Q.The value of the expression $|\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2$ is ________.

The key idea is the relationship between the dot product, cross product magnitude, and the magnitudes of the vectors — essentially the vector triple product identity or the Lagrange identity.

Step 1: Recall the definitions. For any two vectors $\vec{a}$ and $\vec{b}$:

• $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$
• $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$

Step 2: Square both expressions and add them: …

This problem is a classic example of how the dot and cross products together encode the full geometric relationship between two vectors. The dot product gives $|\vec{a}||\vec{b}|\cos\theta$, while the magnitude of the cross product gives $|\vec{a}||\vec{b}||\sin\theta|$. Squaring and adding them eliminates the angle dependence entirely.

Let’s work through it step by step.

1. Recall the definitions

   For any two vectors $\vec{a}$ and $\vec{b}$ with an angle $\theta$ between them:

   • $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$
   • $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}||\sin\theta|$

   These are not just formulas — they capture how much the vectors align (dot) versus how much they are perpendicular (cross).

2. Square both expressions

   • $(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\cos^2\theta$
   • $|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2\sin^2\theta$ (The absolute value on $\sin\theta$ disappears when squared.)

3. Add them

4. Apply the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, so the expression simplifies to: $|\vec{a}|^2|\vec{b}|^2$ …

Method: The Lagrange Identity for $|\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2$

Use this whenever the squared cross- and dot-products of the same pair of vectors are added.

Steps

Step 1: Write each product in magnitude–angle form

$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$ and $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$.

Step 2: Square and add

$|\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2=|\vec{a}|^2|\vec{b}|^2(\sin^2\theta+\cos^2\theta).$ …

Common Mistakes

Mistake 1: Using the cross-product vector instead of its magnitude

Why it's wrong: $\vec{a}\times\vec{b}$ is a vector; the identity needs $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$. Correct approach: square the magnitude so the $\sin^2\theta$ term appears cleanly.

Mistake 2: Forgetting to invoke $\sin^2\theta+\cos^2\theta=1$

Why it's wrong: without it the answer is left angle-dependent. Correct approach: after squaring and adding, the identity collapses everything to $|\vec{a}|^2|\vec{b}|^2$. …