Question

The angle between the vectors with direction ratios proportional to 1, 1, 2 and $\sqrt{3}-1,-\sqrt{3}-1,4$ is ________________.

Answer 1

Let $\overrightarrow{a}$ be a vector parallel to the vector having direction ratios 1, 1 and 2 and $\overrightarrow{b}$ be the vector parallel to the vector having direction ratios $+\sqrt{3}-1,-\sqrt{3}-1,4$

$$\begin{gathered} \therefore \overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k} \\ \overrightarrow{b}=\left(\sqrt{3}-1\right)\hat{i}+\left(-\sqrt{3}-1\right)\hat{j}+4\hat{k} \\ \mathrm{then}\mathrm{cos\theta }=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\left(\hat{i}+\hat{j}+2\hat{k}\right).\left(\left(\sqrt{3}-1\right)\hat{i}-\left(\sqrt{3}+1\right)\hat{j}+4\hat{k}\right)\mathrm{where}\mathrm{\theta }\mathrm{is}\mathrm{angle}\mathrm{between}\overrightarrow{a}\mathrm{and}\overrightarrow{b} \\ \mathrm{i}.\mathrm{e}\mathrm{cos\theta }=\frac{\left(\sqrt{3}-1\right).1-\left(\sqrt{3}+1\right).1+8}{\sqrt{1^2+1^2+2^2}\sqrt{{\left(\sqrt{3}-1\right)}^2+{\left(\sqrt{3}+1\right)}^2+16}} \\ =\frac{6}{\sqrt{6}\times \sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}+16}} \\ =\frac{6}{\sqrt{6}\times \sqrt{8}} \\ \mathrm{i}.\mathrm{e}\mathrm{cos\theta }=\frac{\sqrt{6}}{2\sqrt{2}}=\frac{\sqrt{3}}{2} \\ \mathrm{i}.\mathrm{e}\mathrm{\theta }=\frac{\mathrm{\pi }}{3} \end{gathered}$$