Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where $|\vec{a}| = 4, |\vec{b}| = 3, \theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^2 + 4(\vec{a} \cdot \vec{b})^2$ is equal to

### Analyzing the Setup We are given two vectors, $\vec{a}$ and $\vec{b}$, with magnitudes $|\vec{a}| = 4$ and $|\vec{b}| = 3$. The angle $\theta$ between them is constrained within the interval $(\frac{\pi}{4}, \frac{\pi}{3})$. We aim to simplify the expression: $$|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^2 + 4(\vec{a} \cdot \vec{b})^2$$ ### The Art of Expansion We begin by focusing on the cross product term: $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})$. Treating this like a binomial expansion, we distribute the terms: $$\vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}$$ Applying the laws of vector algebra, we know that the cross product of any vector with itself is the zero vector ($\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$). Furthermore, since the cross product is anti-commutative, we have $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. Substituting these identities, the expression simplifies as follows: $$\vec{0} + \vec{a} \times \vec{b} - (-(\vec{a} \times \vec{b})) - \vec{0} = 2(\vec{a} \times \vec{b})$$ ### The Bridge to Trigonometry Returning to our main expression, we substitute the simplified cross product result: $$|2(\vec{a} \times \vec{b})|^2 + 4(\vec{a} \cdot \vec{b})^2 = 4|\vec{a} \times \vec{b}|^2 + 4(\vec{a} \cdot \vec{b})^2$$ Factoring out the constant $4$, we obtain: $$4(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2)$$ Recalling the definitions $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$, we substitute these into the equation: $$4((|\vec{a}||\vec{b}|\sin\theta)^2 + (|\vec{a}||\vec{b}|\cos\theta)^2)$$ ### The Grand Finale Factoring out $|\vec{a}|^2|\vec{b}|^2$, we are left with the fundamental trigonometric identity: $$4|\vec{a}|^2|\vec{b}|^2(\sin^2\theta + \cos^2\theta) = 4|\vec{a}|^2|\vec{b}|^2(1)$$ The angle $\theta$ vanishes entirely, rendering the given range a distraction. Substituting the magnitudes $|\vec{a}|=4$ and $|\vec{b}|=3$: $$4(4^2)(3^2) = 4(16)(9) = 64 \times 9$$ The final value of the expression is **576**.

Wait, why does $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})$ simplify to $2(\vec{a} \times \vec{b})$?

When you expand using the distributive property, you get $\vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}$. Since $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$, and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$, the expression becomes $\vec{0} + \vec{a} \times \vec{b} - (-(\vec{a} \times \vec{b})) - \vec{0}$, which is $2(\vec{a} \times \vec{b})$.

JEE Main 2022 (25 June Shift 1)

LEVELJEE Main

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

Q: Why can't we just pick an angle like $60^\circ$ since the problem gives a range?

The range $\theta \in (\frac{\pi}{4}, \frac{\pi}{3})$ is a classic JEE distractor. Because the final expression simplifies to a form independent of $\theta$, the specific value of the angle does not matter. Always look for identities that might eliminate variables!

Q: Is there a faster way to solve this?

Yes, Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. Recognizing this identity immediately allows you to bypass the expansion and trigonometric substitution, saving precious time in the exam.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Visualize the vectors

a

and

b

as two arrows originating from the same point. The expression

∣ (a - b) \times (a + b) ∣^{2}

looks intimidating, but treat the cross product like a binomial expansion. What happens when you cross a vector with itself?

Hint 2

The Arsenal: Remember the fundamental properties of the cross product: it is distributive, anti-commutative (

b \times a = - (a \times b)

), and the cross product of any vector with itself is the zero vector. Also, keep the identity

∣ a \times b ∣^{2} + (a \cdot b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2}

in your back pocket.

Hint 3

The Execution: Expand the cross product term

(a - b) \times (a + b)

carefully. You will find it simplifies to

2 (a \times b)

. Now, substitute this into the original expression and use the definitions of dot and cross products to see how the angle

θ

disappears.

