Question

Find the angle between the following pairs of lines:

(i)$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$

(ii $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$

Answer 1

(i) Let $\overrightarrow b_1$ and $\overrightarrow b_2$ be the vectors parallel to the pair of lines, $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$, respectively.

∴ $\overrightarrow b_1=2\hat i+5\hat j-3\hat k$ and $\overrightarrow b_2=-\hat i+8\hat j+4\hat k$

$\mid \overrightarrow b_1\mid =\sqrt{(2)^2+(5)^2+(-3)^2}=\sqrt {38}$

$\mid \overrightarrow b_2\mid =\sqrt{(-1)^2+(8)^2+(4)^2}=\sqrt {81}=9$

$\mid \overrightarrow b_1.\mid \overrightarrow b_2$ = $(2\hat i+5\hat j-3\hat k)$.$(-\hat i+8\hat j+4\hat k)$

= 2(-1)+5×8+(-3).4
=-2+40-12
=26

The angle, Q, between the given pair of lines is given by the relation,

cos Q=$\begin{vmatrix}\frac{\overrightarrow b_1.\overrightarrow b_2}{\|\overrightarrow b_1 \| \overrightarrow b_2 \|} \end{vmatrix}$

$\Rightarrow$ cos Q=$\frac{26}{9\sqrt{38}}$

$\Rightarrow$ Q=$\cos^{-1} \bigg(\frac{26}{9\sqrt{38}}\bigg)$

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(ii) Let $\overrightarrow b_1,\overrightarrow b_2$ be the vectors parallel to the given pair of lines, $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$, respectively.

$\overrightarrow b_1=2\hat i+2\hat j+\hat k$ and $\overrightarrow b_2=4\hat i+\hat j+8\hat k$

∴|$\overrightarrow b_1$|=$\sqrt{(2)^2+(2)^2+(1)^2}=\sqrt9=3$

|$\overrightarrow b_2$|=$\sqrt{4^2+1^2+8^2}=\sqrt{81}=9$

$\overrightarrow b_1.\overrightarrow b_2$=(2\hat i+2\hat j+\hat k).$$(4\hat i+\hat j+8\hat k)

=2×4+2×1+1×8
=8+2+8
=18

If Q is the angle between the given pair of lines,

then cos Q=$\begin{vmatrix}\frac{\overrightarrow b_1.\overrightarrow b_2}{\|\overrightarrow b_1 \| \overrightarrow b_2 \|} \end{vmatrix}$

$\Rightarrow$ cos Q=$\frac{18}{3*9}=\frac{2}{3}$

$\Rightarrow$ Q= $\cos^{-1}\bigg(\frac{2}{3}\bigg)$