Q.If angle between two unit vectors $\vec{a}$ and $\vec{b}$ be $\theta$, then prove that $\sin\dfrac{\theta}{2} = \dfrac{1}{2}|\vec{a} - \vec{b}|$.

The problem asks us to find the magnitude of one of the vectors given a relationship involving their dot product and another relationship between their magnitudes. The core idea here is to correctly expand the dot product expression and then use the given magnitude relationship to form an equation that can be solved.

The expression $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})$ is analogous to the algebraic identity $(x+y)(x-y) = x^2 - y^2$. In vector algebra, the dot product behaves similarly, but we must remember that $\vec{x} \cdot \vec{x} = |\vec{x}|^2$. This property is crucial for simplifying the expression into terms of vector magnitudes.

1. Expand the dot product expression. We use the distributive property of the dot product, which works just like multiplication in scalar algebra:

2. Simplify using dot product properties.

   Recall two fundamental properties of the dot product:

   • The dot product of a vector with itself gives the square of its magnitude: $\vec{x} \cdot \vec{x} = |\vec{x}|^2$.
   • The dot product is commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.

   Applying these properties to our expanded expression:

The middle terms, $-\vec{a} \cdot \vec{b}$ and $+\vec{a} \cdot \vec{b}$, cancel each other out.
So, the expression simplifies to:

> [!IMPORTANT]
> This is a very common and useful identity in vector algebra:
> $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$

We are given that $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198$.
Therefore, we have the equation:

3. Substitute the given relationship between magnitudes. …

The key idea here is simple but often flipped: a vector is defined by its terminal minus initial coordinates. If you mix up which point is which, you'll get the opposite sign — a classic trap in vector problems.

Let’s break it down.

1. Understand what the question asks We are given terminal point A $(2, -3, 5)$ and initial point B $(3, -4, 7)$. The vector from B to A is written as $\overrightarrow{BA}$ (or sometimes $\vec{v}$ with tail at B and head at A). The formula is:

2. Apply the formula component-wise

   Subtract B’s coordinates from A’s:

   • $x$-component: $2 - 3 = -1$
   • $y$-component: $-3 - (-4) = -3 + 4 = 1$
   • $z$-component: $5 - 7 = -2$

3. Write the vector in unit vector form

   The components $(-1, 1, -2)$ correspond to:

4. Match with the options Looking at the choices:
   • (A) $\hat{i} - \hat{j} + 2\hat{k}$
   • (B) $\hat{i} + \hat{j} + 2\hat{k}$
   • (C) $-\hat{i} - \hat{j} - 2\hat{k}$ …