Straight Line in Space - Celestial Tutors

Different forms of Straight line in space

Any straight line in space can be written in three forms – vector equation, Cartesian form and parametric form

Vector equation of a line passing through a fixed point with position vector $\dpi{120} \overrightarrow{a}$ and parallel to a given vector $\dpi{120} \overrightarrow{b}$ is given as,

$\dpi{150} \mathbf{\overrightarrow{r}=\overrightarrow{a}+t\overrightarrow{b}}$ where t is a scalar.

Here a is the point lying on the line and b is the direction of line.

Vector equation of a line passing through two points:

If a line is passing through two points $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ then its vector equation is written as,

$\dpi{150} \mathbf{\overrightarrow{r}=\overrightarrow{a}+t\left (\overrightarrow{b} -\overrightarrow{a} \right )}$

Cartesian(symmetric) equations of a line

Cartesian equations of a line passing through two given points (x1,y1,z1) and (x2,y2,z2) is given by

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

This is also called symmetric equations of line.

Parametric form of a line

Set of parametric equations of a line can be obtained from Cartesian form itself by setting each equation equal to any parameter say ‘t’.

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

Then parametric equations are given as,

$\dpi{120} \mathbf{\frac{x-x_{1}}{x_{2}-x_{1}}=t \Rightarrow x=x_{1}+t(x_{2}-x_{1})}$

$\dpi{120} \mathbf{\frac{y-y_{1}}{y_{2}-y_{1}}=t \Rightarrow y=y_{1}+t(y_{2}-y_{1})}$

$\dpi{120} \mathbf{\frac{z-z_{1}}{z_{2}-z_{1}}=t \Rightarrow z=z_{1}+t(z_{2}-z_{1})}$

Example1.Find the vector equation of the line passing through points A(3,4,-7) and B(1,-1,6). Also find the Cartesian equation and parametric equations of line.

Solution: Given position vectors of two points are,

a = 〈3,4-7〉 b =〈1,-1,6〉

Vector equation of line is given as,

$\dpi{120} \overrightarrow{r}=\overrightarrow{a}+t\left (\overrightarrow{b} -\overrightarrow{a} \right )$

$\dpi{120} \overrightarrow{r}=\left \langle 3,4,-7 \right \rangle +t\left (\left \langle 1,-1,6 \right \rangle-\left \langle 3,4,-7 \right \rangle \right )$

=〈3,4,-7〉 + t〈1-3,-1-4,6+7〉

= 〈3,4,-7〉+ t〈-2,-5,13〉

Cartesian(symmetric) form is given as,

$\dpi{120} \frac{x-3}{1-3}=\frac{y-4}{-1-4}=\frac{z-(-7)}{6+7}$

$\dpi{120} \frac{x-3}{-2}=\frac{y-4}{-5}=\frac{z+7}{13}$

Parametric equations are given as,

$\dpi{120} \frac{x-3}{-2}=\frac{y-4}{-5}=\frac{z+7}{13}= t$

x = -2t +3

y = -5t+4

z = 13t-7

Example2. Determine if the line that passes through the point (0,−3,8) and is parallel to the line given by x=10+3t, y=12t and z= −3−t passes through the xz-plane. If it does, then give the coordinates of that point.

Solution: Line is given in parametric form. Direction vector of given line are the coefficients of t 〈3,12,-1〉.

Any line parallel to this given line will have same direction vector. So the vector form of new parallel line passing through point (0,-3,8) is given as

$\dpi{120} \overrightarrow{r}=\left \langle 0,-3,8 \right \rangle +t\left \langle 3,12,-1 \right \rangle$

=〈 3t, -3+12t, 8-t〉

If this line passes through xz plane then y coordinate must be 0, therefore

-3+12t = 0

t =1/4

So, the line does pass through the xz-plane. To get the complete coordinates of the point all we need to do is plug t=1/4 into equations. We’ll use the vector form

$\dpi{120} \overrightarrow{r}=\left \langle 3t,-3+12t,8-t \right \rangle=\left \langle 3\left ( \frac{1}{4} \right ),-3+12\left ( \frac{1}{4} \right ),8-\frac{1}{4} \right \rangle$

$\dpi{120} =\left \langle \frac{3}{4},0,\frac{31}{4} \right \rangle$

This is the point where new line is passing through xz plane.

Example3.The Cartesian(symmetric) equations of a line are 6x-2= 3y+1=2z-2. Find the vector equation of the line.

Solution: In symmetric form of line , coefficients of x,y and z must be unity. So to get proper form of symmetric equations ,we make the coefficients of x,y and z as unity.

6x-2 =3y+1 =2z-2

6(x- 2/6 ) = 3(y+1/3 ) = 2(z-1)

Dividing all sides by 6,

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

This shows line is passing through point (1/3 ,-1/3 ,1) and has direction vector as (1,2,3). So vector form of this line is given as,

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

Example4. Find the direction cosines of given line . Also find its vector equation.

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

Solution: Given symmetric equation can be rewritten as,

$\dpi{120} \frac{x-2}{2}=\frac{y-\frac{5}{2}}{\frac{-3}{2}} = \frac{z+1}{0}$

This shows this line is passing through point (2,5/2 ,-1) and has direction ratio proportional to 2,-3/2,0.

