The value of c which makes the angle 45° between $\vec{a} = \hat{i} + c\hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$ is
- A. 1
- B. $1/\sqrt{2}$
- C. $-1/\sqrt{2}$
- D. 0
Answer: D. 0
Explanation: $\cos 45 = \frac{\vec{a}\cdot\vec{b}}{|a||b|} = \frac{1+c}{\sqrt{1+c²}\sqrt{2}} = \frac{1}{\sqrt{2}}$. $1+c = \sqrt{1+c²}$. Square both sides: $1+2c+c² = 1+c² \Rightarrow 2c=0 \Rightarrow c=0$.