Let AB be the line through the points (4,7,8), and (2,3,4), and CD be the line through the points (-1,-2,1), and (1,2,5).
The direction ratios, a1, b1, c1 of AC are (2-4), (3-7), and (4-8) i.e., -2, -4, and -4.
The direction ratios, a2, b2, c2 of CD are (1-(-1)), (2-(-2)), and (5-1) i.e., 2, 4, and 4.
AB will parellel to CD, if \(\frac{a_1}{a_2}\)=\(\frac{b_1}{b_2}\)=\(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\)
=\(-\frac{2}{2}\)
=-1
\(\frac{b_1}{b_2}\)
=\(-\frac{4}{4}\)
=-1
\(\frac{c_1}{c_2}\)
=\(-\frac{4}{4}\)
=-1
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
Thus, AB is parallel to CD.
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
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)