Visualized Solution

Visualizing Vectors $a$ and $b$

Given: $∣ a ∣ = 4$ and $∣ b ∣ = 3$
Angle between them: $θ \in (\frac{π}{4}, \frac{π}{3})$
Goal: Evaluate $∣ (a - b) \times (a + b) ∣^{2} + 4 (a \cdot b)^{2}$

Analyzing the Cross Product Term

Let's focus on the first part: $(a - b) \times (a + b)$
This represents the cross product of two new vectors formed by addition and subtraction.

Expanding the Cross Product

Expand using the distributive property:
$(a - b) \times (a + b) = a \times a + a \times b - b \times a - b \times b$

Applying Vector Cross Product Properties

Property 1: The cross product of a vector with itself is zero.
$a \times a = 0$ and $b \times b = 0$
Property 2: Anti-commutativity of cross product.
$b \times a = - (a \times b)$

Simplifying the Cross Product Term

Substitute the properties back into the expansion:
$= 0 + a \times b - (- a \times b) - 0$
$= 2 (a \times b)$

Reconstructing the Original Expression

Substitute the simplified term back into the original expression:
Original: $∣ (a - b) \times (a + b) ∣^{2} + 4 (a \cdot b)^{2}$
Becomes: $∣2 (a \times b) ∣^{2} + 4 (a \cdot b)^{2}$
Squaring the constant: $4∣ a \times b ∣^{2} + 4 (a \cdot b)^{2}$

Recalling Dot and Cross Product Definitions

Magnitude of Cross Product: $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$
Dot Product: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$
Here, $θ$ is the angle between vectors $a$ and $b$ .

Substituting Definitions into the Expression

Substitute the definitions into our expression:
$4 (∣ a ∣∣ b ∣ sin θ)^{2} + 4 (∣ a ∣∣ b ∣ cos θ)^{2}$
Distribute the squares:
$4∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ + 4∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ$

Factoring Out Common Terms

Notice the common factor in both terms: $4∣ a ∣^{2} ∣ b ∣^{2}$
Factor it out:
$= 4∣ a ∣^{2} ∣ b ∣^{2} (sin^{2} θ + cos^{2} θ)$

Applying the Fundamental Trigonometric Identity

Recall the Pythagorean identity: $sin^{2} θ + cos^{2} θ = 1$
Substitute this into the factored expression:
$= 4∣ a ∣^{2} ∣ b ∣^{2} (1)$
$= 4∣ a ∣^{2} ∣ b ∣^{2}$

Substituting the Given Magnitudes

We are given: $∣ a ∣ = 4$ and $∣ b ∣ = 3$
Substitute these values into our simplified expression:
$= 4 (4)^{2} (3)^{2}$

Final Arithmetic Calculation

Calculate the squares: $4^{2} = 16$ and $3^{2} = 9$
Multiply the terms:
$= 4 \times 16 \times 9$
$= 64 \times 9$
$= 576$

Conclusion and Lagrange's Identity

Final Answer: $576$
Key Concept: The simplification relies on Lagrange's Identity:
$∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$
The given angle $θ \in (\frac{π}{4}, \frac{π}{3})$ was extra information designed to distract!

00:00 / 00:00

The Sigma Insight: Vector (Cross) Product

Solution Diagram

Analyzing the Setup

We are given two vectors,

a

and

b

, with magnitudes

∣ a ∣ = 4

and

∣ b ∣ = 3

. The angle

θ

between them is constrained within the interval

(\frac{π}{4}, \frac{π}{3})

We aim to simplify the expression:

∣ (a - b) \times (a + b) ∣^{2} + 4 (a \cdot b)^{2}

The Art of Expansion

We begin by focusing on the cross product term:

(a - b) \times (a + b)

