Question

If the angle between the vectors $\hat{i}+\hat{k}\mathrm{and}\hat{i}-\hat{j}+\alpha \overrightarrow{k}is\frac{\mathrm{\pi }}{3},\mathrm{then}\alpha =\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Given:
The angle between the vectors $\hat{i}+\hat{k}\mathrm{and}\hat{i}-\hat{j}+\alpha \overrightarrow{k}\mathrm{is}\frac{\mathrm{\pi }}{3}$


$$\begin{gathered} \mathrm{Let}\overrightarrow{a}=\hat{i}+\hat{k}\mathrm{and}\overrightarrow{b}=\hat{i}-\hat{j}+\alpha \overrightarrow{k} \\ \mathrm{Angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{is}\theta =\frac{\mathrm{\pi }}{3} \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2} \\ =\sqrt{2} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2+{\left(\alpha \right)}^2} \\ =\sqrt{2+{\alpha }^2} \\ \overrightarrow{a}·\overrightarrow{b}=\left(1\right)\left(1\right)+\left(0\right)\left(-1\right)+\left(1\right)\left(\alpha \right) \\ =1+\alpha \\ \mathrm{Now}, \\ \overrightarrow{a}·\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ \Rightarrow 1+\alpha =\sqrt{2}\sqrt{2+{\alpha }^2}\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow 1+\alpha =\sqrt{4+2{\alpha }^2}\times \frac{1}{2} \\ \Rightarrow 2+2\alpha =\sqrt{4+2{\alpha }^2} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(2+2\alpha \right)}^2=4+2{\alpha }^2 \\ \Rightarrow 4+4{\alpha }^2+8\alpha =4+2{\alpha }^2 \\ \Rightarrow 2{\alpha }^2+8\alpha =0 \\ \Rightarrow 2\alpha \left(\alpha +4\right)=0 \\ \Rightarrow \alpha =0,-4 \end{gathered}$$

Hence, $\alpha =\underline{0,-4}$.