Let a = a1i + a2j + a3k and b = b1i + b2j + b3k be two vectors such that |a| = 1; a ⋅ b = 2 and |b|

Q: Why is $\vec{b} \cdot (\vec{a} \times \vec{b})$ always zero? I keep trying to calculate components to prove it.

That is a fantastic instinct to verify, but remember the geometric definition of the cross product! The vector $\vec{a} \times \vec{b}$ is, by definition, orthogonal to both $\vec{a}$ and $\vec{b}$. Since the dot product of any two perpendicular vectors is zero, $\vec{b} \cdot (\vec{a} \times \vec{b})$ must be zero. You don't need components; you just need the definition!

Q: Why can't I just assume $\vec{a} = \hat{i}$ and $\vec{b} = 2\hat{i} + 2\sqrt{3}\hat{j}$ to make the math easier?

While that specific choice might satisfy the given conditions, it is a dangerous habit for JEE Advanced. By picking specific components, you are restricting the generality of the vectors. Always rely on vector identities like Lagrange's Identity or the properties of dot/cross products. They work for *all* vectors, ensuring your solution is robust and mathematically sound.

Q: Do I need to worry about the sign of the angle? My result for $\cos \theta$ is negative.

Absolutely not! A negative cosine value simply means the angle $\theta$ is obtuse (between $90^\circ$ and $180^\circ$). The $\cos^{-1}$ function is perfectly defined for negative values in the range $[-1, 1]$. Embrace the negative sign—it tells you that $\vec{b}$ and $\vec{c}$ are pointing generally away from each other.

Question

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ be two vectors such that $|\vec{a}| = 1$; $\vec{a} \cdot \vec{b} = 2$ and $|\vec{b}| = 4$. If $\vec{c} = 2(\vec{a} 	imes \vec{b}) - 3\vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to :

Options:

(A) $\cos^{-1}(\frac{2}{\sqrt{3}})$

(B) $\cos^{-1}(-\frac{1}{\sqrt{3}})$

(C) $\cos^{-1}(-\frac{\sqrt{3}}{2})$

(D) $\cos^{-1}(\frac{2}{3})$

Sigma Percentile · Accepted Answer

### The Dance of Vectors: A Journey into 3D Space

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of vector algebra.

When you look at a problem involving $\vec{a}$, $\vec{b}$, and a complex combination like $\vec{c} = 2(\vec{a} 	imes \vec{b}) - 3\vec{b}$, it is easy to feel overwhelmed. You might be tempted to start breaking these vectors into components, hunting for $a_1, a_2, a_3$.

But stop. Take a breath. In JEE Advanced, the most elegant path is rarely the one that requires the most brute force. Let us solve this with grace.

### Phase 1: The Geometry of Orthogonality

We are tasked with finding the angle $	heta$ between $\vec{b}$ and $\vec{c}$. The universal key to finding an angle between two vectors is the dot product formula:

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

To find this, we need two things: the dot product $\vec{b} \cdot \vec{c}$ and the magnitude $|\vec{c}|$.

Let us tackle the numerator first. We substitute the expression for $\vec{c}$ into the dot product:

$$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2(\vec{a} 	imes \vec{b}) - 3\vec{b})$$

Distributing $\vec{b}$, we get:

$$\vec{b} \cdot \vec{c} = 2(\vec{b} \cdot (\vec{a} 	imes \vec{b})) - 3(\vec{b} \cdot \vec{b})$$

Here is the magic. Recall that $\vec{a} 	imes \vec{b}$ is a vector perpendicular to the plane of $\vec{a}$ and $\vec{b}$. Therefore, it is perpendicular to $\vec{b}$ itself.

The dot product of any two perpendicular vectors is zero. Thus, $\vec{b} \cdot (\vec{a} 	imes \vec{b}) = 0$.

Our expression simplifies instantly to $\vec{b} \cdot \vec{c} = -3|\vec{b}|^2$. Given $|\vec{b}| = 4$, we have:

$$\vec{b} \cdot \vec{c} = -3(16) = -48$$

The numerator is conquered!

### Phase 2: The Power of Lagrange's Identity

Now, we need the denominator: the magnitude of $\vec{c}$. We start by squaring it:

$$|\vec{c}|^2 = |2(\vec{a} 	imes \vec{b}) - 3\vec{b}|^2$$

Expanding this using the algebraic identity for the square of a vector magnitude, we get:

$$|\vec{c}|^2 = 4|\vec{a} 	imes \vec{b}|^2 + 9|\vec{b}|^2 - 12(\vec{a} 	imes \vec{b}) \cdot \vec{b}$$

Again, that last term vanishes because of orthogonality! We are left with:

$$|\vec{c}|^2 = 4|\vec{a} 	imes \vec{b}|^2 + 9|\vec{b}|^2$$

But what is $|\vec{a} 	imes \vec{b}|^2$? This is where Lagrange's Identity becomes our best friend:

$$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$

Substituting our known values, $|\vec{a}| = 1$, $|\vec{b}| = 4$, and $\vec{a} \cdot \vec{b} = 2$, we get:

$$|\vec{a} 	imes \vec{b}|^2 = (1)^2(4)^2 - (2)^2 = 16 - 4 = 12$$

### Phase 3: The Final Calculation

We are in the home stretch. Plugging our value of $12$ back into the magnitude equation:

$$|\vec{c}|^2 = 4(12) + 9(4)^2 = 48 + 144 = 192$$

Taking the square root, we find:

$$|\vec{c}| = \sqrt{192} = 8\sqrt{3}$$

Now, we return to our original angle formula:

$$\cos 	heta = \frac{-48}{4 	imes 8\sqrt{3}} = \frac{-48}{32\sqrt{3}}$$

Simplifying this fraction by dividing the numerator and denominator by $16$, we get:

$$\cos 	heta = \frac{-3}{2\sqrt{3}}$$

Since $3 = (\sqrt{3})^2$, this simplifies beautifully to:

$$\cos 	heta = -\frac{\sqrt{3}}{2}$$

Thus, the angle is **$	heta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}ight) = 150^\circ$**.

You have navigated the complexity, utilized the identities, and arrived at the truth. This is the essence of physics and mathematics—not just calculation, but the art of simplification.

Select Answer:

Visualized Solution

Given Vectors and Properties

The Angle Formula

Calculating $b \cdot c$

Property of Orthogonality

Final Value of $b \cdot c$

Finding $∣ c ∣^{2}$

Lagrange's Identity

Calculating $∣ c ∣$

Substituting into $cos θ$

Final Angle $θ$

The Sigma Insight: Vector (Cross) Product

The Dance of Vectors

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Given Vectors and Properties

The Angle Formula

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

Property of Orthogonality

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b⋅c

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

Lagrange's Identity

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

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

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