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 \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
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 \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Solution and Explanation
(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=\(\cos^{-1} \bigg(\frac{26}{9\sqrt{38}}\bigg)\)
(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= \(\cos^{-1}\bigg(\frac{2}{3}\bigg)\)
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Concepts Used:
Angle between Two Lines
The two straight lines, whenever intersects, form two sets of angles. The angles so formed after the intersection are;
- A pair of acute angle
- Another pair of an obtuse angle
The absolute values of angles created depend on the slopes of the intersecting lines.
It is also worth taking note, that the angle so formed by the intersection of two lines cannot be calculated if any of the lines is parallel to the y-axis as the slope of a line parallel to the y-axis is an indeterminate.
Read More: Angle Between Two Lines