If $|\vec{a}|=4, |\vec{b}|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi/6$ then $(\vec{a} \times \vec{b})^2$ is equal to Options: (A) 48 (B) 16 (C) $\vec{a}$ (D) none of these

### Analyzing the Setup We are given two vectors, $\vec{a}$ and $\vec{b}$, with magnitudes $|\vec{a}|=4$ and $|\vec{b}|=2$. The angle between these vectors is $\theta = \pi/6$. Our objective is to determine the value of $(\vec{a} \times \vec{b})^2$. ### The Geometric Approach The magnitude of the cross product is defined by the area of the parallelogram spanned by the vectors: $$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$$ Since the square of a vector is equivalent to the square of its magnitude, we seek: $$(\vec{a} \times \vec{b})^2 = (|\vec{a}| |\vec{b}| \sin \theta)^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$$ ### Final Calculation Substituting the given values $|\vec{a}|=4$, $|\vec{b}|=2$, and $\theta = \pi/6$: $$|\vec{a}|^2 = 16, \quad |\vec{b}|^2 = 4, \quad \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ Multiplying these components together: $$(\vec{a} \times \vec{b})^2 = 16 \times 4 \times \frac{1}{4} = 16$$ ### Verification via Lagrange's Identity We can verify this result using **Lagrange's Identity**, which states: $$(\vec{a} \times \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$ First, calculate the dot product: $$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 4 \times 2 \times \cos\left(\frac{\pi}{6}\right) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}$$ Squaring the dot product yields $(4\sqrt{3})^2 = 16 \times 3 = 48$. Applying the identity: $$(\vec{a} \times \vec{b})^2 = (16 \times 4) - 48 = 64 - 48 = 16$$ The final result is **16**.

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Animated Solution for Mathematics - Vector Algebra: If $∣ a ∣ = 4, ∣ b ∣ = 2$ and the angle between $a$ and $b$ is $π /6$ then $(a \times b)^{2}$ is equal to

Q: Can I use Lagrange's Identity here? Is it faster?

Lagrange's Identity, which states $(\vec{a} \times \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$, is a powerful tool! It is not necessarily 'faster' in this specific problem, but it is an excellent way to verify your answer. If you calculate the dot product first, you will find it leads to the exact same result, giving you 100% confidence in your solution.

Select Answer:

Visualized Solution

Visualizing the Vectors $a$ and $b$

We are given two vectors, $a$ and $b$ , in a three-dimensional space.
The magnitude of vector $a$ is $∣ a ∣ = 4$ .
The magnitude of vector $b$ is $∣ b ∣ = 2$ .
The angle between them is $θ = \frac{π}{6}$ (or $3 0^{\circ}$ ).

Understanding the Cross Product $a \times b$

The cross product $a \times b$ is a vector perpendicular to both $a$ and $b$ .
Its direction is given by the right-hand rule.
The magnitude is defined as: $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$

Squaring the Cross Product $(a \times b)^{2}$

The square of a vector is equal to the square of its magnitude: $v^{2} = ∣ v ∣^{2}$
Therefore, $(a \times b)^{2} = ∣ a \times b ∣^{2}$
Expanding this using the magnitude formula: $(a \times b)^{2} = (∣ a ∣∣ b ∣ sin θ)^{2}$

Expanding the Expression

Applying the exponent to each term: $(a \times b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ$
Now, we are ready to substitute the given values into this expanded formula.

Substituting the Magnitudes

Substitute $∣ a ∣ = 4$ and $∣ b ∣ = 2$ into the expression:
$(a \times b)^{2} = (4)^{2} \times (2)^{2} \times sin^{2} (\frac{π}{6})$

Calculating $∣ a ∣^{2}$ and $∣ b ∣^{2}$

Calculate the square of $∣ a ∣$ : $4^{2} = 16$
Calculate the square of $∣ b ∣$ : $2^{2} = 4$
The product of these squares is: $16 \times 4 = 64$

Evaluating $sin^{2} (\frac{π}{6})$

We know that: $sin (\frac{π}{6}) = \frac{1}{2}$
Squaring this value: $sin^{2} (\frac{π}{6}) = (\frac{1}{2})^{2} = \frac{1}{4}$

Final Multiplication

Substitute the calculated values back into the expression:
$(a \times b)^{2} = 64 \times \frac{1}{4}$
Simplifying the fraction: $64 \times \frac{1}{4} = 16$

Alternative Method: Lagrange's Identity

We can also use Lagrange's Identity: $(a \times b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$
Since $a \cdot b = ∣ a ∣∣ b ∣ cos θ = 4 \times 2 \times cos (\frac{π}{6}) = 43$
$(a \cdot b)^{2} = (43)^{2} = 48$
$(a \times b)^{2} = 64 - 48 = 16$
Both methods yield the same result. The correct option is 16.