. Treating this like a binomial expansion, we distribute the terms:

a \times a + a \times b - b \times a - b \times b

Applying the laws of vector algebra, we know that the cross product of any vector with itself is the zero vector (

a \times a = 0

and

b \times b = 0

). Furthermore, since the cross product is anti-commutative, we have

b \times a = - (a \times b)

Substituting these identities, the expression simplifies as follows:

0 + a \times b - (- (a \times b)) - 0 = 2 (a \times b)

The Bridge to Trigonometry

Returning to our main expression, we substitute the simplified cross product result:

∣2 (a \times b) ∣^{2} + 4 (a \cdot b)^{2} = 4∣ a \times b ∣^{2} + 4 (a \cdot b)^{2}

Factoring out the constant

4

, we obtain:

4 (∣ a \times b ∣^{2} + (a \cdot b)^{2})

Recalling the definitions

∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ

and

a \cdot b = ∣ a ∣∣ b ∣ cos θ

, we substitute these into the equation:

4 ((∣ a ∣∣ b ∣ sin θ)^{2} + (∣ a ∣∣ b ∣ cos θ)^{2})

The Grand Finale

Factoring out

∣ a ∣^{2} ∣ b ∣^{2}

, we are left with the fundamental trigonometric identity:

4∣ a ∣^{2} ∣ b ∣^{2} (sin^{2} θ + cos^{2} θ) = 4∣ a ∣^{2} ∣ b ∣^{2} (1)

The angle

θ

vanishes entirely, rendering the given range a distraction. Substituting the magnitudes

∣ a ∣ = 4

and

∣ b ∣ = 3

4 (4^{2}) (3^{2}) = 4 (16) (9) = 64 \times 9

The final value of the expression is 576.

Common Traps & FAQs

Why can't we just pick an angle like

6 0^{\circ}

since the problem gives a range?

The range

θ \in (\frac{π}{4}, \frac{π}{3})

is a classic JEE distractor. Because the final expression simplifies to a form independent of

θ

, the specific value of the angle does not matter. Always look for identities that might eliminate variables!

Wait, why does

(a - b) \times (a + b)

simplify to

2 (a \times b)

When you expand using the distributive property, you get

a \times a + a \times b - b \times a - b \times b

. Since

a \times a = 0

and

b \times b = 0

, and

b \times a = - (a \times b)

, the expression becomes

0 + a \times b - (- (a \times b)) - 0

, which is

2 (a \times b)

Is there a faster way to solve this?

Yes, Lagrange's Identity:

∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}

. Recognizing this identity immediately allows you to bypass the expansion and trigonometric substitution, saving precious time in the exam.

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

Enter Numerical Value:

Visualized Solution

Visualizing Vectors $a$ and $b$

Analyzing the Cross Product Term

Expanding the Cross Product

Applying Vector Cross Product Properties

Simplifying the Cross Product Term

Reconstructing the Original Expression

Recalling Dot and Cross Product Definitions

Substituting Definitions into the Expression

Factoring Out Common Terms

Applying the Fundamental Trigonometric Identity

Substituting the Given Magnitudes

Final Arithmetic Calculation

Conclusion and Lagrange's Identity

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Art of Expansion

The Bridge to Trigonometry

The Grand Finale

Similar Questions

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

If $∣ a ∣ = 4, ∣ b ∣ = 2$ and the angle between $a$ and $b$ is $π /6$ then $(a \times b)^{2}$ is equal to

Let $a, b, c$ be three vectors such that $∣ a ∣ = 31, 4∣ b ∣ = ∣ c ∣ = 2$ and $2 (a \times b) = 3 (c \times a)$ . If the angle between $b$ and $c$ is $2 π /3$ , then $(\frac{a \times c}{a \cdot b})^{2}$ is equal to

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to

Let $a$ and $b$ be two vectors such that $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ and the angle between $b$ and $c$ is $α$ , then $192 sin^{2} α$ is equal to _______.