So its direction cosines are,

$\dpi{120} \frac{2}{\sqrt{2^{2}+\left (\frac{-3}{2} \right )^{2}+0^{0}}},\frac{\frac{-3}{2}}{\sqrt{2^{2}+\left (\frac{-3}{2} \right )^{2}+0^{0}}},\frac{0}{\sqrt{2^{2}+\left (\frac{-3}{2} \right )^{2}+0^{0}}}$

$\dpi{120} =\frac{2}{\sqrt{\frac{25}{4}}},\frac{\frac{-3}{2}}{\sqrt{\frac{25}{4}}},\frac{0}{\sqrt{\frac{25}{4}}}\Rightarrow \frac{2}{\frac{5}{2}},\frac{\frac{-3}{2}}{\frac{5}{2}},\frac{0}{\frac{5}{2}}\Rightarrow \frac{4}{5},\frac{-3}{5},0$

The line is passing through point having position vector 〈2, 5/2, -1〉 and direction vector as 〈2, -3/2 ,0〉 so its vector equation is written as,

$\dpi{120} \overrightarrow{r}=\left \langle 2,\frac{5}{2},-1 \right \rangle +t\left \langle 2,\frac{-3}{2},0 \right \rangle$

$\dpi{120} \overrightarrow{r}=2i+\frac{5}{2}j-k +t\left ( 2i-\frac{3}{2}j+0k\right )$

Example5.If the points A(-1,3,2) , B(-4,2,-2), C(5,5,k) are collinear then find value of k.

Solution: The Cartesian equations of a line passing through points A(-1,3,2) and B(-4,2,-2) are given as,

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

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

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

Since points A,B and C are given collinear so point C (5,5,k) must satisfy this equation,

$\dpi{120} \frac{5+1}{3}=\frac{5-3}{1}=\frac{k-2}{4}$

2 =2 = (k-2)/4

After solving 2=(k-2)/4 we get k= 10

Angle between two lines

Let the vector equations of two lines is given as,

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

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

And θ is the angle between them then,

$\dpi{150} \mathbf{cos\theta =\frac{b_{1}.b_{2}}{\left | b_{1} \right |\left | b_{2} \right |}}$

where b1 and b2 are the direction vectors of two lines.

Example6. Find the angle between the lines

$\dpi{120} {\color{Red} \overrightarrow{r_{1}}=\left \langle 3,2,-4 \right \rangle+t\left \langle 3,-2,0 \right \rangle}$

Solution: The direction vectors of given lines are :

b1=〈3,-2,0〉 and b2 =〈1, 3/2, 2〉

$\dpi{120} \left | b_{1} \right |=\sqrt{3^{2}+(-2)^{2}+0^{2}}=\sqrt{13}$

$\dpi{120} \left | b_{2} \right |=\sqrt{1^{2}+(\frac{3}{2})^{2}+2^{2}}=\sqrt{\frac{29}{4}}$

b1.b2= 〈3,-2,0〉. 〈1, 3/2, 2〉 = 3-3+0 = 0

$\dpi{120} {cos\theta =\frac{b_{1}.b_{2}}{\left | b_{1} \right |\left | b_{2} \right |}}=\frac{0}{\sqrt{13}\sqrt{29/4}}$

=> θ = π/2

Perpendicularity of two lines

If two lines are perpendicular then dot product of their direction vectors would be 0 and vice versa.

Example7. Find the value of k if the given lines are perpendicular to each other.

$\dpi{120} {\color{Red} \frac{1-x}{3}=\frac{7y-14}{2k}=\frac{z-3}{2}}$ and $\dpi{120} {\color{Red} \frac{7-7x}{3k}=\frac{y-5}{1}=\frac{6-z}{5}}$

Solution: First we need to write these equations in proper symmetric form making coefficients of x,y and z as unity.

$\dpi{120} \frac{x-1}{-3}=\frac{y-2}{\frac{2k}{7}}=\frac{z-3}{2}$ and $\dpi{120} \frac{x-1}{\frac{-3k}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$

Direction vectors of these lines are

b1=〈-3, 2k/7 ,2 〉 and b2 =〈-3k/7 ,1,-5〉

Since lines are given perpendicular so dot product of their direction vectors would be 0.

$\dpi{120} \overrightarrow{b_{1}}.\overrightarrow{b_{2}}=0\Rightarrow \left \langle -3,\frac{2k}{7},2 \right \rangle.\left \langle \frac{-3k}{7},1,-5 \right \rangle=0$

$\dpi{120} \frac{9k}{7}+\frac{2k}{7}-10=0$

11k/7 = 10

k = 70/11

Practice problems:

Find the vector equation as well as Cartesian equation of the line passing through two points A(1,2,-1) and B(2,1,1).
Show that the points with position vectors 〈 -2,3,0〉 , 〈1,2,3〉 and 〈7,0,-1〉 are collinear.
Show that the following lines are perpendicular to each other.

$\dpi{120} \frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\dpi{120} \frac{x-1}{-3}=\frac{y-2}{\frac{2k}{7}}=\frac{z-3}{2}$

If the coordinates of points A,B,C and D are (1,2,3) ,(4,5,7), (-4,3,-6) and (2,9,2) respectively . then find the angle between lines AB and CD.

Answers:

1) $\dpi{120} \overrightarrow{r}=\left \langle 1,2,-1 \right \rangle+t\left \langle 1,-1,2 \right \rangle , \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2}$

4) 0

Different forms of Straight line in space

Angle between two lines

Perpendicularity of two lines

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