Parallel and Perpendicular lines in space

Parallel Lines in space

Two lines having vector equations as

$\dpi{120} \overrightarrow{r_{1}}=\overrightarrow{a_{1}}+t\overrightarrow{b_{1}}$ and

$\dpi{120} \overrightarrow{r_{2}}=\overrightarrow{a_{2}}+t\overrightarrow{b_{2}}$
are parallel if their direction vectors b1 and b2 are parallel

i.e $\dpi{120} \mathbf{\overrightarrow{b_{1}}=k \overrightarrow{b_{2}}}$ for some scalar k.

Example1. Find the vector equation of a line passing through a point (2,-1,3) and parallel to the line r = (i+j) + t(2i+j-2k).

Solution: We know that any line parallel to given line will have same direction vectors as of the given line.

So we have to find the vector equation of a line which is passing through (2,-1,3) and have direction vector as 〈2,1,-2〉.

r = 〈2,-1,3〉 + t 〈2,1,-2〉

r = 〈 2+2t, -1+t , 3-2t〉

Parametric equations of line are:

x=2+2t, y=-1+t, z=3-2t

and Cartesian(symmetric) equation is given as…

$\dpi{120} \frac{x-2}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$

Line passing through a point and perpendicular to two given lines.

Suppose a line is passing through point P and perpendicular to two lines

$\dpi{120} \overrightarrow{r_{1}}=\overrightarrow{a_{1}}+t\overrightarrow{b_{1}}$ and

$\dpi{120} \overrightarrow{r_{2}}=\overrightarrow{a_{2}}+t\overrightarrow{b_{2}}$

then its equation would be given as,

$\dpi{150} \mathbf{\overrightarrow{r}= \overrightarrow{p}+t\left ( \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}} \right )}$

Example2. A line passes through point (2,-1,3) and is perpendicular to the line = (i+j-k) + t(2i-2j+k) and = (2i-j-3k) + t(i+2j+2k). Find the equation of this line.

Solution:

$\dpi{120} \overrightarrow{b_{1}}\times \overrightarrow{b_{2}}=\begin{vmatrix} i & j & k\\ 2& -2 &1 \\ 1& 2& 2 \end{vmatrix}$

= i(-4-2) –j(4-1)+k(4+2)

= -6i-3j+6k

So the vector equation of perpendicular line would be,

$\dpi{120} \mathbf{\overrightarrow{r}= \overrightarrow{p}+t\left ( \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}} \right )}$

$\dpi{120} \mathbf{\overrightarrow{r}= \left \langle 2,-1,3 \right \rangle+t\left ( -6 -3,6 \right )}$

Intersection of two lines:

Following algorithm is used to find intersection of two lines.

Write the general points lying on both lines using parameters s and t.
If the lines intersect , then they must have a common point. So set the corresponding parts of both lines equal to each other.
Solve the three equations for parameters s and t. If these values satisfy the third equation then the lines intersect otherwise not.

Example3.Show that the lines r=〈1,1,-1〉 + s〈3,-1,0〉 and r = 〈4,0,-1〉 + t 〈2,0,3〉 intersect each other. Also find their point of intersection.

Solution: The general points lying on first line is given as,

r=〈1,1,-1〉 + s〈3,-1,0〉 => (1+3s , 1-s, -1)

General points lying on second line is given as,

r = 〈4,0,-1〉 + t 〈2,0,3〉 => (4+2t,0,-1+3t)

If two lines intersect then they must have some common point. So,

1+3s = 4+2t 1-s = 0 -1=-1+3t

3s-2t = 3 s =1

Substituting s=1 into 3s-2t=3, we get…

3(1)-2t = 3

-2t = 0 => t=0

These values of s and t must satisfy the third equation.

Substituting t=0 into -1+3t = -1

We get, -1 = -1

So these lines intersect at s=1 and t=0. Using these values in line equations , we get coordinates of intersecting point as,

(1+3s , 1-s, -1) = (4,0,-1)

(4+2t,0,-1+3t) = (4,0,-1)

Example4. Find point of intersection of following two lines.

$\dpi{120} {\color{Red} L_{1}: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}}$

$\dpi{120} {\color{Red} L_{2}: \frac{x-4}{5}=\frac{y-1}{2}= z}$

Solution: General points lying on line $\dpi{120} L_{1}: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=s$ are given as,

(2s+1 , 3s+2 ,4s+3)

Points on line $\dpi{120} L_{2}: \frac{x-4}{5}=\frac{y-1}{2}= z=t$ are given as,

(5t+4,2t+1,t)

If two lines intersect then they must have a common point. So,

2s+1=5t+4 3s+2=2t+1 4s+3 = t

2s-5t = 3 3s-2t = -1 4s-t = -3

Solving first two equations, we get

s = -1 and t= -1

Substituting these values into third equation, we get

4s-t =-3 => 4(-1)-(-1) = -3

-4+1 = -3

-3 = -3

The values of s and t satisfy third equation too so these lines intersect and coordinates of intersecting point are,

s =-1, (2s+1 , 3s+2 ,4s+3)= (2(-1)+1 , 3(-1)+2 ,4(-1)+3)

= (-1,-1,-1)

t= -1 (5t+4,2t+1,t) = (5(-1)+4,2(-1)+1, -1)

= (-1,-1,-1)

Check whether given lines are parallel, perpendicular or Skew.