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 ∣ = 2

. The angle between these vectors is

θ = π /6

Our objective is to determine the value of

(a \times b)^{2}

The Geometric Approach

The magnitude of the cross product is defined by the area of the parallelogram spanned by the vectors:

∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ

Since the square of a vector is equivalent to the square of its magnitude, we seek:

(a \times b)^{2} = (∣ a ∣∣ b ∣ sin θ)^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ

Final Calculation

Substituting the given values

∣ a ∣ = 4

∣ b ∣ = 2

, and

θ = π /6

∣ a ∣^{2} = 16, ∣ b ∣^{2} = 4, sin^{2} (\frac{π}{6}) = (\frac{1}{2})^{2} = \frac{1}{4}

Multiplying these components together:

(a \times b)^{2} = 16 \times 4 \times \frac{1}{4} = 16

Verification via Lagrange's Identity

We can verify this result using Lagrange's Identity, which states:

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

First, calculate the dot product:

a \cdot b = ∣ a ∣∣ b ∣ cos θ = 4 \times 2 \times cos (\frac{π}{6}) = 8 \times \frac{3}{2} = 43

Squaring the dot product yields

(43)^{2} = 16 \times 3 = 48

. Applying the identity:

(a \times b)^{2} = (16 \times 4) - 48 = 64 - 48 = 16

The final result is 16.

Common Traps & FAQs

Wait, why do we treat

(a \times b)^{2}

as a scalar? Isn't the cross product a vector?

That is a fantastic observation! You are absolutely right that

a \times b

is a vector. However, in vector algebra, the notation

v^{2}

is shorthand for the dot product of the vector with itself,

v \cdot v

, which is equal to

∣ v ∣^{2}

. Since the dot product of any two vectors results in a scalar, the square of a vector is always a scalar quantity.

Can I use Lagrange's Identity here? Is it faster?

Lagrange's Identity, which states

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

, is a powerful tool! It is not necessarily 'faster' in this specific problem, but it is an excellent way to verify your answer. If you calculate the dot product first, you will find it leads to the exact same result, giving you 100% confidence in your solution.

I keep getting confused with the sine and cosine values. How do I avoid silly mistakes?

The best way to avoid these 'silly' mistakes is to draw a quick 30-60-90 triangle on the corner of your rough sheet. For

θ = π /6

(30 degrees), remember that

sin (3 0^{\circ}) = 1/2

. Writing this down explicitly before plugging it into your equation acts as a mental anchor, preventing those common calculation slips.

Animated Solution for Mathematics - Vector Algebra: If $∣ a ∣ = 4, ∣ b ∣ = 2$ and the angle between $a$ and $b$ is $π /6$ then $(a \times b)^{2}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Vectors $a$ and $b$

Understanding the Cross Product $a \times b$

Squaring the Cross Product $(a \times b)^{2}$

Expanding the Expression

Substituting the Magnitudes

Calculating $∣ a ∣^{2}$ and $∣ b ∣^{2}$

Evaluating $sin^{2} (\frac{π}{6})$

Final Multiplication

Alternative Method: Lagrange's Identity

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Geometric Approach

Final Calculation

Verification via Lagrange's Identity

Similar Questions

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 $∣ a ∣ = 2, ∣ b ∣ = 3$ 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

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

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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

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 $a$ and $b$ be two vector such that $∣ a ∣ = 14, ∣ b ∣ = 6$ and $∣ a \times b ∣ = 48$ . Then $(a \cdot b)^{2}$ is equal to

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If $∣ a ∣ = 2, ∣ b ∣ = 5$ and $∣ a \times b ∣ = 8$ , then $∣ a \cdot b ∣$ is equal to:

Let $a$ and $b$ be the vectors along the diagonal of a parallelogram having area $22$ . Let the angle between $a$ and $b$ be acute. $∣ a ∣ = 1$ and $∣ a \cdot b ∣ = ∣ a \times b ∣$ . If $c = 22 (a \times b) - 2 b$ , then an angle between $b$ and $c$ is:

Animated Solution for Mathematics - Vector Algebra: .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

Select Answer:

Visualized Solution

Visualizing the 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

Understanding the Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a×b

Squaring the Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a×b)2

Expanding the Expression

Substituting the Magnitudes

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a∣2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣b∣2

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2(6π​)

Final Multiplication

Alternative Method: Lagrange's Identity

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Geometric Approach

Final Calculation

Verification via Lagrange's Identity

Similar Questions

.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 _______.

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Animated Solution for Mathematics - Vector Algebra: If $∣ a ∣ = 4, ∣ b ∣ = 2$ and the angle between $a$ and $b$ is $π /6$ then $(a \times b)^{2}$ is equal to

Visualizing the Vectors $a$ and $b$

Understanding the Cross Product $a \times b$

Squaring the Cross Product $(a \times b)^{2}$

Calculating $∣ a ∣^{2}$ and $∣ b ∣^{2}$

Evaluating $sin^{2} (\frac{π}{6})$

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 $∣ a ∣ = 2, ∣ b ∣ = 3$ 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

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

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

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 $a$ and $b$ be two vector such that $∣ a ∣ = 14, ∣ b ∣ = 6$ and $∣ a \times b ∣ = 48$ . Then $(a \cdot b)^{2}$ is equal to

If $∣ a ∣ = 2, ∣ b ∣ = 5$ and $∣ a \times b ∣ = 8$ , then $∣ a \cdot b ∣$ is equal to: