Let a and b be the vectors of the same magnitude such that (|a+b|+|a-b|)/(|a+b|-|a-b|) = √2 + 1. The

Q: Why did we use Componendo and Dividendo? Can't we just solve it directly?

You absolutely could, but it would be algebraic suicide! Componendo and Dividendo is a classic JEE trick for handling ratios of the form $\frac{x+y}{x-y}$. It allows us to isolate the variables we need without getting bogged down in messy cross-multiplication. Always look for this pattern when you see a ratio of sums and differences.

Q: Why did we assume $|\vec{a}|^2 = |\vec{b}|^2 = k$?

This is a technique called 'algebraic hygiene.' By replacing the repeated magnitude terms with a single variable $k$, we reduce the cognitive load and prevent transcription errors. It makes the equations cleaner and helps you focus on the vector relationships rather than the notation.

Q: Is there a geometric way to solve this without all the algebra?

Yes! You could use the law of cosines or the properties of the rhombus diagonals directly. However, the algebraic method we used is more robust and less prone to visualization errors under exam pressure. Both paths lead to the same truth, but the algebraic path is a reliable workhorse.

Question

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|} = \sqrt{2} + 1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is :

Options:

(A) $2 + 4\sqrt{2}$

(B) $1 + \sqrt{2}$

(C) $2 + \sqrt{2}$

(D) $4 + 2\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane, holding two vectors, $\vec{a}$ and $\vec{b}$. The problem states that they have the same magnitude, which is a crucial piece of information.

When you place these two vectors tail-to-tail, they form the adjacent sides of a rhombus. Because the diagonals of a rhombus are represented by the vector sum $\vec{a}+\vec{b}$ and the vector difference $\vec{a}-\vec{b}$, this geometric insight serves as our starting point.

Let us define their squared magnitude as $k$, such that:

$$|\vec{a}|^2 = |\vec{b}|^2 = k$$

This simple substitution will be our best friend as we navigate the algebra ahead.

### The Algebraic Sword: Componendo and Dividendo

Now, consider the provided expression:

$$\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|} = \sqrt{2} + 1$$

This is a classic setup for the **Componendo and Dividendo** rule. This rule states that if $\frac{x}{y} = \frac{p}{q}$, then $\frac{x+y}{x-y} = \frac{p+q}{p-q}$.

By applying this to our equation, the cross terms in the numerator and denominator vanish. This leaves us with a much cleaner ratio:

$$\frac{|\vec{a}+\vec{b}|}{|\vec{a}-\vec{b}|} = 1 + \sqrt{2}$$

### The Bridge to Dot Products

We need to find the ratio $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$. To get there, we must bridge the gap between magnitudes and the dot product.

Squaring both sides of our simplified ratio is the logical next step. Squaring $(1+\sqrt{2})$ yields $3 + 2\sqrt{2}$. Thus, we have:

$$|\vec{a}+\vec{b}|^2 = (3+2\sqrt{2})|\vec{a}-\vec{b}|^2$$

Recall the fundamental vector identity:

$$|\vec{x} \pm \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 \pm 2\vec{a} \cdot \vec{b}$$

Substituting our $k$ value, the numerator becomes $2k + 2\vec{a} \cdot \vec{b}$ and the denominator becomes $2k - 2\vec{a} \cdot \vec{b}$.

### The Final Victory

We now solve the equation:

$$k + \vec{a} \cdot \vec{b} = (3+2\sqrt{2})(k - \vec{a} \cdot \vec{b})$$

Expand the right side carefully and group the terms involving $\vec{a} \cdot \vec{b}$ on one side and the terms involving $k$ on the other. This allows us to isolate the ratio $\frac{\vec{a} \cdot \vec{b}}{k}$:

$$\frac{\vec{a} \cdot \vec{b}}{k} = \frac{1}{\sqrt{2}}$$

Finally, we return to the expression we were asked to solve: $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$. Expanding this, we get:

$$\frac{2k + 2\vec{a} \cdot \vec{b}}{k} = 2 + 2\left(\frac{\vec{a} \cdot \vec{b}}{k}\right)$$

Substituting our value of $\frac{1}{\sqrt{2}}$, we arrive at the final result:

**$$2 + \sqrt{2}$$**

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be the vectors of the same magnitude such that $\frac{∣ a + b ∣ + ∣ a - b ∣}{∣ a + b ∣ - ∣ a - b ∣} = 2 + 1$ . Then $\frac{∣ a + b ∣ ^{2}}{∣ a ∣ ^{2}}$ is :

Select Answer:

Visualized Solution

Visualizing the Vectors

Applying Componendo and Dividendo

Simplifying the Ratio

Squaring for Magnitudes

Expanding using Vector Properties

Solving for the Dot Product

Final Simplification of the Ratio

Calculating the Final Answer

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Algebraic Sword

The Bridge to Dot Products

The Final Victory

Similar Questions

Let $a, b$ and $c$ be three unit vectors such that $∣ a - b ∣^{2} + ∣ a - c ∣^{2} = 8$ . Then $∣ a + 2 b ∣^{2} + ∣ a + 2 c ∣^{2}$ is equal to

Let $u, v, w$ be such that $∣ u ∣ = 1, ∣ v ∣ = 2, ∣ w ∣ = 3$ . If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v, w$ are perpendicular to each other then $∣ u - v + w ∣$ equals

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Let $a, b, c \in R$ be such that $a^{2} + b^{2} + c^{2} = 1$ . If $a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})$ , where $θ = \frac{π}{9}$ , then the angle between the vectors $a \hat{i} + b \hat{j} + c \hat{k}$ and $b \hat{i} + c \hat{j} + a \hat{k}$ is :

If $x$ and $y$ be two non-zero vectors such that $∣ x + y ∣ = ∣ x ∣$ and $2 x + λ y$ is perpendicular to $y$ , then the value of $λ$ is

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

Let $a$ and $b$ be two vectors such that $∣2 a + 3 b ∣ = ∣3 a + b ∣$ and the angle between $a$ and $b$ is $6 0^{\circ}$ . If $\frac{1}{8} a$ is a unit vector, then $∣ b ∣$ is equal to :