Let a and b be two unit vectors such that the angle between them is π/4. If θ is the angle between t

Q: Why can we just ignore the cross product term when calculating the dot product $\vec{u} \cdot \vec{v}$?

Because $\vec{u}$ is a linear combination of $\hat{a}$ and $\hat{b}$, meaning it lies in the plane spanned by them. The cross product $\hat{a} \times \hat{b}$ is, by definition, perpendicular to that plane. Any vector in the plane dotted with a vector perpendicular to the plane is zero. It is a beautiful geometric shortcut that saves you from unnecessary calculations!

Q: The algebra for the denominator looks terrifying. Is there a trick?

It looks intimidating, but it is just a product of two binomials. Expand it carefully. You will get $18 + 11\sqrt{2}$. When you rationalize the final expression, the $164$ in the denominator will appear naturally. Trust the process and keep your steps organized!

Q: Why do we calculate $|\vec{v}|^2$ using the Pythagorean theorem instead of just expanding the dot product?

You could expand the dot product, but using the Pythagorean theorem is faster and less prone to errors. Since the in-plane part and the perpendicular part are orthogonal, $|\vec{v}|^2 = |\text{in-plane}|^2 + |\text{perpendicular}|^2$. It saves you from dealing with cross-terms that would eventually vanish anyway.

Question

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{4}$. If $	heta$ is the angle between the vectors $(\hat{a} + \hat{b})$ and $(\hat{a} + 2\hat{b} + 2(\hat{a} 	imes \hat{b}))$, then the value of $164 \cos^2 	heta$ is equal to :

Options:

(A) $90 + 27\sqrt{2}$

(B) $45 + 18\sqrt{2}$

(C) $90 + 3\sqrt{2}$

(D) $54 + 90\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a vector problem; we are exploring the architecture of space itself.

When you look at the unit vectors $\hat{a}$ and $\hat{b}$, do not just see symbols. See two unit-length arrows originating from the origin, locked in a dance at an angle of $\frac{\pi}{4}$.

Our goal is to find the angle $	heta$ between two composite vectors, $\vec{u}$ and $\vec{v}$. This is a classic JEE Advanced challenge—it tests your ability to bridge the gap between abstract algebra and geometric intuition.

### The Foundation

We begin by defining our vectors. We have $\vec{u} = \hat{a} + \hat{b}$. This vector is the diagonal of the parallelogram formed by $\hat{a}$ and $\hat{b}$, and it lies entirely within the plane defined by these two vectors.

Then we have $\vec{v} = \hat{a} + 2\hat{b} + 2(\hat{a} 	imes \hat{b})$. The term $2(\hat{a} 	imes \hat{b})$ is the "outlier." It is the vector that breaks the symmetry of the plane, pointing perpendicularly into the third dimension.

### The Power of Orthogonality

To find $\cos 	heta$, we need the dot product $\vec{u} \cdot \vec{v}$ and the magnitudes $|\vec{u}|$ and $|\vec{v}|$. Let us tackle the dot product first:

$$\vec{u} \cdot \vec{v} = (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} 	imes \hat{b}))$$

Because $\vec{u}$ lies in the plane of $\hat{a}$ and $\hat{b}$, and $(\hat{a} 	imes \hat{b})$ is perpendicular to that plane, their dot product is zero. The cross product term vanishes entirely.

We are left with:

$$\vec{u} \cdot \vec{v} = (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b}) = |\hat{a}|^2 + 3(\hat{a} \cdot \hat{b}) + 2|\hat{b}|^2$$

Substituting $|\hat{a}| = 1$, $|\hat{b}| = 1$, and $\hat{a} \cdot \hat{b} = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, we find:

$$\vec{u} \cdot \vec{v} = 1 + \frac{3}{\sqrt{2}} + 2 = 3 + \frac{3}{\sqrt{2}}$$

### The Magnitude Calculation

For $\vec{u}$, the magnitude squared is:

$$|\vec{u}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) = 1 + 1 + 2\left(\frac{1}{\sqrt{2}}ight) = 2 + \sqrt{2}$$

For $\vec{v}$, we use the Pythagorean theorem because the in-plane component $(\hat{a} + 2\hat{b})$ and the perpendicular component $2(\hat{a} 	imes \hat{b})$ are orthogonal:

$$|\vec{v}|^2 = |\hat{a} + 2\hat{b}|^2 + |2(\hat{a} 	imes \hat{b})|^2$$

Calculating the components:

$$|\hat{a} + 2\hat{b}|^2 = |\hat{a}|^2 + 4|\hat{b}|^2 + 4(\hat{a} \cdot \hat{b}) = 1 + 4 + 4\left(\frac{1}{\sqrt{2}}ight) = 5 + 2\sqrt{2}$$

$$|2(\hat{a} 	imes \hat{b})|^2 = 4|\hat{a}|^2|\hat{b}|^2\sin^2\left(\frac{\pi}{4}ight) = 4(1)(1)\left(\frac{1}{2}ight) = 2$$

Thus, $|\vec{v}|^2 = 5 + 2\sqrt{2} + 2 = 7 + 2\sqrt{2}$.

### The Grand Finale

We use the definition $\cos^2 	heta = \frac{(\vec{u} \cdot \vec{v})^2}{|\vec{u}|^2 |\vec{v}|^2}$. Substituting our values:

$$\cos^2 	heta = \frac{\left(3 + \frac{3}{\sqrt{2}}ight)^2}{(2 + \sqrt{2})(7 + 2\sqrt{2})}$$

Expanding the numerator:

$$\left(3 + \frac{3}{\sqrt{2}}ight)^2 = 9 + \frac{9}{2} + 2(3)\left(\frac{3}{\sqrt{2}}ight) = 13.5 + 9\sqrt{2} = \frac{27 + 18\sqrt{2}}{2}$$

Expanding the denominator:

$$(2 + \sqrt{2})(7 + 2\sqrt{2}) = 14 + 4\sqrt{2} + 7\sqrt{2} + 4 = 18 + 11\sqrt{2}$$

After algebraic simplification, the expression for $164 \cos^2 	heta$ yields the final result:

**$90 + 27\sqrt{2}$**

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Visualized Solution

Defining the Unit Vectors

Dot and Cross Products

Defining Vector $u$ and its Magnitude

Defining Vector $v$

Magnitude Square of $v$

Dot Product $u \cdot v$

Setting up $cos^{2} θ$

Simplifying the Expression

Rationalizing the Denominator

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Foundation

The Power of Orthogonality

The Magnitude Calculation

The Grand Finale

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Select Answer:

Visualized Solution

Defining the Unit Vectors

Dot and Cross Products

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u and its Magnitude

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

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

Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u⋅v

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2θ