Let |a| = 2, |b| = 3 and the angle between the vectors a and b be π/4. Then |(a + 2b) × (2a - 3b)|²

Q: Wait, why do we reject the negative root here? Or rather, why does the magnitude become positive?

Great question! Remember that the magnitude of a vector, denoted by $|\vec{v}|$, is by definition a non-negative scalar. When we have an expression like $|-7(\vec{a} \times \vec{b})|^2$, we are squaring the entire term. The property $|k\vec{v}| = |k| |\vec{v}|$ ensures that the constant factor $-7$ becomes $|-7| = 7$. Squaring that gives us $49$. The magnitude is always positive, so we don't need to worry about negative roots.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure vector algebra problem, not a physics problem involving energy. In JEE, you will often encounter problems that look like physics (using vectors) but are actually testing your mastery of vector identities. Stick to the algebraic properties of cross products—they are your most reliable tools here.

Q: Does the order of vectors really matter in the cross product?

It is the most critical detail! In scalar multiplication, $a \times b = b \times a$, but in vector cross products, $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. If you accidentally swap the order without changing the sign, your final answer will be completely wrong. Always treat the cross product as order-sensitive.

Question

Let $|\vec{a}| = 2, |\vec{b}| = 3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\pi/4$. Then $|(\vec{a} + 2\vec{b}) 	imes (2\vec{a} - 3\vec{b})|^2$ is equal to

Options:

(A) 441

(B) 482

(C) 841

(D) 882

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the magnitudes $|\vec{a}| = 2$ and $|\vec{b}| = 3$, with the angle between them being $	heta = \frac{\pi}{4}$. Our objective is to evaluate the expression $|(\vec{a} + 2\vec{b}) 	imes (2\vec{a} - 3\vec{b})|^2$.

When dealing with the cross product of two binomials, we expand them similarly to standard algebra. However, we must strictly respect the **non-commutative property** of the cross product by maintaining the order of the vectors.

### The Algebraic Expansion

Expanding the expression $(\vec{a} + 2\vec{b}) 	imes (2\vec{a} - 3\vec{b})$ term by term yields:

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

This results in four distinct terms that we can simplify using vector identities.

### The Power of Properties

We simplify the expression to $2(\vec{a} 	imes \vec{a}) - 3(\vec{a} 	imes \vec{b}) + 4(\vec{b} 	imes \vec{a}) - 6(\vec{b} 	imes \vec{b})$. Recall that the cross product of any vector with itself is the zero vector, $\vec{0}$, because $\sin(0) = 0$.

Consequently, $\vec{a} 	imes \vec{a} = \vec{0}$ and $\vec{b} 	imes \vec{b} = \vec{0}$. The expression collapses to:

$$-3(\vec{a} 	imes \vec{b}) + 4(\vec{b} 	imes \vec{a})$$

Since $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$, we substitute this into our equation:

$$-3(\vec{a} 	imes \vec{b}) - 4(\vec{a} 	imes \vec{b}) = -7(\vec{a} 	imes \vec{b})$$

### The Numerical Payoff

We now calculate the squared magnitude of our result: $|-7(\vec{a} 	imes \vec{b})|^2$. Using the property $|k\vec{v}|^2 = k^2 |\vec{v}|^2$, we obtain:

$$|-7(\vec{a} 	imes \vec{b})|^2 = 49 |\vec{a} 	imes \vec{b}|^2$$

Recalling the definition $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta$, we square both sides to get:

$$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta$$

Substituting the known values $|\vec{a}|^2 = 4$, $|\vec{b}|^2 = 9$, and $\sin^2(\frac{\pi}{4}) = \frac{1}{2}$:

$$49 	imes 4 	imes 9 	imes \frac{1}{2} = 49 	imes 18 = 882$$

The final result of the expression is **882**.

Animated Solution for Mathematics - Vector Algebra: Let $∣ a ∣ = 2, ∣ b ∣ = 3$ and the angle between the vectors $a$ and $b$ be $π /4$ . Then $∣ (a + 2 b) \times (2 a - 3 b) ∣^{2}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Vectors $a$ and $b$

Expanding the Cross Product

Applying Distributive Property

Property of Self-Cross Product

Simplifying the Expression

Anti-commutative Property

Final Simplified Vector

Squaring the Magnitude

Formula for Cross Product Magnitude

Substituting the Values

Atomic Calculation

Final Multiplication

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Algebraic Expansion

The Power of Properties

The Numerical Payoff

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∣a∣=2,∣b∣=3 and the angle between the vectors a and b be π/4. Then ∣(a+2b)×(2a−3b)∣2 is equal to