Question

Find the angle between the pairs of lines with direction ratios proportional to
(i) 5, −12, 13 and −3, 4, 5
(ii) 2, 2, 1 and 4, 1, 8
(iii) 1, 2, −2 and −2, 2, 1
(iv) a, b, c and b − c, c − a, a − b.

Answer 1

(i) 5, −12, 13 and −3, 4, 5

$$\begin{gathered} \mathrm{Let}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_2}\mathrm{be}\mathrm{vectors}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{two}\mathrm{given}\mathrm{lines}. \\ \mathrm{Then},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{given}\mathrm{lines}\mathrm{is}\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_2}. \\ \mathrm{Now}, \\ \overrightarrow{m_{\mathit{1}}}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{to}5,-12,13 \\ \overrightarrow{m_2}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{to}-3,4,5 \\ \therefore \overrightarrow{m_1}=5\hat{i}-12\hat{j}+13\hat{k} \\ \overrightarrow{m_2}=-3\hat{i}+4\hat{j}+5\hat{k} \\ \mathrm{Let}\theta \mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines}. \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{m_1}.\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|} \\ =\frac{\left(5\hat{i}-12\hat{j}+13\hat{k}\right).\left(-3\hat{i}+4\hat{j}+5\hat{k}\right)}{\sqrt{5^2+{\left(-12\right)}^2+{13}^2}\sqrt{{\left(-3\right)}^2+4^2+5^2}} \\ =\frac{-15-48+65}{13\sqrt{2}\times 5\sqrt{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=\frac{1}{65} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{65}\right) \end{gathered}$$

(ii) 2, 2, 1 and 4, 1, 8

$$\begin{gathered} \mathrm{Let}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_2}\mathrm{be}\mathrm{vectors}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{two}\mathrm{lines}. \\ \mathrm{Then},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines}\mathrm{is}\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_{2.}} \\ \mathrm{Now}, \\ \overrightarrow{m_{\mathit{1}}}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{to}2,2,1 \\ \overrightarrow{m_2}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{t}o4,1,8 \\ \therefore \overrightarrow{m_1}=2\hat{i}-2\hat{j}+\hat{k} \\ \overrightarrow{m_2}=4\hat{i}+\hat{j}+8\hat{k} \\ \mathrm{Let}\theta \mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines}. \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{m_1}.\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|} \\ =\frac{\left(2\hat{i}+2\hat{j}+\hat{k}\right).\left(4\hat{i}+\hat{j}+8\hat{k}\right)}{\sqrt{2^2+2^2+1^2}\sqrt{4^2+1^2+8^2}} \\ =\frac{8+2+8}{3\times 9} \\ =\frac{2}{3} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{3}\right) \end{gathered}$$

(iii) 1, 2, −2 and −2, 2, 1

$$\begin{gathered} \mathrm{Let}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_2}\mathrm{be}\mathrm{vectors}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{two}\mathrm{given}\mathrm{lines}. \\ \mathrm{Then},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{given}\mathrm{lines}\mathrm{is}\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_2.} \\ \mathrm{Now}, \\ \overrightarrow{m_{\mathit{1}}}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{to}1,2,-2 \\ \overrightarrow{m_2}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{t}o-2,2,1 \\ \therefore \overrightarrow{m_1}=\hat{i}+2\hat{j}-2\hat{k} \\ \overrightarrow{m_2}=-2\hat{i}+2\hat{j}+\hat{k} \\ \mathrm{Let}\theta \mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines}. \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{m_1}.\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|} \\ =\frac{\left(\hat{i}+2\hat{j}-2\hat{k}\right).\left(-2\hat{i}+2\hat{j}+\hat{k}\right)}{\sqrt{1^2+2^2+{\left(-2\right)}^2}\sqrt{{\left(-2\right)}^2+2^2+1^2}} \\ =\frac{-2+4-2}{3\times 3} \\ =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$

(iv) a, b, c and b − c, c − a, a − b.

$$\begin{gathered} \mathrm{Let}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_2}\mathrm{be}\mathrm{vectors}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{two}\mathrm{lines}. \\ \mathrm{Then},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{lines}\mathrm{is}\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{m_1}\mathrm{and}\overrightarrow{m_2}. \\ \mathrm{Now}, \\ \overrightarrow{m_{\mathit{1}}}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{to}a,b,c \\ \overrightarrow{m_2}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{having}\mathrm{direction}\mathrm{ratios}\mathrm{proportional}\mathrm{t}ob-c,c-a,a-b \\ \therefore \overrightarrow{m_1}=a\hat{i}+b\hat{j}+c\hat{k}\mathrm{and}\overrightarrow{m_2}=\left(b-c\right)\hat{i}+\left(c-a\right)\hat{j}+\left(a-b\right)\hat{k} \\ \mathrm{Let}\theta \mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{lines}. \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{m_1}.\overrightarrow{m_2}}{\left|\overrightarrow{m_1}\right|\left|\overrightarrow{m_2}\right|} \\ =\frac{\left(a\hat{i}+b\hat{j}+c\hat{k}\right).\left\{\left(b-c\right)\hat{i}+\left(c-a\right)\hat{j}+\left(a-b\right)\hat{k}\right\}}{\sqrt{a^2+b^2+c^2}\sqrt{{\left(b-c\right)}^2+{\left(c-a\right)}^2+{\left(a-b\right)}^2}} \\ =\frac{ab-ac+bc-ba+ca-cb}{\sqrt{a^2+b^2+c^2}\sqrt{{\left(b-c\right)}^2+{\left(c-a\right)}^2+{\left(a-b\right)}^2}} \\ =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$