If |a| = 2, |b| = 5 and |a × b| = 8, then |a · b| is equal to:

Q: Wait, why do we use Lagrange's Identity instead of finding the angle $\theta$ first?

You absolutely could find $\theta$! Using $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$, you would find $\sin\theta = 8 / (2 \times 5) = 0.8$. Then you could find $\cos\theta$ and calculate the dot product. However, Lagrange's Identity is a more elegant, direct, and 'JEE-style' shortcut that avoids trigonometric inverse functions and potential sign errors. It's all about efficiency!

Q: Is the dot product always positive?

Great question! The dot product $\vec{a} \cdot \vec{b}$ can be positive, negative, or zero depending on the angle between the vectors. However, the question asks for the magnitude $|\vec{a} \cdot \vec{b}|$, which is always non-negative. When we take the square root of $(\vec{a} \cdot \vec{b})^2$, we are looking for the absolute value, so we only consider the positive root.

Q: What if the vectors were perpendicular?

If the vectors were perpendicular, the angle $\theta$ would be $90^\circ$. In that case, the dot product would be zero, and the magnitude of the cross product would be equal to the product of the magnitudes of the vectors ($|\vec{a}| |\vec{b}|$). Lagrange's Identity would still hold perfectly: $0^2 + (|\vec{a}| |\vec{b}|)^2 = |\vec{a}|^2 |\vec{b}|^2$. It's a universal tool!

Question

If $|\vec{a}| = 2, |\vec{b}| = 5$ and $|\vec{a} 	imes \vec{b}| = 8$, then $|\vec{a} \cdot \vec{b}|$ is equal to:

Options:

(A) 6

(B) 4

(C) 3

(D) 5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Vectors $\vec{a}$ and $\vec{b}$ are not merely arrows; they represent a geometric relationship defined by their magnitudes and the angle $	heta$ between them. In the context of JEE Advanced, these vectors serve as the fundamental language of space and physics.

When two vectors $\vec{a}$ and $\vec{b}$ originate from the same point, they define a parallelogram. The magnitude of their cross product, $|\vec{a} 	imes \vec{b}|$, represents the area of this parallelogram, which is given as $8$.

The dot product, $\vec{a} \cdot \vec{b}$, represents the projection of one vector onto another. These two operations are intrinsically linked by the fundamental trigonometric identity:
$$\sin^2 	heta + \cos^2 	heta = 1$$

### The Bridge: Lagrange's Identity

To connect these operations, we utilize the powerful **Lagrange's Identity**. This identity is a cornerstone of vector algebra, testing the ability to recognize the underlying structure of vector relationships.

The identity is expressed as:
$$|\vec{a} 	imes \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$

Think of this as the vector version of the Pythagorean theorem. It states that the square of the area (cross product) plus the square of the projection (dot product) is always equal to the product of the squares of the individual magnitudes.

### The Calculation

We are given the magnitudes $|\vec{a}| = 2$ and $|\vec{b}| = 5$, along with the cross product magnitude $|\vec{a} 	imes \vec{b}| = 8$. Substituting these values into Lagrange's Identity, we obtain:

$$8^2 + (\vec{a} \cdot \vec{b})^2 = (2)^2 \cdot (5)^2$$

Calculating the squares, we arrive at:

$$64 + (\vec{a} \cdot \vec{b})^2 = 4 \cdot 25$$

$$64 + (\vec{a} \cdot \vec{b})^2 = 100$$

Isolating the dot product term, we find:

$$(\vec{a} \cdot \vec{b})^2 = 100 - 64$$

$$(\vec{a} \cdot \vec{b})^2 = 36$$

Taking the square root of both sides, we determine the final result:

$$|\vec{a} \cdot \vec{b}| = 6$$

### The Takeaway

By utilizing **Lagrange's Identity**, we bypassed the need to calculate the angle $	heta$ or engage in complex trigonometry. This approach demonstrates the essence of JEE Advanced problem-solving: identifying the most efficient path through the forest of equations.

Always remember that every vector problem is a geometric story waiting to be solved with precision and elegance.

Animated Solution for Mathematics - Vector Algebra: If $∣ a ∣ = 2, ∣ b ∣ = 5$ and $∣ a \times b ∣ = 8$ , then $∣ a \cdot b ∣$ is equal to:

Select Answer:

Visualized Solution

Identify Given Parameters

Cross Product as Area

Lagrange's Identity

Raw Setup (Substitution)

Atomic Compute: Squaring

Atomic Compute: Multiplication

Isolating the Dot Product

Final Result: $∣ a \cdot b ∣$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Bridge

The Calculation

The Takeaway

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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; } If ∣a∣=2,∣b∣=5 and ∣a×b∣=8, then ∣a⋅b∣ is equal to: