For λ 0, let θ be the angle between the vectors a = i + λj - 3k and b = 3i - j + 2k. If the vector

Q: Why can't we just use the angle formula $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ right away?

That is a great instinct, but look at the variables! We have an unknown $\lambda$ in vector $\vec{a}$. Without knowing $\lambda$, we cannot calculate the dot product or the magnitude of $\vec{a}$. The perpendicularity condition is the 'key' provided by the problem to unlock the value of $\lambda$ first.

Q: Why does $(\vec{a} + \vec{b}) \perp (\vec{a} - \vec{b})$ imply it's a rhombus?

In geometry, a parallelogram with perpendicular diagonals is a special case known as a rhombus. In a rhombus, all sides are equal in length. Therefore, the condition $(\vec{a} + \vec{b}) \perp (\vec{a} - \vec{b})$ is mathematically equivalent to saying the adjacent sides of the parallelogram, $\vec{a}$ and $\vec{b}$, have equal magnitudes ($|\vec{a}| = |\vec{b}|$).

Q: Does the negative sign in $\cos \theta = -\frac{5}{14}$ matter?

It matters conceptually! A negative cosine value tells us that the angle $\theta$ between the vectors is obtuse (greater than $90^\circ$). However, for the final calculation, we are asked to find $(14 \cos \theta)^2$. Squaring the result will eliminate the negative sign, so don't let it worry you.

Question

For $\lambda > 0$, let $	heta$ be the angle between the vectors $\vec{a} = \hat{i} + \lambda\hat{j} - 3\hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$. If the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular, then the value of $(14 \cos 	heta)^2$ is equal to

Options:

(A) 50

(B) 40

(C) 25

(D) 20

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: Unlocking the Rhombus

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful vector problem. Often, when we see vectors with unknown components like $\vec{a} = \hat{i} + \lambda\hat{j} - 3\hat{k}$, our first instinct is to panic.

But I want you to take a deep breath. In the JEE Advanced arena, vectors are not just lists of numbers; they are geometric entities living in space. Let us walk through this journey together.

### Phase 1: The Geometric Insight

We are given two vectors, $\vec{a}$ and $\vec{b}$. The problem introduces a condition: the sum $\vec{a} + \vec{b}$ and the difference $\vec{a} - \vec{b}$ are mutually perpendicular.

If you visualize $\vec{a}$ and $\vec{b}$ as adjacent sides of a parallelogram, then $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are precisely the diagonals of that parallelogram.

When the diagonals of a parallelogram are perpendicular, the parallelogram is a rhombus. This is a powerful realization! It implies that the adjacent sides must be equal in length, or mathematically, $|\vec{a}| = |\vec{b}|$.

### Phase 2: The Algebraic Expansion

If two vectors are perpendicular, their dot product must be zero. So, we set up our equation:

$$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0$$

Now, let us expand this using the distributive property of the dot product:

$$\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 0$$

Because the dot product is commutative (meaning $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$), the middle terms cancel out perfectly! We are left with:

$$|\vec{a}|^2 - |\vec{b}|^2 = 0 \implies |\vec{a}|^2 = |\vec{b}|^2$$

This confirms our geometric intuition. The magnitudes are equal.

### Phase 3: Solving for the Unknown

Now, we calculate the squared magnitudes using the components provided:

$$|\vec{a}|^2 = 1^2 + \lambda^2 + (-3)^2 = 10 + \lambda^2$$

$$|\vec{b}|^2 = 3^2 + (-1)^2 + 2^2 = 9 + 1 + 4 = 14$$

Equating them, we find $10 + \lambda^2 = 14$, which simplifies to $\lambda^2 = 4$. Since the problem explicitly states $\lambda > 0$, we discard the negative root and find $\lambda = 2$. Our vector $\vec{a}$ is now fully defined: $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$.

### Phase 4: The Final Calculation

We are almost there. We need to find the value of $(14 \cos 	heta)^2$. The angle $	heta$ is defined by the standard dot product formula:

$$\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$

First, let us compute the dot product $\vec{a} \cdot \vec{b}$ with our known $\lambda = 2$:

$$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5$$

We already know $|\vec{a}|^2 = 14$ and $|\vec{b}|^2 = 14$, so $|\vec{a}| = \sqrt{14}$ and $|\vec{b}| = \sqrt{14}$. Plugging these into our formula:

$$\cos 	heta = \frac{-5}{\sqrt{14} \cdot \sqrt{14}} = -\frac{5}{14}$$

Finally, we calculate the target expression:

$$(14 \cos 	heta)^2 = \left(14 \cdot \left(-\frac{5}{14}ight)ight)^2 = (-5)^2 = 25$$

And there we have it! The **final answer is 25**. Notice how the complexity melted away once we understood the geometric relationship between the diagonals.

Select Answer:

Visualized Solution

Visualizing the Vectors

The Perpendicularity Condition

Dot Product of Perpendicular Vectors

Expanding the Dot Product

Geometric Insight: The Rhombus

Calculating Magnitudes

Solving for $λ$

The Angle Formula

Calculating the Dot Product

Evaluating $a \cdot b$ and $cos θ$

Final Calculation

The Sigma Insight: Scalar (Dot) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

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