Question

Find the angle between the line $\overrightarrow{r}=\left(2\hat{i}+3\hat{j}+9\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}+4\hat{k}\right)$ and the plane $\overrightarrow{r}·\left(\hat{i}+\hat{j}+\hat{k}\right)=5.$

Answer 1

$$\begin{gathered} \text{We know that the angle θ between the line}\overrightarrow{r}\text{=}\overrightarrow{a}\text{+ λ}\overrightarrow{b}\text{and the plane}\overrightarrow{r}\text{.}\overrightarrow{n}\text{=}\mathit{\text{d}}\text{is given by} \\ \sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}. \\ \text{Here,} \\ \overrightarrow{b}=2\hat{i}+3\hat{j}+4\hat{k}\text{and}\overrightarrow{n}=\hat{i}+\hat{j}+\hat{k} \\ \text{So,}\sin \theta =\frac{\left(2\hat{i}+3\hat{j}+4\hat{k}\right).\left(\hat{i}+\hat{j}+\hat{k}\right)}{\left|2\hat{i}+3\hat{j}+4\hat{k}\right|\left|\hat{i}+\hat{j}+\hat{k}\right|}=\frac{2+3+4}{\sqrt{4+9+16}\sqrt{1+1+1}}=\frac{9}{\sqrt{29}\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{29}} \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{3\sqrt{3}}{\sqrt{29}}\right) \end{gathered}$$