Kerala KEAM 5 June 2024 Solved Paper

Step 1: The given expression is: $$| \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} | ^2.$$First, expand the square of the vector: $$| \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} | ^2 = (\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) \cdot (\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}).$$Using the distributive property of the dot product: $$= \vec{a} \cdot \vec{a} - 2 (\vec{a} \cdot \vec{b}) (\vec{a} \cdot \vec{b}) + (\vec{a} \cdot \vec{b})^2 \vec{b} \cdot \vec{b}.$$Since $\vec{a}$ and $\vec{b}$ are unit vectors, we have $\vec{a} \cdot \vec{a} = 1$ and $\vec{b} \cdot \vec{b} = 1$. So the expression becomes: $$1 - 2 (\vec{a} \cdot \vec{b})^2 + (\vec{a} \cdot \vec{b})^2.$$Thus, we get: $$1 - (\vec{a} \cdot \vec{b})^2.$$Now, using the formula for the dot product of two unit vectors: $$\vec{a} \cdot \vec{b} = \cos \theta.$$Therefore, the expression becomes: $$1 - \cos^2 \theta.$$Using the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$, we can simplify this to: $$\sin^2 \theta.$$Thus, the correct answer is option (B). Quick Tip: For unit vectors $\vec{a}$ and $\vec{b}$, the dot product $\vec{a} \cdot \vec{b} = \cos \theta$, and this identity simplifies many vector-related expressions.