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

Let $a$ and $b$ be two vectors such that $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}, a \cdot b = 3$ and $∣ a \times b ∣^{2} = 75$ . Then $∣ a ∣^{2}$ is equal to

Let $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ , $b = \hat{i} - 2 \hat{j} - 2 \hat{k}$ and $c = - \hat{i} + 4 \hat{j} + 3 \hat{k}$ . If $d$ is a vector perpendicular to both $b$ and $c$ , and $a \cdot d = 18$ , then $∣ a \times d ∣^{2}$ is equal to

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣, ∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $3 0^{\circ}$ , then $∣ (a \times b) \times c ∣ =$

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

Enter Numerical Value:

Visualized Solution

Visualizing Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Analyzing the Cross Product Term

Expanding the Cross Product

Applying Vector Cross Product Properties

Simplifying the Cross Product Term

Reconstructing the Original Expression

Recalling Dot and Cross Product Definitions

Substituting Definitions into the Expression

Factoring Out Common Terms

Applying the Fundamental Trigonometric Identity

Substituting the Given Magnitudes

Final Arithmetic Calculation

Conclusion and Lagrange's Identity

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Art of Expansion

The Bridge to Trigonometry

The Grand Finale

Similar Questions

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∣a∣=4,∣b∣=2 and the angle between a and b is π/6 then (a×b)2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be three vectors such that ∣a∣=31​,4∣b∣=∣c∣=2 and 2(a×b)=3(c×a). If the angle between b and c is 2π/3, then (a⋅ba×c​)2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+j^​−2k^,b=i^+j^​ and c=a×b. Let d be a vector such that ∣d−a∣=11​,∣c×d∣=3 and the angle between c and d is 4π​. Then a⋅d is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a and b be two vectors such that ∣a∣=1,∣b∣=4 and a⋅b=2. If c=(2a×b)−3b and the angle between b and c is α, then 192sin2α is equal to _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a and b be two vectors such that ∣a+b∣2=∣a∣2+2∣b∣2,a⋅b=3 and ∣a×b∣2=75. Then ∣a∣2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+3j^​+4k^, b=i^−2j^​−2k^ and c=−i^+4j^​+3k^. If d is a vector perpendicular to both b and c, and a⋅d=18, then ∣a×d∣2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+j^​−2k^ and b=i^+j^​. If c is a vector such that a⋅c=∣c∣,∣c−a∣=22​ and the angle between (a×b) and c is 30∘, then ∣(a×b)×c∣=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a^,b^ be unit vectors. If c be a vector such that the angle between a^ and c is 12π​, and b^=c+2(c×a^), then ∣6c∣2 is equal to

Visualizing Vectors $a$ and $b$

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

If $∣ a ∣ = 4, ∣ b ∣ = 2$ and the angle between $a$ and $b$ is $π /6$ then $(a \times b)^{2}$ is equal to

Let $a, b, c$ be three vectors such that $∣ a ∣ = 31, 4∣ b ∣ = ∣ c ∣ = 2$ and $2 (a \times b) = 3 (c \times a)$ . If the angle between $b$ and $c$ is $2 π /3$ , then $(\frac{a \times c}{a \cdot b})^{2}$ is equal to

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to

Let $a$ and $b$ be two vectors such that $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ and the angle between $b$ and $c$ is $α$ , then $192 sin^{2} α$ is equal to _______.

Let $a$ and $b$ be two vectors such that $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}, a \cdot b = 3$ and $∣ a \times b ∣^{2} = 75$ . Then $∣ a ∣^{2}$ is equal to

Let $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ , $b = \hat{i} - 2 \hat{j} - 2 \hat{k}$ and $c = - \hat{i} + 4 \hat{j} + 3 \hat{k}$ . If $d$ is a vector perpendicular to both $b$ and $c$ , and $a \cdot d = 18$ , then $∣ a \times d ∣^{2}$ is equal to

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣, ∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $3 0^{\circ}$ , then $∣ (a \times b) \times c ∣ =$

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