Question

Find the angle between the given planes.
(i) $\overrightarrow{r}·\left(2\hat{i}-3\hat{j}+4\hat{k}\right)=1\mathrm{and}\overrightarrow{r}·\left(-\hat{i}+\hat{j}\right)=4$

(ii) $\overrightarrow{r}·\left(2\hat{i}-\hat{j}+2\hat{k}\right)=6\mathrm{and}\overrightarrow{r}·\left(3\hat{i}+6\hat{j}-2\hat{k}\right)=9$

(iii) $\overrightarrow{r}·\left(2\hat{i}+3\hat{j}-6\hat{k}\right)=5\mathrm{and}\overrightarrow{r}·\left(\hat{i}-2\hat{j}+2\hat{k}\right)=9$

Answer 1

$$\begin{gathered} \left(i\right)\text{We know that the angle between the planes}\overrightarrow{r}.\overrightarrow{n_1}=d_{\mathit{1}}\mathit{,}\mathit{}\overrightarrow{r}\mathit{.}\mathit{}\overrightarrow{n_{\mathit{2}}}=d_2\text{is given by} \\ \cos \theta =\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} \\ \text{Here,}\overrightarrow{n_1}=2\hat{i}-3\hat{j}+4\hat{k};\overrightarrow{n_2}=-\hat{i}+\hat{j}+0\hat{k} \\ \text{So,}\cos \theta =\frac{\left(2\hat{i}-3\hat{j}+4\hat{k}\right).\left(-\hat{i}+\hat{j}+0\hat{k}\right)}{\left|2\hat{i}-3\hat{j}+4\hat{k}\right|\left|-\hat{i}+\hat{j}+0\hat{k}\right|}=\frac{-2-3}{\sqrt{4+9+16}\sqrt{1+1+0}}=\frac{-5}{\sqrt{29}\sqrt{2}}=\frac{-5}{\sqrt{58}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-5}{\sqrt{58}}\right) \end{gathered}$$

$$\begin{gathered} \left(ii\right)\text{We know that the angle between the planes}\overrightarrow{r}.\overrightarrow{n_1}=d_{\mathit{1}}\mathit{,}\mathit{}\overrightarrow{r}\mathit{.}\mathit{}\overrightarrow{n_{\mathit{2}}}=d_2\text{is given by} \\ \cos \theta =\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} \\ \text{Here,}\overrightarrow{n_1}=2\hat{i}-\hat{j}+2\hat{k};\overrightarrow{n_2}=3\hat{i}+6\hat{j}-2\hat{k} \\ \text{So,}\cos \theta =\frac{\left(2\hat{i}-\hat{j}+2\hat{k}\right).\left(3\hat{i}+6\hat{j}-2\hat{k}\right)}{\left|2\hat{i}-\hat{j}+2\hat{k}\right|\left|3\hat{i}+6\hat{j}-2\hat{k}\right|}=\frac{6-6-4}{\sqrt{4+1+4}\sqrt{9+36+4}}=\frac{-4}{\left(3\right)\left(7\right)}=\frac{-4}{21} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-4}{21}\right) \end{gathered}$$

$$\begin{gathered} \left(iii\right)\text{We know that the angle between the planes}\overrightarrow{r}.\overrightarrow{n_1}=d_{\mathit{1}}\mathit{,}\mathit{}\overrightarrow{r}\mathit{.}\mathit{}\overrightarrow{n_{\mathit{2}}}=d_2\text{is given by} \\ \cos \theta =\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} \\ \text{Here,}\overrightarrow{n_1}=2\hat{i}+3\hat{j}-6\hat{k};\overrightarrow{n_2}=\hat{i}-2\hat{j}+2\hat{k} \\ \text{So,}\cos \theta =\frac{\left(2\hat{i}+3\hat{j}-6\hat{k}\right).\left(\hat{i}-2\hat{j}+2\hat{k}\right)}{\left|2\hat{i}+3\hat{j}-6\hat{k}\right|\left|\hat{i}-2\hat{j}+2\hat{k}\right|}=\frac{2-6-12}{\sqrt{4+9+36}\sqrt{1+4+4}}=\frac{-16}{\left(7\right)\left(3\right)}=\frac{-16}{21} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-16}{21}\right) \end{gathered}$$