Let a, b be unit vectors. If c be a vector such that the angle between a and c is π/12, and b = c +

Q: Wait, why does the dot product $\vec{c} \cdot (\vec{c} \times \hat{a})$ vanish? Is this always true?

Yes, it is a fundamental geometric truth! The vector $\vec{c} \times \hat{a}$ is, by definition, perpendicular to the plane containing $\vec{c}$ and $\hat{a}$. Since it is perpendicular to the plane, it must be perpendicular to every vector lying within that plane, including $\vec{c}$ itself. Thus, their dot product is zero.

Q: I'm stuck on the trigonometry. How do I handle $\sin^2(\frac{\pi}{12})$ without a calculator?

Great question! Use the half-angle formula: $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$. Here, $\theta = \frac{\pi}{12}$, so $2\theta = \frac{\pi}{6}$ (or $30^\circ$). Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, you get $\sin^2(\frac{\pi}{12}) = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$. It's much cleaner than memorizing the exact value of $\sin(15^\circ)$!

Q: Why can't we just find $\vec{c}$ directly? Do we need to know its components?

You absolutely do not need the components! In JEE Advanced, we often deal with 'vector magnitudes' rather than specific coordinate representations. By working with the squared magnitude $|\vec{c}|^2$, we bypass the need to know the individual components of $\vec{c}$, which would be impossible to determine anyway since $\vec{c}$ is not uniquely fixed.

Q: Is $|6\vec{c}|^2$ the same as $6|\vec{c}|^2$?

Be careful! The scalar $6$ comes out of the magnitude squared as $6^2 = 36$. So, $|6\vec{c}|^2 = 36|\vec{c}|^2$. Always remember to square the scalar constant when pulling it out of the magnitude operator.

Q: What if I forgot the identity $|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2\vec{x} \cdot \vec{y}$?

You can always derive it! Just write it as $(\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y})$ and use the distributive property of the dot product. It's a quick 5-second proof that saves you from memorizing formulas.

Question

Let $\hat{a}, \hat{b}$ be unit vectors. If $\vec{c}$ be a vector such that the angle between $\hat{a}$ and $\vec{c}$ is $\frac{\pi}{12}$, and $\hat{b} = \vec{c} + 2(\vec{c} 	imes \hat{a})$, then $|6\vec{c}|^2$ is equal to

Options:

(A) 6(3-\sqrt{3})

(B) 3+\sqrt{3}

(C) 6(3+\sqrt{3})

(D) 6(\sqrt{3}+1)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are peeling back the layers of vector geometry.

We are given a unit vector $\hat{b}$ defined by the relation $\hat{b} = \vec{c} + 2(\vec{c} 	imes \hat{a})$. At first glance, this looks like a messy algebraic equation, but it is a beautiful geometric construction.

We have a vector $\vec{c}$ and a vector perpendicular to it, $2(\vec{c} 	imes \hat{a})$, being added together. This orthogonality is the key to unlocking the entire problem.

### The Power of Squaring

When we see a vector equation involving magnitudes, our first instinct should be to square both sides. Because the magnitude squared of a vector is simply the dot product of that vector with itself, we use the identity $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$.

Given that $\hat{b}$ is a unit vector, we know $|\hat{b}|^2 = 1$. Let's apply this to our relation:

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

Expanding this using the identity $|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$, we get:

$$1 = |\vec{c}|^2 + |2(\vec{c} 	imes \hat{a})|^2 + 2 \vec{c} \cdot (2(\vec{c} 	imes \hat{a}))$$

### The Elegance of Cancellation

Here is where the magic happens. Look at the term $2 \vec{c} \cdot (2(\vec{c} 	imes \hat{a}))$.

Because the cross product $\vec{c} 	imes \hat{a}$ is inherently perpendicular to $\vec{c}$, their dot product is zero. The entire interaction term vanishes!

We are left with a much simpler, cleaner equation:

$$1 = |\vec{c}|^2 + 4|\vec{c} 	imes \hat{a}|^2$$

### Bridging Trigonometry and Vectors

Now, we must address the cross product magnitude. We know that $|\vec{c} 	imes \hat{a}| = |\vec{c}| |\hat{a}| \sin 	heta$.

Since $\hat{a}$ is a unit vector ($|\hat{a}| = 1$) and the angle $	heta = \frac{\pi}{12}$, we have $|\vec{c} 	imes \hat{a}| = |\vec{c}| \sin(\frac{\pi}{12})$. Substituting this back into our equation, we get:

$$1 = |\vec{c}|^2 + 4(|\vec{c}| \sin \frac{\pi}{12})^2 = |\vec{c}|^2 (1 + 4 \sin^2 \frac{\pi}{12})$$

To solve this, we need the value of $\sin^2(\frac{\pi}{12})$. Using the identity $\sin^2(	heta) = \frac{1 - \cos(2	heta)}{2}$, we find:

$$\sin^2 \frac{\pi}{12} = \frac{1 - \cos(\frac{\pi}{6})}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$$

### The Final Calculation

Substituting this back into our equation:

$$1 = |\vec{c}|^2 (1 + 4 \cdot \frac{2 - \sqrt{3}}{4}) = |\vec{c}|^2 (1 + 2 - \sqrt{3}) = |\vec{c}|^2 (3 - \sqrt{3})$$

Thus, $|\vec{c}|^2 = \frac{1}{3 - \sqrt{3}}$. Our target is $|6\vec{c}|^2$, which is $36|\vec{c}|^2$. Therefore:

$$|6\vec{c}|^2 = \frac{36}{3 - \sqrt{3}}$$

To rationalize, we multiply the numerator and denominator by the conjugate $(3 + \sqrt{3})$:

$$\frac{36(3 + \sqrt{3})}{9 - 3} = \frac{36(3 + \sqrt{3})}{6} = 6(3 + \sqrt{3})$$

The final result is **$18 + 6\sqrt{3}$**.

Animated Solution for Mathematics - Vector Algebra: Let $\overset{a}{^}, \hat{b}$ be unit vectors. If $c$ be a vector such that the angle between $\overset{a}{^}$ and $c$ is $\frac{π}{12}$ , and $\hat{b} = c + 2 (c \times \overset{a}{^})$ , then $∣6 c ∣^{2}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Vectors

The Vector Relation

Squaring the Magnitude

Expanding the Expression

The Orthogonality Property

Cross Product Magnitude

Factoring the Equation

Calculating $sin^{2} \frac{π}{12}$

Substituting the Value

Solving for $∣ c ∣^{2}$

Setting Up the Final Target

Rationalizing the Denominator

Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of Squaring

The Elegance of Cancellation

Bridging Trigonometry and Vectors

The Final Calculation

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