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 \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
(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= \(\cos^{-1}\bigg(\frac{2}{3}\bigg)\)
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)\)
and \(\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)\)
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\) and
\(\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\)
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0,0,0) 3x-4y+12z=3
(b) (3,-2,1) 2x-y+2z+3=0
(c) (2,3,-5) x+2y-2z=9
(d) (-6,0,0) 2x-3y+6z-2=0
determine whether the given planes are parallel or perpendicular,and in case they are neither, find the angles between them. (a)7x+5y+6z+30=0 and 3x-y-10z+4=0
(b)2x+y+3z-2=0 and x-2y+5=0
(c)2x-2y+4z+5=0 and 3x-3y+6z-1=0
(d)2x-y+3z-1=0 and 2x-y+3z+3=0
(e)4x+8y+z-8=0 and y+z-4=0
The two straight lines, whenever intersects, form two sets of angles. The angles so formed after the intersection are;
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