If vectors $\vec{a} = \frac{\vec{i} + \vec{j}}{\sqrt{2}}, \vec{b} = \frac{-\vec{i} + \vec{j}}{\sqrt{2}}$ and $\vec{c} = \vec{k}$ then the value of $(\vec{r} \cdot \vec{a})^2 + (\vec{r} \cdot \vec{b})^2 + (\vec{r} \cdot \vec{c})^2$ is equal to:
Step-by-Step Solution
Key Concept: Sum of squared direction cosines for orthonormal basis equals the magnitude squared of the vector.
Any vector $\vec{r}$ can be decomposed as $\vec{r} = (\vec{r} \cdot \vec{a})\vec{a} + (\vec{r} \cdot \vec{b})\vec{b} + (\vec{r} \cdot \vec{c})\vec{c}$ where $\vec{a}, \vec{b}, \vec{c}$ are orthonormal. Taking dot product with itself: $(\vec{r} \cdot \vec{a})^2 + (\vec{r} \cdot \vec{b})^2 + (\vec{r} \cdot \vec{c})^2 = |\vec{r}|^2$, which follows from orthonormality.
Correct Answer: 1