B(t) = g1(t)i + g2(t)j, t ∈ [0, 1], where f1, f2, g1, g2 are continuo

Q: Why do we need the Intermediate Value Theorem? Can't we just see that they swapped positions?

Intuition is a wonderful starting point, but in JEE Advanced, we must bridge the gap between intuition and rigorous proof. Simply observing that they swapped positions tells us they *must* have crossed, but the Intermediate Value Theorem (IVT) provides the mathematical bedrock that proves this crossing is inevitable, not just a coincidence.

Q: Does the continuity of the functions $f_1, f_2, g_1, g_2$ really matter?

It is the most important part! Without the guarantee of continuity, the function $h(t)$ could theoretically 'jump' from a negative value to a positive value without ever touching zero. Continuity ensures the path is unbroken, making the zero-crossing a mathematical certainty.

Q: What if $h(t)$ was zero at the endpoints $t=0$ or $t=1$?

That would actually make the problem even simpler! If $h(0)=0$ or $h(1)=0$, the vectors are parallel at the very start or the very end. The problem asks us to show they are parallel for *some* $t$, so finding a zero at the boundary is just as valid as finding one in the middle.

Question

Let $\vec{A}(t) = f_1(t)\hat{i} + f_2(t)\hat{j}$ and $\vec{B}(t) = g_1(t)\hat{i} + g_2(t)\hat{j}, t \in [0, 1]$, where $f_1, f_2, g_1, g_2$ are continuous functions. If $\vec{A}(t)$ and $\vec{B}(t)$ are nonzero vectors for all $t$ and $\vec{A}(0) = 2\hat{i} + 3\hat{j}, \vec{A}(1) = 6\hat{i} + 2\hat{j}, \vec{B}(0) = 3\hat{i} + 2\hat{j}$ and $\vec{B}(1) = 2\hat{i} + 6\hat{j}$. Then show that $\vec{A}(t)$ and $\vec{B}(t)$ are parallel for some $t$.

Sigma Percentile · Accepted Answer

### The Dance of the Vectors: A Journey Through Continuity

Imagine, for a moment, that you are standing on a vast, flat plane. In front of you, two arrows—let us call them $\vec{A}(t)$ and $\vec{B}(t)$—are performing a silent, graceful dance. As time $t$ flows from $0$ to $1$, these vectors stretch, shrink, and rotate.

At the beginning, at $t=0$, $\vec{A}(0) = 2\hat{i} + 3\hat{j}$ and $\vec{B}(0) = 3\hat{i} + 2\hat{j}$. They are distinct, pointing in different directions.

But as the clock ticks toward $t=1$, they transform into $\vec{A}(1) = 6\hat{i} + 2\hat{j}$ and $\vec{B}(1) = 2\hat{i} + 6\hat{j}$. Notice something fascinating? They have swapped their relative orientations. It is as if they have passed through each other.

But how can we prove, with absolute mathematical certainty, that at some precise moment in this dance, they were perfectly parallel? This is not just a geometry problem; it is a story about the power of continuity.

### The Geometry of Parallelism

To solve this, we must first define what it means for two vectors to be parallel. In a two-dimensional plane, two vectors $\vec{u} = x_1\hat{i} + y_1\hat{j}$ and $\vec{v} = x_2\hat{i} + y_2\hat{j}$ are parallel if they point in the same or opposite directions.

Geometrically, this means the area of the parallelogram they span is zero. Algebraically, this is equivalent to saying their cross product magnitude vanishes. We look for the condition where the determinant of their components is zero:

$$x_1 y_2 - x_2 y_1 = 0$$

This is our target. If we can find a time $t$ where this expression equals zero, we have found our moment of alignment.

### Constructing the Bridge: The Function $h(t)$

Let us define a function $h(t)$ that tracks this determinant over time:

$$h(t) = f_1(t)g_2(t) - f_2(t)g_1(t)$$

Here, $f_1(t)$ and $f_2(t)$ are the components of $\vec{A}(t)$, and $g_1(t)$ and $g_2(t)$ are the components of $\vec{B}(t)$. This function $h(t)$ is our "parallelism detector."

When $h(t) > 0$, the vectors are oriented one way; when $h(t) < 0$, they are oriented the other. If $h(t) = 0$, they are parallel.

### The Calculation: Testing the Boundaries

Now, let us test our detector at the start and the end of the interval. At $t=0$, we substitute the given values:

$$h(0) = (2)(2) - (3)(3) = 4 - 9 = -5$$

Our detector reads $-5$. It is negative. Now, let us jump to the end, at $t=1$:

$$h(1) = (6)(6) - (2)(2) = 36 - 4 = 32$$

Our detector now reads $32$. It is positive. We have moved from a negative value to a positive value. But what happened in between?

### The Hero of the Story: The Intermediate Value Theorem

This is where the magic of calculus enters the scene. The problem states that $f_1, f_2, g_1, g_2$ are continuous functions. Because the sum, difference, and product of continuous functions are also continuous, our function $h(t)$ is guaranteed to be continuous on the interval $[0, 1]$.

Think of $h(t)$ as a path drawn on a piece of paper. We know the path starts at $-5$ and ends at $32$. To get from a negative number to a positive number without lifting your pen—without any breaks or jumps—you **must** cross the zero line.

This is the essence of the **Intermediate Value Theorem (IVT)**. It tells us that for any value $k$ between $h(0)$ and $h(1)$, there exists at least one $t \in (0, 1)$ such that $h(t) = k$. Since $0$ is between $-5$ and $32$, there must be a time $t$ where $h(t) = 0$.

### Conclusion: The Inevitable Alignment

And there it is. The moment $h(t) = 0$, the vectors $\vec{A}(t)$ and $\vec{B}(t)$ are parallel. We didn't need to solve for the exact time $t$ or find the explicit equations of motion.

We only needed to understand the nature of the movement. By analyzing the boundaries and trusting the continuity of the system, we have proven that these two vectors, in their journey from $t=0$ to $t=1$, were destined to align.

You have just used the power of topology and calculus to solve a problem of geometry. Keep this perspective—that math is not just about numbers, but about the behavior of systems—and you will conquer any problem the JEE throws your way.

Visualized Solution

Visualizing the Vectors at $t = 0$

Visualizing the Vectors at $t = 1$

The Condition for Parallelism

Defining the Function $h (t)$

Raw Setup for $h (0)$

Atomic Compute for $h (0)$

Raw Setup for $h (1)$

Atomic Compute for $h (1)$

The Logic Bridge of Continuity

Applying the Intermediate Value Theorem

The Final Conclusion

The Sigma Insight: Vector (Cross) Product

The Dance of the Vectors

The Geometry of Parallelism

Constructing the Bridge

The Calculation

The Hero of the Story

Conclusion

Similar Questions

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If $a, b, c$ and $d$ are distinct vectors such that $a \times c = b \times d$ and $a \times b = c \times d$ . Prove that $(a - d) \cdot (b - c) \neq = 0$ i.e. $a \cdot b + d \cdot c \neq = d \cdot b + a \cdot c$ .

Visualized Solution

Visualizing the Vectors at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=0

Visualizing the Vectors at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=1

The Condition for Parallelism

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(t)

Raw Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(0)

Atomic Compute for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(0)

Raw Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(1)

Atomic Compute for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(1)