Two vectors x and y have equal magnitude. The magnitude of (x - y) is n times the magnitude of (x +

Q: Why is the formula for $|\mathbf{x} - \mathbf{y}|$ having a minus sign?

The vector $\mathbf{x} - \mathbf{y}$ can be thought of as $\mathbf{x} + (-\mathbf{y})$. The angle between $\mathbf{x}$ and $-\mathbf{y}$ is $180^\circ - \theta$. Since $\cos(180^\circ - \theta) = -\cos\theta$, the formula becomes $x^2 + y^2 - 2xy\cos\theta$.

Q: Can we solve this problem by assuming $\mathbf{x}$ and $\mathbf{y}$ along the axes?

You could assume $\mathbf{x}$ is along the x-axis, but you cannot assume $\mathbf{y}$ is along the y-axis because the angle $\theta$ between them is unknown and is exactly what we need to find!

Q: What does it mean physically if $n=1$?

If $n=1$, then $|\mathbf{x} - \mathbf{y}| = |\mathbf{x} + \mathbf{y}|$. This implies the diagonals of the parallelogram are equal, which only happens when the parallelogram is a rectangle. Since the sides are equal, it's a square, meaning the angle is $90^\circ$.

Q: Why did we square the initial equation?

The magnitude of a vector sum involves a square root (e.g., $\sqrt{x^2 + y^2 + 2xy\cos\theta}$). Squaring both sides removes the square roots, making the algebraic manipulation much easier.

Q: Can we cancel $2x^2$ from both sides? What if $x=0$?

If $x=0$, then both vectors are zero vectors, and the angle between them is undefined. In physics problems like this, it's implicitly assumed that the vectors are non-zero, so dividing by $x^2$ is perfectly valid.

Question

Two vectors $\mathbf{x}$ and $\mathbf{y}$ have equal magnitude. The magnitude of $(\mathbf{x} - \mathbf{y})$ is $n$ times the magnitude of $(\mathbf{x} + \mathbf{y})$. The angle between $\mathbf{x}$ and $\mathbf{y}$ is

Options:

(A) $\cos^{-1}\left(\frac{-n^2-1}{n^2-1}\right)$

(B) $\cos^{-1}\left(\frac{1-n^2}{1+n^2}\right)$

(C) $\cos^{-1}\left(\frac{n^2+1}{-n^2-1}\right)$

(D) $\cos^{-1}\left(\frac{n^2+1}{n^2-1}\right)$

Sigma Percentile · Accepted Answer

The problem of finding the angle between two vectors given a relationship between their sum and difference is a classic in kinematics and vector algebra. It tests your geometric intuition and your algebraic discipline. Let's dive into the fascinating physics and math behind this problem!

### Visualizing the Vectors

Imagine you are standing at the origin of a coordinate system. You draw two vectors, $\mathbf{x}$ and $\mathbf{y}$, starting from where you stand.

The problem states that these two vectors have the exact same magnitude. If you were to complete the parallelogram formed by these two vectors, you wouldn't just get any parallelogram—you would get a **rhombus**.

In this rhombus, the diagonal starting from the origin represents the vector sum, $\mathbf{x} + \mathbf{y}$. The other diagonal, connecting the tips of the two vectors, represents the vector difference, $\mathbf{x} - \mathbf{y}$.

### The Master Equation

We are given a very specific constraint: the magnitude of the difference is $n$ times the magnitude of the sum.

Mathematically, this is written as:
$$|\mathbf{x} - \mathbf{y}| = n |\mathbf{x} + \mathbf{y}|$$

Dealing with magnitudes directly often involves square roots, which can be messy. To make our lives easier, we square both sides of the equation:
$$|\mathbf{x} - \mathbf{y}|^2 = n^2 |\mathbf{x} + \mathbf{y}|^2$$

### The Power of the Parallelogram Law

Now, we bring in the heavy artillery: the **Parallelogram Law of Vector Addition**.

The squared magnitude of the sum of two vectors is given by:
$$|\mathbf{x} + \mathbf{y}|^2 = x^2 + y^2 + 2xy\cos	heta$$

Similarly, the squared magnitude of their difference is:
$$|\mathbf{x} - \mathbf{y}|^2 = x^2 + y^2 - 2xy\cos	heta$$

Substituting these into our squared constraint equation, we get:
$$x^2 + y^2 - 2xy\cos	heta = n^2(x^2 + y^2 + 2xy\cos	heta)$$

### Algebraic Simplification

This equation looks a bit intimidating, but remember our initial condition: the vectors have equal magnitudes! This means we can substitute $y = x$ everywhere in the equation.

Let's see the magic happen:
$$x^2 + x^2 - 2x^2\cos	heta = n^2(x^2 + x^2 + 2x^2\cos	heta)$$

Combining the like terms, we get:
$$2x^2 - 2x^2\cos	heta = n^2(2x^2 + 2x^2\cos	heta)$$

Notice how every single term has a $2x^2$ in it? Assuming our vectors are non-zero (which is a safe assumption in such problems), we can divide the entire equation by $2x^2$.

This beautifully simplifies our equation to:
$$1 - \cos	heta = n^2(1 + \cos	heta)$$

### The Final Angle

We are now in the home stretch. Our goal is to isolate $\cos	heta$. Let's expand the right side:
$$1 - \cos	heta = n^2 + n^2\cos	heta$$

Next, we group all the terms containing $\cos	heta$ on one side, and the constant terms on the other:
$$1 - n^2 = \cos	heta + n^2\cos	heta$$

Factoring out $\cos	heta$ on the right side gives:
$$1 - n^2 = \cos	heta(1 + n^2)$$

Finally, we divide by $(1 + n^2)$ to completely isolate $\cos	heta$:
$$\cos	heta = \frac{1 - n^2}{1 + n^2}$$

To find the angle $	heta$, we simply take the inverse cosine:
$$	heta = \cos^{-1}\left(\frac{1 - n^2}{1 + n^2}ight)$$

And there we have it! A beautiful geometric relationship distilled into a clean algebraic result.

Animated Solution for Physics - Kinematics: Two vectors $x$ and $y$ have equal magnitude. The magnitude of $(x - y)$ is $n$ times the magnitude of $(x + y)$ . The angle between $x$ and $y$ is

Select Answer:

Visualized Solution

$∣ x ∣ = ∣ y ∣$

$∣ x - y ∣ = n ∣ x + y ∣$

$∣ x - y ∣^{2} = n^{2} ∣ x + y ∣^{2}$

$y = x$

$\div 2 x^{2}$

$cos θ$

$θ = cos^{- 1} (\frac{1 - n ^{2}}{1 + n ^{2}})$

The Sigma Insight: Vector Addition, Subtraction, and Resolution

Visualizing the Vectors

The Master Equation

The Power of the Parallelogram Law

Algebraic Simplification

The Final Angle

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Animated Solution for Physics - Kinematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two vectors x and y have equal magnitude. The magnitude of (x−y) is n times the magnitude of (x+y). The angle between x and y is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣x∣=∣y∣

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣x−y∣=n∣x+y∣

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣x−y∣2=n2∣x+y∣2

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ÷2x2

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=cos−1(1+n21−n2​)