Here is a video example to understand this concept.

Example 5. Here are two cool lines:

${\color{Red} L_1: x= 6+4t, y=4-3t , z=-7+t}$

${\color{Red} L_2 : \frac{x+1}{8}=\frac{y-1}{-6}=\frac{z+5}{2}}$

Are these lines parallel, intersecting or skew ?

Perpendicular Distance of a point from a line.

Perpendicular is the shortest distance of a point from the given line. The idea behind perpendicular distance is the projection of perpendicular (drawn from point to the line) on the given line.

This distance can be found using two ways.

First method (Using dot product) :

Let $\dpi{120} \overrightarrow{r} = \overrightarrow{a} +t \overrightarrow{b}$ be the given line. And P be the point whose distance is required from the line.

Write the position vector of a general point on the given line and assume it as point L.
Obtain $\dpi{120} \overrightarrow{PL}$ which would be in form of parameter t.
Set dot product $\dpi{120} \overrightarrow{PL}.\overrightarrow{b}=0$ and solve for parameter t.
Plug in t value into $\dpi{120} \overrightarrow{PL}$ and find $\dpi{120} \left |\overrightarrow{PL} \right |$

Here $\dpi{120} \left |\overrightarrow{PL} \right |$ is the perpendicular (shortest distance) from point P to line L.

Example5: Find shortest distance(length of perpendicular) of point P(0,2,3) from the line

$\dpi{120} {\color{Red} \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}}$

Solution: let L be any point lying on this given line and general form of this point is

$\dpi{120} \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=t$

Step 1) Vector form of this line is r = 〈-3,1,-4〉 + t 〈 5,2,3〉

x= 5t-3, y=2t+1 , z=3t-4

So point L, in general form is given as,

L = (5t-3,2t+1,3t-4)

Step 2)

$\dpi{120} \overrightarrow{PL}$ = L-P = 〈5t-3-0, 2t+1-2 , 3t-4-3〉

= 〈5t-3, 2t-1 , 3t-7〉

Step 3) Since PL is perpendicular to given line so,

$\dpi{120} \overrightarrow{PL}.\overrightarrow{b}=\left \langle 5t-3, 2t-1, 3t-7 \right \rangle . \left \langle 5,2,3 \right \rangle$

= 5(5t-3)+2(2t-1)+3(3t-7) = 0

Solving it, we get t =1

Step 4) Substituting t=1 into PL we get,

$\dpi{120} \overrightarrow{PL}= \left \langle 2,1,-4 \right \rangle$

$\dpi{120} \left |\overrightarrow{PL} \right |= \sqrt{2^{2}+(-1)^{2}+4^{2}}=\sqrt{21} units$

Second method (using cross product):

For this method we use following steps.

Get a point lying on the given line , Say it L.
Find vector LP .
Find cross product of vector LP and direction vector of line, b and use the formula

$\dpi{120} \mathbf{d= \frac{\left | \overrightarrow{LP }\times \overrightarrow{b}\right |}{\left |\overrightarrow{b} \right |}}$

Example5: Find shortest distance(length of perpendicular) of point P(0,2,3) from the line

$\dpi{120} {\color{Red} \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}}$

Solution: Using Cartesian form of line,

$\dpi{120} \frac{x-x_{1}}{b_{1}}=\frac{y-y_{1}}{b_{2}}=\frac{z-z_{1}}{b_{3}}$

(x1 ,y1 ,z1 ) is the point lying on this line and 〈b1,b2,b3〉 is the direction vector of this line.

Step 1) Let L = (-3,1,-4) is the point on this line and b = 〈5,2,3〉 is the direction vector of this line.

Step 2) LP = (0,2,3)-(-3,1,-4) = (3,1,7)

Step 3)

$\dpi{120} \overrightarrow{LP}\times \overrightarrow{b }= \begin{vmatrix} i & j & k\\ 3 & 1 & 7 \\ 5& 2& 3 \end{vmatrix}$

= i(3-14)-j(9-35)+k(6-5)

= -11i+26j+k

$\dpi{120} \left |\overrightarrow{LP}\times \overrightarrow{b} \right |=\sqrt{(-11)^{2}+26^{2}+1^{2}}= \sqrt{798}$

$\dpi{120} \left |\overrightarrow{b} \right |=\sqrt{5^{2}+2^{2}+3^{2}}=\sqrt{38}$

Distance $\dpi{120} d = \frac{\sqrt{798}}{\sqrt{38}}=\sqrt{21}units$

Practice problems:

Find the vector equation of a line passing through a point (2,-1,3) and parallel to the line r = (i-2j+k) + t(2i+3j-5k).
Show that the lines

$\dpi{120} L_{1}: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and

$\dpi{120} L_{2}: \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$

intersect each other. Also find their point of intersection.

Find the length of perpendicular drawn from the point (5,4,-1) to the line r= i + t(2i+9j+5k).

Answers:

r = (2i-j+3k) + t(2i+3j-5k)
(1/2 ,-1/2, -3/2)
√(2109/110)

Parallel Lines in space

Line passing through a point and perpendicular to two given lines.

Intersection of two lines:

Check whether given lines are parallel, perpendicular or Skew.

Perpendicular Distance of a point from a line.

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