For any two vectors u and v, prove that (a) (u . v)² + |u x v|² = |u|² |v|² and (b) (1 + |u|²)(1 + |

Q: Wait, why does the cross product term $(\vec{u} + \vec{v}) \cdot (\vec{u} \times \vec{v})$ become zero?

This is the beauty of vector geometry! The vector $(\vec{u} \times \vec{v})$ is defined to be perpendicular to both $\vec{u}$ and $\vec{v}$. Since it is perpendicular to both, it is perpendicular to any linear combination of them, including their sum $(\vec{u} + \vec{v})$. The dot product of any two perpendicular vectors is always zero.

Q: Can I just use coordinates like $\vec{u} = (u_1, u_2, u_3)$ to prove this?

You absolutely can, and it is a valid method! However, using coordinate geometry often leads to messy algebraic expansion. The geometric approach we used here is more elegant and helps you build an intuition for how vectors behave in space, which is a crucial skill for JEE Advanced.

Q: Is Lagrange's Identity always true for any two vectors?

Yes, it is a fundamental identity in vector algebra. It holds true regardless of the dimension, though it is most commonly applied in 3D space. It essentially acts as a bridge between the dot product (scalar projection) and the cross product (vector area).

Q: Why did we expand the RHS instead of the LHS in part (b)?

In proofs, it is almost always easier to expand the more complex-looking side and simplify it down to the simpler side. The RHS has more terms, which gives us more 'material' to work with and manipulate using the identities we already know.

Q: What if the angle $\theta$ is $0$ or $\pi$? Does the identity still hold?

Great observation! If $\theta = 0$ or $\pi$, the vectors are collinear. In this case, the cross product magnitude $|\vec{u} \times \vec{v}|$ becomes zero, and the dot product becomes $\pm |\vec{u}||\vec{v}|$. If you plug these into the identity, you will find that it still holds perfectly. The math is robust!

Question

For any two vectors $\vec{u}$ and $\vec{v}$, prove that (a) $(\vec{u} \cdot \vec{v})^2 + |\vec{u} 	imes \vec{v}|^2 = |\vec{u}|^2 |\vec{v}|^2$ and (b) $(1 + |\vec{u}|^2)(1 + |\vec{v}|^2) = (1 - \vec{u} \cdot \vec{v})^2 + |\vec{u} + \vec{v} + (\vec{u} 	imes \vec{v})|^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Harmony

Imagine you are standing in a 3D space with two vectors, $\vec{u}$ and $\vec{v}$, originating from the same point. Let $	heta$ be the angle between them.

We know the dot product, $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos 	heta$, measures how much these vectors 'agree' with each other. Conversely, the cross product magnitude, $|\vec{u} 	imes \vec{v}| = |\vec{u}| |\vec{v}| \sin 	heta$, measures the area of the parallelogram they span.

Now, consider the expression $(\vec{u} \cdot \vec{v})^2 + |\vec{u} 	imes \vec{v}|^2$. If we substitute our definitions, we obtain:

$$ (|\vec{u}| |\vec{v}| \cos 	heta)^2 + (|\vec{u}| |\vec{v}| \sin 	heta)^2 $$

The common factor $|\vec{u}|^2 |\vec{v}|^2$ is waiting to be pulled out. Once we factor it, we are left with $(\cos^2 	heta + \sin^2 	heta)$.

Since this is the fundamental identity of trigonometry, which equals $1$, we arrive at the elegant **Lagrange's Identity**:

$$ (\vec{u} \cdot \vec{v})^2 + |\vec{u} 	imes \vec{v}|^2 = |\vec{u}|^2 |\vec{v}|^2 $$

This is the Pythagorean theorem manifesting in vector space. It confirms that the square of the projection and the square of the perpendicular area always sum to the square of the product of the magnitudes.

### The Grand Expansion

Now, let us tackle the second expression: $(1 - \vec{u} \cdot \vec{v})^2 + |\vec{u} + \vec{v} + (\vec{u} 	imes \vec{v})|^2$. In JEE Advanced, complexity is often just a mask for simplicity.

Let us focus on the second term. We treat $\vec{A} = (\vec{u} + \vec{v})$ and $\vec{B} = (\vec{u} 	imes \vec{v})$ as two distinct entities. Using the expansion $|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2\vec{A} \cdot \vec{B}$, we get:

$$ |\vec{u} + \vec{v}|^2 + |\vec{u} 	imes \vec{v}|^2 + 2(\vec{u} + \vec{v}) \cdot (\vec{u} 	imes \vec{v}) $$

Here is the 'Aha!' moment. The vector $(\vec{u} 	imes \vec{v})$ is, by definition, perpendicular to the plane containing $\vec{u}$ and $\vec{v}$.

Therefore, it is perpendicular to their sum $(\vec{u} + \vec{v})$. The dot product $2(\vec{u} + \vec{v}) \cdot (\vec{u} 	imes \vec{v})$ vanishes into thin air, leaving us with:

$$ |\vec{u} + \vec{v}|^2 + |\vec{u} 	imes \vec{v}|^2 $$

### The Final Convergence

Now, we expand the remaining pieces. Expanding $(1 - \vec{u} \cdot \vec{v})^2$ gives us $1 - 2\vec{u} \cdot \vec{v} + (\vec{u} \cdot \vec{v})^2$.

Expanding $|\vec{u} + \vec{v}|^2$ gives us $|\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$. When we combine everything, the $-2\vec{u} \cdot \vec{v}$ and $+2\vec{u} \cdot \vec{v}$ terms cancel out perfectly.

We are left with:

$$ 1 + |\vec{u}|^2 + |\vec{v}|^2 + (\vec{u} \cdot \vec{v})^2 + |\vec{u} 	imes \vec{v}|^2 $$

Look closely at the last two terms. They are exactly what we solved in Part (a). Replacing them with $|\vec{u}|^2 |\vec{v}|^2$, we get:

$$ 1 + |\vec{u}|^2 + |\vec{v}|^2 + |\vec{u}|^2 |\vec{v}|^2 $$

With one final act of factoring by grouping, we pull out $(1 + |\vec{u}|^2)$ to reach the final result:

$$ (1 + |\vec{u}|^2)(1 + |\vec{v}|^2) $$

You have just proven that these vectors, no matter how they are oriented, obey a beautiful, rigid logic. Keep that curiosity alive, and you will conquer any problem the exam throws at you.

Animated Solution for Mathematics - Vector Algebra: For any two vectors $u$ and $v$ , prove that (a) $(u \cdot v)^{2} + ∣ u \times v ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2}$ and (b) $(1 + ∣ u ∣^{2}) (1 + ∣ v ∣^{2}) = (1 - u \cdot v)^{2} + ∣ u + v + (u \times v) ∣^{2}$ .

Visualized Solution

Visualizing the Vectors

The Cross Product Vector

Defining the Dot Product

Defining the Cross Product Magnitude

Squaring and Adding Terms

Factoring and Trigonometric Identity

Conclusion of Part (a)

Starting Part (b) - Expanding RHS

Applying Vector Magnitude Property

The Perpendicularity Trick

Expanding the Remaining Squares

Combining All Expanded Terms

Applying Lagrange's Identity

Final Factorization

Conclusion and Takeaway

The Sigma Insight: Vector (Cross) Product

Analyzing the Geometric Harmony

The Grand Expansion

The Final Convergence

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