Topic: Engineering Mathematics
The value of c which makes the angle $45^{\circ}$ between $\vec{a}=i+cj$ and $\vec{b}=i+j$ is
- A. 1
- B. $\frac{1}{\sqrt{2}}$
- C. $\frac{-1}{\sqrt{2}}$
- D. 0
Answer: D. 0
Explanation: $\vec{a}\cdot\vec{b}=|\vec{a}|.|\vec{b}|\cos~\theta$\n$\cos~\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$\n$\cos~45^{\circ}=\frac{1+c}{\sqrt{2}\cdot\sqrt{1+c^{2}}}$\n$\frac{1}{\sqrt{2}}=\frac{1+c}{\sqrt{2}\cdot\sqrt{1+c^{2}}}$\n$(1+c)^{2}=1+c^{2}$\n$\Rightarrow c^{2}+2c+1=c^{2}+1$\n$c=0$